New Syllabus Mathematics 8 Singapore Math Worktext-1

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Philippine Copyright 2016 by Rex Book Store, Inc.

RBS Mathematics Series The New Syllabus Mathematics 8 Revised Edition 2016 ISBN 978-971-23-8552-0 Classification: Worktext (82-MB-00141-0B) Published, copyrighted 2016, and distributed by Rex Book Store, Inc. (RBSI) with main office at 856 Nicanor Reyes Sr. St., Sampaloc, Manila/Tel. Nos.: 735-1364, 736-0567 RBSI Branches: LUZON •MORAYTA: 856 N. Reyes Sr. St., Sampaloc, Manila / Tel. Nos.: 736-0169, 733-6746; Telefax: 736-4191 •RECTO: 2161-65 Freedom Building, C.M. Recto Avenue, Sampaloc, Manila / Tel. Nos.: 522-4521, 522-4305, 5224107, 733-8637 •RECTO (La Consolacion): Mendiola, Manila • MAKATI: Unit UG-2, Star Centrum Bldg., Sen. Gil Puyat Ave., Makati City / Tel. No.: 818-5363; Telefax: 893-3744 •ROCKWELL: 1st Floor, Ateneo Professional School, Rockwell Center, Bel-Air, Makati City / Tel. No.: 729-2015 •CUBAO: Unit 10 UGF, Doña Consolacion Bldg., Gen. Santos Ave., Araneta Center, Cubao, Quezon City /Telefax: 911-1070 •ORTIGAS: G/F East Tower, Philippine Stock Exchange Center, Exchange Road, Ortigas Center, Pasig City / Tel. No.: (02) 650-4347 •CAVITE: Block 4, Lot 20 Don Gregorio Heights 2, Zone 1-A Aguinaldo Highway, Dasmariñas, Cavite / Telefax: (046) 416-1824 •CAVITE (Tanza): (Display Area) Block 5, Lot 6, City View 4 and 5, Brgy. Tanauan, Tanza, Cavite •NAGA: 1-1A Geronimo ******ebook converter DEMO Watermarks*******

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PREFACE New Syllabus Mathematics (NSM) is a series of worktexts specially designed to provide valuable learning experiences to engage the hearts and minds of students in the learning process. Included in the worktexts are Investigation, Class Discussion, Thinking Time, Journal Writing, Performance Task and Problems in Real-World Contexts to support the teaching and learning of Mathematics. Every chapter begins with a chapter opener which motivates students in learning the topic. Interesting stories about mathematicians, real-life examples and applications are used to arouse students’ interest and curiosity so that they can appreciate the beauty of Mathematics in their surroundings. The use of ICT helps students to visualize and manipulate mathematical objects more easily, thus making the learning of Mathematics more interactive. Ready-to-use interactive ICT templates are available at http://www.shinglee.com.sg/StudentResources/.

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KEY FEATURES CHAPTER OPENER Each chapter begins with a chapter opener to arouse students’ interest and curiosity in learning the topic.

LEARNING OBJECTIVES Learning objectives help students to be more aware of what they are about to study so that they can monitor their own progress.

RECAP Relevant prerequisites will be revisited at the beginning of the chapter or at appropriate junctures so that students can build upon their prior knowledge, thus creating meaningful links to their existing schema.

WORKED EXAMPLE This shows students how to apply what they have learned to solve related problems and how to present their working clearly. A suitable heading is included in brackets to distinguish between the different Worked Examples.

PRACTICE NOW At the end of each Worked Example, a similar question will be provided for immediate practice. Where appropriate, this includes further questions of progressive difficulty.

SIMILAR QUESTIONS ******ebook converter DEMO Watermarks*******

A list of similar questions in the Exercise is given here to help teachers choose questions that their students can do on their own.

EXERCISE The questions are classified into three levels of difficulty – Basic, Intermediate, and Advanced.

SUMMARY At the end of each chapter, a succinct summary of the key concepts is provided to help students consolidate what they have learned.

REVIEW EXERCISE This is included at the end of each chapter for the consolidation of learning of concepts.

CHALLENGE YOURSELF Optional problems are included at the end of each chapter to challenge and stretch high-ability students to their fullest potential.

REVISION EXERCISE This is included after every few chapters to help students assess their learning. Learning experiences have been infused into Investigation, Class Discussion, Thinking Time, Journal Writing, and Performance Task.

Investigation

Activities are included to guide students to investigate and discover ******ebook converter DEMO Watermarks*******

important mathematical concepts so that they can construct their own knowledge meaningfully.

Class Discussion Questions are provided for students to discuss in class, with the teacher acting as the facilitator. The questions will assist students to learn new knowledge, think mathematically, and enhance their reasoning and oral communication skills.

Thinking Time Key questions are also included at appropriate junctures to check if students have grasped various concepts and to create opportunities for them to further develop their thinking.

Journal Writing Opportunities are provided for students to reflect on their learning and to communicate mathematically. It can also be used as a formative assessment to provide feedback to students to improve on their learning.

Performance Task Mini projects are designed to develop research and presentation skills in the students.

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MARGINAL NOTES

This contains important information that students should know.

This guides students on how to approach a problem.

This includes information that may be of interest to students.

This contains certain mathematical concepts or rules that students have learned previously.

This contains puzzles, fascinating facts and interesting stories about Mathematics as enrichment for students.

This guides students to search on the Internet for valuable information or interesting online games for their independent and self-directed learning.

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Contents CHAPTER 1 Factorization of Polynomials 1.1

Factorization

1.2

Factorization of Quadratic Expressions

1.3

Factorization of Algebraic Expressions

1.4

Factorization Using Special Algebraic Identities

1.5

Factorization by Grouping

1.6 Factorization Using the Sum and Difference of Two Cubes Summary Review Exercise 1

CHAPTER 2 Rational Algebraic Expressions 2.1

Algebraic Fractions

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2.2

Multiplication and Division of Algebraic Fractions

2.3

Addition and Subtraction of Algebraic Fractions

2.4 Manipulation of Algebraic Formulae Summary Review Exercise 2

CHAPTER 3 Graphs of Linear Equations 3.1

Cartesian Coordinates

3.2

Graphs of Linear Equations in Two Variables

3.3 Slope of a Straight Line Summary Review Exercise 3

CHAPTER 4 Linear Graphs and Simultaneous Linear Equations 4.1

Horizontal and Vertical Lines

4.2 Graphs of Linear Equations in the form ax + by = k and y = mx + c ******ebook converter DEMO Watermarks*******

4.3 Solving Simultaneous Linear Equations Using Graphical Method 4.4 Solving Simultaneous Linear Equations Using Algebraic Methods 4.5 Applications of Simultaneous Equations in Real-World Contexts Summary Review Exercise 4 Revision Exercise A

CHAPTER 5 Linear Inequalities in Two Variables 5.1

Linear Inequalities in Two Variables

5.2 Application of Systems of Linear Inequalities in Two Variables in Real-World Contexts Summary Review Exercise 5

CHAPTER 6 Functions and Linear Graphs 6.1

Relations

6.2

Functions

6.3 Linear Functions ******ebook converter DEMO Watermarks*******

Summary Review Exercise 6

CHAPTER 7 Geometry Logic Statements 7.1

Statements

7.2

‘If-then’ Statements and ‘If and only if’ Statements

7.3 Converse, Inverse and Contrapositive of ‘If-then’ Statements 7.4 Proving Mathematical Statements Summary Review Exercise 7 Revision Exercise B

CHAPTER 8 Axiomatic Structure of a Mathematical System 8.1

The Axiomatic System

8.2 Axiomatic Structure of Geometry Summary Review Exercise 8 ******ebook converter DEMO Watermarks*******

CHAPTER 9 Triangle Congruence 9.1

Congruent Figures

9.2

Congruence Tests

9.3 Applications of Congruent Triangles Summary Review Exercise 9 Revision Exercise C

CHAPTER 10 Theorems on Triangle Inequalities 10.1

Triangle Inequality Theorems

10.2 Properties of Parallel Lines Summary Review Exercise 10

CHAPTER 11 Probability of Simple Events ******ebook converter DEMO Watermarks*******

11.1

Introduction to Probability

11.2

Sample Space

11.3

Probability of Single Events

11.4

Further Examples on Probability of Single Events

11.5 Simple Combined Events, Possibility Diagrams and Tree Diagrams Summary Review Exercise 11 Revision Exercise D Problems in Real-World Contexts Practice Now Answers Answers

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Chapter One Factorization of Polynomials Various algebraic properties can be used to factorize complicated expressions. What are some examples of such algebraic properties?

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LEARNING OBJECTIVES At the end of this chapter, you should be able to: factorize completely different types of polynomials, solve problems involving factors of polynomials.

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1.1 Factorization In Grade 7, we have learned how to expand linear expressions, e.g. 2(x + y) = 2x + 2y. We shall now learn how to carry out the reverse process, i.e. express a linear expression as a product of its factors. This process is called factorization. Factorization is the process of expressing an algebraic expression as a product of two or more algebraic expressions. It is the reverse of expansion.

To factorize algebraic expressions, we will need to identify the common factors, i.e. common numbers or common variables of the terms. For example, in 4x + 2 = 2(2x + 1), 2 is the common factor of the two terms, in 6pq – 3p = 3p(2q – 1), 3 and p are the common factors of the two terms.

Worked Example

1

(Factorization of Algebraic Expressions) Factorize each of the following expressions completely. a. 6x – 12 b. 4ay – 24az

Always check that no other factors can be taken out from the terms within the ******ebook converter DEMO Watermarks*******

brackets in your answer. For example, in (b), if we write 4ay – 24az = 4(ay – 6az), the factorization is incomplete as a is also a factor of the expression.

Solution: a. 6x – 12 = 6(x – 2) b. 4ay – 24az = 4a(y – 6z)

Factorize each of the following expressions completely. a. –10x + 25 b. 24x + 16 c. 2x + kx d. 2ab + 4abc e. 18a – 54ay + 36az

Exercise 1A Questions 1(a)-(e), 2(a)-(f)

Simplify (x – a)(x – b)(x – c)…(x – z).

Equivalent Expressions ******ebook converter DEMO Watermarks*******

Work in pairs. Some algebraic expressions, which consist of five pairs of equivalent expressions, are given in Table 4.4. An example of a pair of equivalent expressions is 3x – 12 and 3(x – 4) as 3x – 12 = 3(x – 4). Match and justify each pair of equivalent expressions. Table 1.1

BASIC LEVEL 1. Factorize each of the following expressions completely. a. 12x – 9 b. –25y – 35 c. 27b – 36by d. 8ax + 12a – 4az e. 4m – 6my – 18mz INTERMEDIATE LEVEL 2. Factorize each of the following expressions completely. a. 3a − ax b. 8cde − 2cd − de c. ¼ xy – ¼ y d. e. 8ab − 4abc + 12ac − 4a ******ebook converter DEMO Watermarks*******

ADVANCED LEVEL 3. Factorize each of the following expressions completely. a. –39b2 – 13ab b. 5x + 10x(b + c) c. 3xy – 6x(y – z) d. 2x(7 + y) – 14x(y + 2) e. –3a(2 + b) + 18a(b – 1) f. –4y(x – 2) – 12y(3 – x)

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1.2 Factorization of Quadratic Expressions In Grade 7, we have learned how to expand the product of two linear factors to obtain a quadratic expression of the form ax2 + bx + c, where a, b and c are constants and a ≠ 0. In this section, we shall learn how to carry out the reverse process, i.e. factorization.

Factorization of Quadratic Expressions Using Algebra Discs Recall that in Grade 7, when we expand the product of the two linear factors, x + 2 and x + 3, we obtain the quadratic expression, x2 + 5x + 6. Recall also that there are four distinct regions in the rectangular array. The top-left region contains the x2-discs, the bottom-right region contains the 1-discs and the other two regions contain the x-discs.

Therefore, (x + 2)(x + 3) = x(x + 3) + 2(x + 3)                     = x2 + 3x + 2x + 6                     = x2 + 5x + 6. Since factorization is the reverse of expansion, when we factorize a quadratic expression, we will obtain two linear factors. Hence, x2 + 5x + 6 = x2 + 3x + 2x + 6 ******ebook converter DEMO Watermarks*******

       = x(x + 3) + 2(x + 3)        = (x + 2)(x + 3). Algebra discs can also be used to factorize quadratic expressions. To factorize x2 + 5x + 6, we first form a rectangular array with the disc at the top-left region and the six

Next, we put in the five

discs at the bottom-right region.

discs to form the rectangle.

Thus we have:

Hence, the two linear factors are x + 2 and x + 3. Notice that in this case, the constant term 6 is factorized into 2 × 3 and the five discs (5x) are divided into two groups (2x and 3x) to complete the rectangle.

x2 = x × x ******ebook converter DEMO Watermarks*******

x=x×1 Example: x2 + 7x + 12 We have to form a rectangle with the disc at the top-left region and the twelve discs at the bottom-right region. The possible factorizations of 12 are 1 × 12, 2 × 6 and 3 × 4. Consider 1 × 12. We are not able to divide ‘7x’ into two groups to complete the rectangle.

Consider 2 × 6. We are also not able to divide ‘7x’ into two groups to complete the rectangle.

Consider 3 × 4. We are able to divide ‘7x’ into ‘3x’ and ‘4x’ to complete the rectangle.

The factorization can be simply illustrated using a multiplication frame:

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Therefore, x2 + 7x + 12 = (x + 3)(x + 4).

x × x = x2 3 × 4 = 12 3 × x = 3x 4 × x = 4x 3x + 4x = 7x Example: x2 − 7x + 12 We have to form a rectangle with the disc at the top-left region and the twelve discs at the bottom-right region. Then we divide ‘–7x’ into ‘–3x’ and ‘–4x’ to complete the rectangle.

The factorization can be simply illustrated using a multiplication frame:

Therefore, x2 – 7x + 12 = (x – 3)(x – 4).

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x × x = x2 (–3) × (–4) = 12 (–3) × x = –3x (–4) × x = –4x (–3x) + (–4x) = –7x

Factorize each of the following expressions by using algebra discs. Alternatively, you may visit http://www.shinglee.com.sg/StudentResources/ to access the AlgeTools™ software (Go to algedisc/Quadratic Expressions/Activity 4). a. x2 + 6x + 5 b. x2 – 6x + 5 c. x2 + 8x + 12 d. x2 – 8x + 12 e. x2 – 2x – 15

Factorization of Quadratic Expressions Using a Multiplication Frame Without using algebra discs, we can use a multiplication frame to help us factorize quadratic expressions. Consider the expression x2 + 8x + 12.

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Step 1: Write x2 in the top-left corner and 12 in the bottom-right corner of the multiplication frame. Step 2: Consider the factors of x2 and 12. Write them in the first column and the first row. Step 3: Multiply them to complete the multiplication frame and check whether the result matches the given expression. Therefore, x2 + 8x + 12 = (x + 2)(x + 6). Let us apply the above method to factorize another expression x2 – 5x + 4. Consider the factorization of x2 and 4, i.e. x2 = x × x and 4 = (–1) × (–4).

Therefore, x2 – 5x + 4 = (x – 1)(x – 4).

The possible factorizations of 12 are 1 × 12, 2 × 6 and 3 × 4. Consider 1 × 12.

Consider 3 × 4.

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The above cases are rejected as the term in x does not match that in the given expression.

Worked Example

2

(Factorizing Quadratic Expressions of the Form x2 + bx + c) Factorize each of the following expressions completely. a. x2 + 7x + 10 b. x2 – 9x + 14 c. x2 + x – 20 d. x2 – x – 12

For ax2 + bx + c where a, b and c > 0, both of the factors of c must be positive, e.g. in (a), we only have to consider 10 = 1 × 10 and 2 × 5.

It is a good practice to check your answer by expanding the product of the two linear factors to see if it gives the original quadratic expression, e.g. in (a), (x + 2)(x + 5) = x(x + 5) + 2(x + 5) = x2 + 5x + 2x + 10 = x2 + 7x + 10.

For ax2 + bx + c where a and c > 0 but b < 0, both of the factors of c must be negative, e.g. in (b), we only have to consider 14 = (–1) × (–14) and (–2) × (– 7).

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Solution: a. x2 = x × x    10 = 1 × 10 or (–1) × (–10)      = 2 × 5 or (–2) × (–5)

∴ x2 + 7x + 10 = (x + 2)(x + 5) b. x2 = x × x   14 = 1 × 14 or (–1) × (–14)      = 2 × 7 or (–2) × (–7)

∴ x2 – 9x + 14 = (x – 2)(x – 7) c. x2 = x × x   –20 = 1 × (–20) or (–1) × 20      = 2 × (–10) or (–2) × 10      = 4 × (–5) or (–4) × 5

∴ x2 + x – 20 = (x – 4)(x + 5) d. x2 = x × x ******ebook converter DEMO Watermarks*******

  –12 = 1 × (–12) or (–1) × 12      = 2 × (–6) or (–2) × 6      = 3 × (–4) or (–3) × 4

∴ x2 – x – 12 = (x + 3)(x – 4)

Factorize each of the following expressions completely. a. x2 + 8x + 7 b. x2 – 11x + 28 c. x2 + x – 2 d. x2 – 7x – 8

Exercise 1B Questions 1(a)-(h), 4

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Worked Example

3

(Factorizing Quadratic Expressions of the Form ax2 + bx + c) Factorize each of the following expressions completely. a. 2x2 + 7x + 3 b. 3x2 + 7x – 6 c. –x2 – 4x + 32 d. 4x2 – 6x – 4

Not all quadratic expressions can be factorized using the multiplication frame, e.g. x2 + 2x – 1.

Solution: a. 2x2 = 2x × x    3 = 1 × 3 or (–1) × (–3)

∴ 2x2 + 7x + 3 = (2x + 1)(x + 3) b. 3x2 = 3x × x    –6 = 1 × (–6) or (–1) × 6      = 2 × (–3) or (–2) × 3 ******ebook converter DEMO Watermarks*******

∴ 3x2 + 7x – 6 = (3x – 2)(x + 3) c. –x2 = –x × x    32 = 1 × 32 or (–1) × (–32)      = 2 × 16 or (–2) × (–16)      = 4 × 8 or (–4) × (–8)

∴ –x2 – 4x + 32 = (–x + 4)(x + 8) d. 4x2 – 6x – 4 = 2(2x2 – 3x – 2) (extract the common factor 2)    2x2 = 2x × x    –2 = 1 × (–2) or (–1) × 2

∴ 4x2 – 6x – 4 = 2(2x + 1)(x – 2)

Factorize each of the following expressions completely. a. 2x2 + 11x + 12 b. 5x2 – 13x + 6 ******ebook converter DEMO Watermarks*******

c. –2x2 + 9x – 9 d. 9x2 – 33x + 24 e. –3x2 + 5x + 2

Exercise 1B Questions 2(a)-(h), 3(a)-(h), 5(a)-(b)

BASIC LEVEL 1. Factorize each of the following expressions completely. 1. a2 + 9a + 8 2. b2 + 8b + 15 3. c2 – 9c + 20 4. d2 – 16d + 28 5. f2 + 6f – 16 6. h2 + 2h – 120 7. k2 – 4k – 12 8. m2 – 20m – 21 2. Factorize each of the following expressions completely. 1. 3n2 + 10n + 7 2. 4p2 + 8p + 3 3. 6q2 – 17q + 12 ******ebook converter DEMO Watermarks*******

4. 4r2 – 7r + 3 5. 8s2 + 2s – 15 6. 6t2 + 19t – 20 7. 4u2 – 8u – 21 8. 18w2 – w – 39 INTERMEDIATE LEVEL 3. Factorize each of the following expressions completely. a. –a2 + 2a + 35 b. –3b2 + 76b – 25 c. 4c2 + 10c + 4 d. 5d2 – 145d + 600 e. 8f2 + 4f – 60 f. 24h2 – 15h – 9 g. 30 + 14k – 4k2 h. 35m2n + 5mn – 30n 4. The area of a rectangle is (x2 + 8x + 12) cm2. If the length of the rectangle is (x + 6) cm, show that its breadth is (x + 2) cm. ADVANCED LEVEL 5. Factorize each of the following expressions completely. a. b. 0.6r – 0.8qr – 12.8q2r ******ebook converter DEMO Watermarks*******

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1.3 Factorization of Algebraic Expressions Factorization of Algebraic Expressions Using a Multiplication Frame In Section 1.2, we have learned how to use a multiplication frame to factorize quadratic expressions. Worked Example 4 illustrates how we are able to use the multiplication frame to factorize other algebraic expressions.

Worked Example

4

(Factorizing Algebraic Expressions of the Form ax2 + bxy + cy2) Factorize each of the following expressions completely. a. x2 + 2xy – 8y2 b. 2x2 – 7xy + 6y2

Solution: a. x2 = x × x    –8y2 = y × (–8y) or (–y) × 8y       = 2y × (–4y) or (–2y) × 4y

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∴ x2 + 2xy – 8y2 = (x – 2y)(x + 4y) b. 2x2 = 2x × x    6y2 = y × 6y or (–y) × (–6y)      = 2y × 3y or (–2y) × (–3y)

∴ 2x2 – 7xy + 6y2 = (2x – 3y)(x – 2y)

1. Factorize each of the following expressions completely. a. x2 – 2xy – 15y2 b. 6x2 + 11xy + 5y2 c. 9a2 – 6ab + b2 2. Factorize each of the following expressions completely. a. x2y2 + 8xy + 16 b. 3x2y2 – 14xy + 16

Exercise 1C Questions 1(a)-(d), 2(a)-(f), 3(a)-(d)

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BASIC LEVEL 1. Factorize each of the following expressions completely. a. a2 + 3ab + 2b2 b. 2x2 + 5xy + 2y2 c. 2c2 + 7cd + 3d2 d. 3p2 + 10pq + 3q2 INTERMEDIATE LEVEL 2. Factorize each of the following expressions completely. a. a2 + 3ab – 4b2 b. c2 – 4cd – 21d2 c. 2h2 + 7hk – 15k2 d. 3m2 – 16mn – 12n2 e. 3p2 + 15pq + 18q2 f. 2r2t – 9rst + 10s2t ADVANCED LEVEL 3. Factorize each of the following expressions completely. a. x2y2 + 2xy – 15 b. 12x2y2 – 17xy – 40 c. 4x2y2z – 22xyz + 24z d.

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1.4 Factorization Using Special Algebraic Identities In Grade 7, we have learned how to expand certain algebraic expressions using the three special algebraic identities:          (a + b)2 = a2 + 2ab + b2          (a – b)2 = a2 – 2ab + b2 (a + b)(a – b) = a2 – b2 Since factorization is the reverse of expansion, we have: a2 + 2ab + b2 = (a + b)2 a2 – 2ab + b2 = (a – b)2     a2 – b2 = (a + b)(a – b)

We can factorize an expression using these special algebraic identities if the expression can be expressed in one of the forms on the left-hand side of the equations.

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Worked Example

5

(Factorizing Algebraic Expressions of the Form a2 + 2ab + b2) Factorize each of the following expressions completely. a. x2 + 18x + 81 b. 9x2 + 24x + 16

02 = 0 12 = 1 22 = 4 32 = 9 42 = 16 52 = 25 0, 1, 4, 9, 16, 25, … are called perfect squares.

Solution: a. x2 + 18x + 81 = x2 + 2(x)(9) + 92             = (x + 9)2 (apply a2 + 2ab + b2 = (a + b)2, where a = x and b = 9) b. 9x2 + 24x + 16 = (3x)2 + 2(3x)(4) + 42              = (3x + 4)2 (apply a2 + 2ab + b2 = (a + b)2, where a = 3x and b = 4)

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1. Factorize each of the following expressions completely. a. x2 + 12x + 36 b. 4x2 + 20x + 25 c. 9x2 + 6x + 1 2. Factorize each of the following expressions completely. a. 4x2 + 2x + 1/4 b. 9/4x2 + 3x + 1

Exercise 1D Questions 1(a)-(d), 5(a)-(d), 9

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Worked Example

6

(Factorizing Algebraic Expressions of the Form a2 – 2ab + b2) Factorize each of the following expressions completely. a. 49 – 84x + 36x2 b. 9x2 – 30xy + 25y2

Solution: a. 49 – 84x + 36x2 = 72 – 2(7)(6x) + (6x)2              = (7 – 6x)2 (apply a2 – 2ab + b2 = (a – b)2, where a = 7 and b = 6x) b. 9x2 – 30xy + 25y2 = (3x)2 – 2(3x)(5y) + (5y)2               = (3x – 5y)2 (apply a2 – 2ab + b2 = (a – b)2, where a = 3x and                        b = 5y)

1. Factorize each of the following expressions completely. a. 4 – 36x + 81x2 b. 25x2 – 10xy + y2 c. 9x2 – 12xy + 4y2 2. Factorize each of the following expressions completely. a. b. ******ebook converter DEMO Watermarks*******

Exercise 1D Questions 2(a)-(d), 6(a)-(d)

Worked Example

7

(Factorizing Algebraic Expressions of the Form a2 – b2) Factorize each of the following expressions completely. a. 4x2 – 25y2 b. 8x2 – 18y2

Solution: a. 4x2 – 25y2 = (2x)2 – (5y)2           = (2x + 5y)(2x – 5y) (apply a2 – b2 = (a + b)(a – b), where a = 2x and b = 5y) b. 8x2 – 18y2 = 2(4x2 – 9y2) (extract the common factor 2)           = 2[(2x)2 – (3y)2]           = 2(2x + 3y)(2x – 3y) (apply a2 – b2 = (a + b)(a – b), where a = 2x and                         b = 3y)

1. Factorize each of the following expressions completely. a. 36x2 – 121y2 b. x2 – 64 c. –4x2 + 81 ******ebook converter DEMO Watermarks*******

2. Factorize the expression

completely.

3. Factorize the expression 4(x + 1)2 – 49 completely.

Exercise 1D Questions 3(a)-(d), 7(a)-(d), 8(a)-(f), 10(a)-(d)

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Worked Example

8

(Problem involving the Use of the Special Algebraic Identity a2 – b2 = (a + b)(a – b)) Without using a calculator, evaluate 1032 – 9.

Solution: 1. – 9 = 1032 – 32       = (103 + 3)(103 – 3) (apply a2 – b2 = (a + b)(a – b), where a = 103 and b = 3)       = 106 × 100       = 10 600

Without using a calculator, evaluate the following. a. 3042 – 1042 b. 2562 – 1562 c. 6.32 – 3.72 d. 1052 – 25 e. 842 – 256

Exercise 1D Questions 4(a)-(b)

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BASIC LEVEL 1. Factorize each of the following expressions completely. a. a2 + 14a + 49 b. 4b2 + 4b + 1 c. c2 + 2cd + d2 d. 4h2 + 20hk + 25k2 2. Factorize each of the following expressions completely. a. m2 – 10m + 25 b. 169n2 – 52n + 4 c. 81 – 180p + 100p2 d. 49q2 – 42qr + 9r2 3. Factorize each of the following expressions completely. a. s2 – 144 b. 36t2 – 25 c. 225 – 49u2 d. 49w2 – 81x2 4. Without using a calculator, evaluate each of the following. a. 592 – 412 b. 7.72 – 2.32 INTERMEDIATE LEVEL 5. Factorize each of the following expressions completely. ******ebook converter DEMO Watermarks*******

a. 3a2 + 12a + 12 b. 25b2 + 5bc + 1/4c2 c. 16/49d2 + 8/35df + 1/25f 2 d. h4 + 2h2k + k2 6. Factorize each of the following expressions completely. a. 36m2 – 48mn + 16n2 b. 1/3p2 – 2/3pq + 1/3q2 c. 16r2 − rs + 1/64 s2 d. 25 – 10tu + t2u2 7. Factorize each of the following expressions completely. a. 32a2 – 98b2 b. c2 − 1/4 d2 c. 9h2/100 − 16k2 d. m2 – 64n4 8. Factorize each of the following expressions completely. a. (a + 3)2 – 9 b. 16 – 25(b + 3)2 c. c2 – (d + 2)2 d. (2h – 1)2 – 4k2 e. 25m2 – (n – 1)2 f. (p + 1)2 – (p – 1)2 ******ebook converter DEMO Watermarks*******

9. The surface area of each face of a cube is (x2 + 4x + 4) cm2. Find i. the length, ii. the volume of the cube. ADVANCED LEVEL 10. Factorize each of the following expressions completely. a. 4(x – 1)2 – 81(x + 1)2 b. 16x2 + 8x + 1 − 9y2 c. 4x2 – y2 + 4y – 4 d. 13x2 + 26xy + 13y2 – 13

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1.5 Factorization by Grouping Factorization of Algebraic Expressions of the Form ax + ay To factorize algebraic expressions of the form ax + ay, we will need to identify the common factors, i.e. common numbers or common variables of the terms. For example, to factorize 4ay – 24az completely, we extract the common factors 4 and a to get 4a(y – 6z).

Worked Example

(Factorizing Algebraic Expressions of the Form ax + ay) Factorize each of the following expressions completely.

9

a. 3x2 + 9xy b. 2πr2 + 2πrh c. a2b – a2b2 d. c2d3 + c3d2 – c2d2

π is a constant.

Solution: a. 3x2 + 9xy = 3x(x + 3y) ******ebook converter DEMO Watermarks*******

b. 2πr2 + 2πrh = 2πr(r + h) c. a2b – a2b2 = a2b(1 – b) d. c2d3 + c3d2 – c2d2 = c2d2(d + c – 1)

Factorize each of the following expressions completely. a. 8x2y + 4x b. πr2 + πrl c. –a3by + a2y d. 3c2d + 6c2d2 + 3c3 e. p2q – 2pq2 + 4p2q2

Exercise 1E Questions 1(a)-(d), 7

Factorization of Algebraic Expressions of the Form ax + bx + kay + kby If we are given the expression ax – bx + 2ay – 2by, how do we factorize it completely? We shall now learn how to factorize algebraic expressions of the form ax + bx + kay + kby. Sometimes, it is possible to identify the common factors by grouping the terms of an algebraic expression. For example, ax – ay + bx – by = (ax – ay) + (bx – by)                      = a(x – y) + b(x – y)                      = (x – y)(a + b) (extract the common factor (x – y)). ******ebook converter DEMO Watermarks*******

It may be necessary to regroup the terms of an algebraic expression before we are able to identify the common factors. For example, cx + dy + dx + cy = (cx + cy) + (dx + dy)                      = c(x + y) + d(x + y)                      = (x + y)(c + d). Alternatively, cx + dy + dx + cy = (cx + dx) + (cy + dy)                       = x(c + d) + y(c + d)                       = (x + y)(c + d). We may need to change the sign of the factor in a group before we can factorize an algebraic expression. For example, h(x – y) + k(y – x) = h(x – y) – k(x – y)                       = (x – y)(h – k).

y – x = –(x – y)

Worked Example

10

(Factorizing Algebraic Expressions of the Form a(x + y) + b(x + y)) Factorize each of the following expressions completely. a. a(2x + 3) + 2(3 + 2x) b. 2b(5x + 2) – (5x + 2) c. 3c(x – y) – 3d(x – y) d. h(x – 2) + k(2 – x)

In (a), 2x + 3 = 3 + 2x. ******ebook converter DEMO Watermarks*******

In (b), –(5x + 2) = (–1)(5x + 2). In (d), k(2 – x) = –k(x – 2).

Solution: a. a(2x + 3) + 2(3 + 2x) = (2x + 3)(a + 2) b. 2b(5x + 2) – (5x + 2) = (5x + 2)(2b – 1) c. 3c(x – y) – 3d(x – y) = 3[c(x – y) – d(x – y)]                 = 3(x – y)(c – d) d. h(x – 2) + k(2 – x) = h(x – 2) – k(x – 2)                = (x – 2)(h – k)

Factorize each of the following expressions completely. a. 2(x + 1) + a(1 + x) b. 9(x + 2) – b(x + 2) c. 3c(2x – 3) – 6d(2x – 3) d. 7h(4 – x) – (x – 4) e. 2c(3x – 2) – 2d(2 – 3x)

Exercise 1E Questions 2(a)-(d), 4(a)-(d), 6(a)-(b)

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Worked Example

11

(Factorizing Algebraic Expressions of the Form ax + bx + kay + kby) Factorize each of the following expressions completely. a. ax – bx + 2ay – 2by b. 6ax + 12by + 9bx + 8ay c. x2 + xy – 3x – 3y d. 6xy – 15y + 10 – 4x

Solution: a. ax – bx + 2ay – 2by = (ax – bx) + (2ay – 2by) (arrange the terms into two groups)                 = x(a – b) + 2y(a – b) (factorize each group)                 = (a – b)(x + 2y) (factorize the two groups) b. 6ax + 12by + 9bx + 8ay = (6ax + 9bx) + (12by + 8ay)                     (rearrange the terms into two groups)                    = 3x(2a + 3b) + 4y(3b + 2a) (factorize each group)                    = (2a + 3b)(3x + 4y) (factorize the two groups) c. x2 + xy – 3x – 3y = (x2 + xy) – (3x + 3y) (arrange the terms into two groups)               = x(x + y) – 3(x + y) (factorize each group)               = (x + y)(x – 3) (factorize the two groups) d. 6xy – 15y + 10 – 4x = (6xy – 15y) + (10 – 4x) (arrange the terms into two groups)                 = 3y(2x – 5) + 2(5 – 2x) (factorize each group)                 = 3y(2x – 5) – 2(2x – 5)                  (change the sign of the factor in the second group)                 = (2x – 5)(3y – 2) (factorize the two groups)

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Factorize each of the following expressions completely. a. xy + 4x + 3y + 12 b. 3by + 4ax + 12ay + bx c. x3 – x2 – 1 + x d. 6xy – 4x – 2z + 3yz e. x2z – 4y – x2y + 4z

Exercise 1E Questions 3(a)-(d), 5(a)-(h), 8

How are we able to apply the concept of factorization by grouping in factorizing a quadratic expression such as 5x2 – 12x – 9?

Equivalent Expressions Work in pairs. Some algebraic expressions, which consist of a few sets of equivalent expressions, are given in Table 1.2. An example of a set of equivalent expressions is 4ay – 24az and 4a(y – 6z) as 4ay – 24az = 4a(y – 6z). Match and justify each set of equivalent expressions. If your classmate does not obtain the correct answer, explain to him what he has done wrong.

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Table 1.2

BASIC LEVEL 1. Factorize each of the following expressions completely. a. 45x2 – 81xy b. 39xy – 15x2z c. xy2z2 – x2y3 d. –15 x3y – 10 x3 2. Factorize each of the following expressions completely. a. 6a(x – 2y) + 5(x – 2y) b. 2b(x + 3y) – c(3y + x) c. 3d(5x – y) – 4f(5x – y) d. 5h(x + 3y) + 10k(x + 3y) 3. Factorize each of the following expressions completely. a. ax – 5a + 4x – 20 b. ax + bx + ay + by c. x + xy + 2y + 2y2 d. x2 – 3x + 2xy – 6y INTERMEDIATE LEVEL 4. Factorize each of the following expressions completely. ******ebook converter DEMO Watermarks*******

a. (x + y)(a + b) – (y + z)(a + b) b. (c + 2d)2 – (c + 2d)(3c – 7d) c. x(2h – k) + 3y(k – 2h) d. 6x(4m – n) – 2y(n – 4m) 5. Factorize each of the following expressions completely. a. 3ax + 28by + 4ay + 21bx b. 12cy + 20c – 15 – 9y c. dy + fy – fz – dz d. 3x2 + 6xy – 4xz – 8yz e. 2xy – 8x + 12 – 3y f. 5xy – 25x2 + 50x – 10y g. x2y2 – 5x2y – 5xy2 + xy3 h. kx + hy – hx – ky ADVANCED LEVEL 6. Factorize each of the following expressions completely. a. 144p(y – 5x2) – 12q(10x2 – 2y) b. 2(5x + 10y)(2y – x)2 – 4(6y + 3x)(x – 2y) 7. i. Factorize the expression

completely.

ii. Hence, by substituting suitable values of p, q and r, find the value of . 8. i. Factorize the expression x3 + 3x – x2 – 3 completely. ******ebook converter DEMO Watermarks*******

ii. Hence, express (x2 – 3)3 – (2 – x2)2 + 3(x2 – 3) in the form (x4 + Ax2 + B)(x2 + C), where A, B and C are integers.

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1.6 Factorization Using the Sum and Difference of Two Cubes Recall the following three algebraic identities: (a + b)2 = a2 + 2ab + b2 (a − b)2 = a2 − 2ab + b2 a2 − b2 = (a + b)(a − b)

We have also learned that a2 + b2 cannot be factorized into two linear factors. In this section, we will learn how to factorize a3 + b3 and a3 − b3.

(a + b)2 ≠ a2 + b2 (a − b)2 ≠ a2 − b2 (a + b)3 ≠ a3 + b3 (a − b)3 ≠ a3 − b3

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Factorization of a3 ± b3 1. Factorize each of the following expressions completely. i. a3 + a2b + ab2, ii. a2b + ab2 + b3. By subtracting your answer in (ii) from that in (i), factorize a3 – b3. 2. Factorize each of the following expressions completely. i. a3 − a2b + ab2, ii. a2b − ab2 + b3. By adding your answers in (i) and (ii), factorize a3 + b3. From the above investigation, we have two special algebraic identities: Sum of Cubes:

a3 + b3 = (a + b)(a2 − ab + b2)

Difference of Cubes: a3 − b3 = (a − b)(a2 + ab + b2)

Worked Example

12

(Factorization Using the Sum and Difference of Two Cubes) Factorize a. 8x3 + 125y3,

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b. (3x + 1)3 – 8.

Solution: a. 8x3 + 125y3 = (2x)3 + (5y)3           = (2x + 5y)[(2x)2 – (2x)(5y) + (5y)2]           = (2x + 5y)(4x2 – 10xy + 25y2) b. (3x + 1)3 – 8 = (3x + 1)3 – (2)3            = [(3x + 1) – 2][(3x + 1)2 + (3x + 1)(2) + 22]            = (3x – 1)[9x2 + 6x + 1 + 6x + 2 + 4]            = (3x – 1)(9x2 + 12x + 7)

Factorize a. 216p3 + 343q6, b. x3 + 27, c. 3x3 – 192, d. 64 – (2x + 3)3, e. (2p + q)3 + (7q – p)3.

Exercise 1F Questions 1(a)-(d), 2(a)-(d), 3(a)-(d), 4

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BASIC LEVEL 1. Factorize each of the following. a. x3 − 8 b. 8x3 + 1 c. 125 − 27x3 d. 2x3 + 128 INTERMEDIATE LEVEL 2. Factorize each of the following. a. 125x3 − 64y3 b. 27x6 + 8y9 c. 343a3 + 216x6 d. 216a3 – 729x9 ADVANCED LEVEL 3. Factorize each of the following. a. 216 − (3x − 4)3 b. (x + 2)6 + 125 c. (3 – 2x)3 + 8a6 d. (a – 4)3 – (2x + 5)3 4. The difference of the radii of two spherical balls is 2 cm and the difference of their volumes is . Find the radii of the balls. (Take π to be .) ******ebook converter DEMO Watermarks*******

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1. One way in which factorization of algebraic expressions can be done is by extracting common factors from all the terms in the given expressions. It is the reverse of expansion. 2. We can use a multiplication frame to factorize quadratic expressions. For example, consider the expression x2 – 2x – 8.

Step 1: Write x2 in the top-left corner and –8 in the bottom-right corner of the multiplication frame. Step 2: Consider the factors of x2 and –8. Write them in the first column and the first row. Step 3: Multiply them to complete the multiplication frame and check whether the result matches the given expression. Therefore, x2 – 2x – 8 = (x + 2)(x – 4). 3. Special Algebraic Identities:

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These algebraic identities are useful for factorizing algebraic expressions which are of similar forms. (a + b)2 and (a – b)2 are known as perfect squares while (a + b)(a – b) is called the difference of two squares. 4. Factorization by Grouping:

5. Two Special Algebraic Identities:

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1. Factorize each of the following expressions completely. a. 21pq + 14q – 28qr b. 4x – 8(y – 2z) 2. Factorize each of the following expressions completely. a. a2 + 13a + 36 b. b2 – 15b + 56 c. c2 + 14c – 51 d. d2 – 12d – 45 e. q2 – 7q – 18 f. y2 + 5y – 24 3. Factorize each of the following expressions completely. a. 9f2 + 18f – 16 b. 3h2 – 19h – 14 c. 14k2 + 49k + 21 d. 18m2 – 39m + 18 e. ce2 + 5ec – 6c f. pn2 + 3np – 28p ******ebook converter DEMO Watermarks*******

4. Factorize the expression

completely.

5. Factorize each of the following expressions completely. a. x2 + 2xy – 63y2 b. 2x2 + 5xy + 3y2 c. 6x2y2 – 5xy – 4 d. 3z – 8xyz + 4x2y2z 6. Factorize each of the following expressions completely. a. 1 – 121x2 b. x2 + 6xy + 9y2 c. 25x2 – 100xy + 100y2 d. 36y2 – 49(x + 1)2 e. 36x2 – y8 f. 7. Factorize each of the following expressions completely. a. –14xy – 21y2 b. 9xy2 – 36x2y c. (2x – 3y)(a + b) + (x – y)(b + a) d. 5(x – 2y) – (x – 2y)2 e. x2 + 3xy + 2x + 6y f. 3x3 – 2x2 + 3x – 2 g. 4cx – 6cy – 8dx + 12dy h. 5xy – 10x – 12y + 6y2 ******ebook converter DEMO Watermarks*******

8. Factorize the expression x3 + x2 – 4x – 4 completely. 9. Without using a calculator, evaluate each of the following. a. 732 – 272 2. 6592 – 3412 10. Factorize each of the following. a. 8x6 − 343 b. 27a3 + 64y6 c. (2x + 3)3 − (4x − 5)3 d. (3x + 5)3 + (x + 1)3 11. An examination consists of 3 papers. The minimum total score required to pass the examination is (10p + 5q) marks. Miguel scores (p – 3q + 13) marks and (3p + 5q – 4) marks in the first two papers. i. Find Miguel’s total score in the first two papers. ii. Given that Miguel obtained the minimum total score required to pass the examination, find his score in the third paper. iii. Factorize the result in (ii).

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1. Suppose we wish to multiply a 2-digit number with another 2-digit number. If the digits in the tens place are the same and their ones digits add up to ten, there is a shortcut to it! For example,

Try this method to find 58 × 52. Then use a calculator to evaluate 58 × 52. Do you arrive at the same answer? Explain why this works by using algebra. 2. Determine the integer values of n for which n2 – 18n + 45 is a prime number.

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Chapter Two Rational Algebraic Expressions Lenses are used in digital cameras to focus an image on the sensing plate. In the construction of a camera, engineers make use of an important lens formula

, where f, u and v are the focal length, the object distance

and the image distance respectively.

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LEARNING OBJECTIVES At the end of this chapter, you should be able to: illustrate rational algebraic expressions, simplify rational algebraic expressions, perform operations on rational algebraic expressions, solve problems involving rational algebraic expressions

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2.1 Algebraic Fractions Recap (Rational Numbers and Numerical Fractions) In Grade 7, we have learned that a rational number is a number that can be expressed as the ratio of two integers a and b, i.e. in the form , where b ≠ 0. Whole numbers, integers and fractions are rational numbers. We have learned about numerical fractions of the form , where a and b are integers, and b ≠ 0. Examples of numerical fractions are and . In this section, we will learn about rational algebraic expressions or algebraic fractions of the form , where A and/or B are algebraic expressions, and B ≠ 0. Examples of algebraic fractions are and The rules for performing operations on algebraic fractions are the same as those for numerical fractions. One important rule is as follows: The value of a fraction remains unchanged if both its numerator and denominator are multiplied or divided by the same non-zero number or expression, i.e.

and

,

where b, c ≠ 0.

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Worked Example

(Simplification by Dividing by Common Factors) Simplify each of the following.

1 a. b.

We obtain the results shown in Worked Example 1 by dividing the numerators and the denominators by their common factors. The final answers should be in the simplest forms, i.e. the numerators and the denominators have no common factors except 1.

In (a),

.

Solution: a.

b.

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Simplify each of the following. a. b. c. d. e.

Exercise 2A Questions 1(a)-(f), 5(a)-(d)

Worked Example

(Simplification involving Factorization) Simplify each of the following.

2 a. b. c.

Solution: (a)

(extract the common factor a from the numerator and divide the numerator and the denominator by a)

(b) (extract the common factor t from the ******ebook converter DEMO Watermarks*******

denominator and divide the numerator and the denominator by t)

(extract the common factor x from the numerator and the common factor 3 from the denominator, and divide the numerator and the denominator by (x – 3))

(c)

Divisions are usually carried out after both the numerators and the denominators have been completely factorized. Do not divide individual terms of the numerators and the denominators by their common factors. For example, is wrong.

Simplify each of the following. a. b. c. d. e.

Exercise 2A Questions 2(a)-(f) ******ebook converter DEMO Watermarks*******

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Worked Example

(Simplification involving Factorization) Simplify each of the following.

3 a. b. c.

Solution: (a)

(extract the common factor 2m from the numerator and factorize the denominator by using

(b)

(factorize the numerator by using the multiplication frame and extract the common factor 3x from the denominator)

(c)

(factorize each group in the numerator and the denominator respectively) (factorize the two groups in the numerator and the denominator respectively)

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1. Simplify each of the following. a. b. c. 2. Simplify each of the following. a. b.

Exercise 2A Questions 3(a)-(f), 5(e)-(m), 7

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2.2 Multiplication and Division of Algebraic Fractions The procedure for the multiplication and division of algebraic fractions is similar to that of the multiplication and division of numerical fractions, except that now we have to consider the variables. In primary school, we have learned that

.

In general, when we multiply by , we have:

In primary school, we have learned that

.

In general, when we divide by , we have:

Dividing one fraction by another fraction is the same as multiplying the first fraction by the reciprocal of the second fraction. The reciprocal of a fraction is obtained by interchanging the numerator and the denominator of the fraction, e.g. the reciprocal of is .

Worked Example

4

(Multiplication and Division of Algebraic Fractions) Simplify each of the following.

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a. b. c. d.

Solution: a. b.

c.

(factorize h2 – 2h + 1 by using a2 – 2ab +

d. b2 = (a – b)2)

Exercise 2A Questions 4(a)-(d), 6(a)-(l)

1. Simplify each of the following. a. b. c. ******ebook converter DEMO Watermarks*******

d. 2. Simplify

.

BASIC LEVEL 1. Simplify each of the following. a. b. c. d. e. f. 2. Simplify each of the following. a. b. c. d. e. f. ******ebook converter DEMO Watermarks*******

3. Simplify each of the following. a. b. c. d. e. f. 4. Simplify each of the following. a. b. c. d. INTERMEDIATE LEVEL 5. Simplify each of the following. a. b. c. d. e. f. g. ******ebook converter DEMO Watermarks*******

h. i. j. k. l. m. 6. Simplify each of the following. a. b. c. d. e. f. g. h. i. j. k. l. ADVANCED LEVEL ******ebook converter DEMO Watermarks*******

7. Simplify

.

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2.3 Addition and Subtraction of Algebraic Fractions Recap (Simplification of Linear Expressions with Fractional Coefficients) In Grade 7, we have learned how to simplify linear expressions with fractional coefficients. For example,

In this section, we will learn how to add and subtract algebraic fractions.

Addition and Subtraction of Algebraic Fractions Worked Example

(Addition and Subtraction of Algebraic Fractions) Express each of the following as a fraction in its simplest form.

5 a. b. c.

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Solution: a.

b.

c.

Exercise 2B Questions 1(a)-(f), 3(a)-(h)

1. Express each of the following as a fraction in its simplest form. 1. 2. 3. 2. Express each of the following as a fraction in its simplest form. ******ebook converter DEMO Watermarks*******

1. 2. 3.

Worked Example

(Addition and Subtraction of Algebraic Fractions with No Common Factors in the Denominators) Express each of the following as a fraction in its simplest form.

6 a. b.

Solution: a.

b.

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Exercise 2B Questions 2(a)-(h), 4(a)-(f), 5(a)-(d), 6

Express each of the following as a fraction in its simplest form. a. b. c. d. e.

BASIC LEVEL 1. Express each of the following as a fraction in its simplest form. a. b. c. d. e. f. 2. Express each of the following as a fraction in its simplest form. a. ******ebook converter DEMO Watermarks*******

b. c. d. e. f. g. h. INTERMEDIATE LEVEL 3. Express each of the following as a fraction in its simplest form. a. b. c. d. e. f. g. h. 4. Express each of the following as a fraction in its simplest form. a. b. c. d. ******ebook converter DEMO Watermarks*******

e. f. 5. Express each of the following as a fraction in its simplest form. a. b. c. d. ADVANCED LEVEL 6. Express

as a fraction in its simplest form.

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2.4 Manipulation of Algebraic Formulae Changing the Subject of a Formula In Grade 7, we have learned that in general, a formula expresses a rule in algebraic terms by making use of variables to write instructions for performing a calculation. For example, the perimeter P of a rectangle is given by P = 2l + 2b, where l and b represent the length and the breadth of the rectangle respectively. Are we able to find an expression for the length, l, of the rectangle in terms of P and b? (Changing the Subject of a Formula)

Worked Example

7

i. Make l the subject of the formula P = 2l + 2b. ii. Hence, calculate the value of l when P = 132 and b = 30.

Solution: i.

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ii.

Exercise 2C Questions 1(a)-(d), 7(a)-(d)

1. i. Make a the subject of the formula v = u + at. ii. Hence, find the value of a when t = 4, u = 10 and v = 50. 2. The simple interest $I payable on an investment is given by , where $P is the principal, R% is the interest rate on the investment per annum and T is the number of years that the investment is held. i. Make T the subject of the formula

.

ii. Hence, find the number of years that an initial investment of $50 000 must be held in a bank that pays simple interest at a rate of 2% per annum to earn an interest of $4000. (Changing the Subject of a Formula)

Worked Example

8

i. Make x the subject of the formula

.

ii. Hence, calculate the value of x when y = −2.

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To change the subject of a formula involving algebraic fractions, we carry out the steps as follows: Step 1: Eliminate the fractions. Step 2: Manipulate the equation such that all the terms with the unknown which we need to express as the subject are on the left-hand side (LHS) of the equation. Step 3: Factorize the expression on the LHS of the equation, if necessary. Step 4: Divide both sides of the equation such that only the subject of the formula remains on the LHS.

Solution: i.

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ii.

Exercise 2C Questions 2(a)-(d), 8(a)-(d), 10, 15

1. i. Make x the subject of the formula

.

ii. Hence, find the value of x when y = −3. 2. i. Make k the subject of the formula

.

ii. Hence, find the value of k when a = 1, b = –2, p = 3 and x = 9. (Changing the Subject of a Formula)

Worked Example

9

i. Make x the subject of the formula ii. Hence, calculate the value of x when a = 4, b = 3 and k = −1.

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Solution: i.

ii.

Exercise 2C Questions 3(a)-(d), 9(a)-(d), 11, 16-17, 20

1. i. Make x the subject of the formula

.

ii. Hence, find the value of x when a = −5, b = 4 and y = 2. 2. i. Make x the subject of the formula

.

ii. Hence, find the value of x when a = −1, b = 2, k = 1 and p = 5. 3. i. Make a the subject of the formula

.

ii. Hence, find the value of a when b = 2 and c = –3. ******ebook converter DEMO Watermarks*******

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Finding the Value of an Unknown in a Formula

Finding the Value of an Unknown in a Formula In solving problems, we often need to change the subject of a formula to make it easier to find the value of an unknown. For example, if the volume V of a cylinder is 780 cm3 and its height h is 10 cm, how can we find the base radius r of the cylinder? We first make r the subject of the formula V = πr2h before substituting in the values of V and h. 1. Can we substitute the values of V and h into the formula V = πr2h before solving the equation to find the value of r? 2. Are the values of r obtained the same? Explain your answer. From the class discussion, we observe that to find the value of an unknown, we are able to substitute the given values into a formula without changing the subject of the formula.

Worked Example

10

(Finding the Value of an Unknown in a Formula Without Changing the Subject of the Formula) Given that , calculate i. the value of y when x = 1, ii. the value of x when y = 2.

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Solution: a.

b.

Exercise 2C Questions 4-5, 12-13, 19

1. Given that

x, find

i. the value of y when x = 5, ii. the value of x when y = 4. 2. Given that

, find

i. the value of y when

,

ii. the value of x when y = 2. ******ebook converter DEMO Watermarks*******

, find the value of b when a = 3.

3. Given that

, find the value of x when y = 4 and z = 3.

4. Given that

Equations involving Algebraic Fractions We shall learn how to solve equations involving algebraic fractions using the method of changing the subject of a formula.

Worked Example

(Solving Equations involving Algebraic Fractions) Solve each of the following equations.

11 a. b.

Solution: a.

b.

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Exercise 2C Questions 6(a)-(g), 14(a)-(e), 18

Solve each of the following equations. a. b. c. d. e.

BASIC LEVEL 1. In each of the following cases, make the letter in the bracket the subject of the formula. a. b. c. d. 2. In each of the following cases, make the letter in the bracket the subject of the formula. a. ******ebook converter DEMO Watermarks*******

b. c. d. 3. In each of the following cases, make the letter in the bracket the subject of the formula. a. b. c. d. 4. Given that

, find the values of x when a = 2, b = 7 and c = 5.

5. Given that

, find

a. the value of a when b = 7 and c = 2, b. the value of c when a = 4 and b = 9. 6. Solve each of the following equations. a. b. c. d. e. f. g. INTERMEDIATE LEVEL ******ebook converter DEMO Watermarks*******

7. In each of the following cases, make the letter in the bracket the subject of the formula. a. b. c. d. 8. In each of the following cases, make the letter in the bracket the subject of the formula. a. b. c. d. 9. In each of the following cases, make the letter in the bracket the subject of the formula. a. b. c. d. 10. Given that

,

i. make h the subject of the formula, ii. find the value of h when V = 245 and r = 7. 11. Given that

,

i. make b the subject of the formula, ******ebook converter DEMO Watermarks*******

ii. find the value of b when a = 2 and c = 5. 12. Given that

, find

a. the value of p when m = 5, n = 7, x = 4 and y = −2, b. the value of n when m = 14, p = 9, x = 2 and y = 3, c. the values of y when m = 5, n = 4, p = 15 and x = 42. 13. Given that

find

a. the value of A when π = 3.142, h = 15 and r = 7, b. the value of h when π = 3.142, A = 15 400 and r = 14. 14. Solve each of the following equations. a. b. c. d. e. 15. In optics, the focal length, f cm, of a lens, can be calculated using the formula , where u cm and v cm are the object distance and the image distance from the center of the lens respectively. i. Make v the subject of the formula

.

ii. Hence, find the image distance when the focal length of the lens is 20 cm and the object distance is 30 cm. 16. The time taken, T seconds, for a pendulum to complete one oscillation is given by the formula , where l m is the length of the pendulum and g is taken to be 10 m s−2. i. Make l the subject of the formula

.

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ii. Hence, find the length of the pendulum if it takes 12 seconds to complete 20 oscillations. 17. The amount of energy, E joules (J), stored in an object with a mass of m kg is given by the formula , where h m is the height of the object above the ground, v m s−1 is the velocity of the object and g is taken to be 10 m s−2. i. Given that v ≥ 0, make v the subject of the formula

.

ii. Hence, find the velocity of an object with a mass of 0.5 kg, if it is 2 m above the ground and has 100 J of energy.

ADVANCED LEVEL 18. Given that

, find the value of .

19. Given that

, find

1. the value of y when a = 13, b = 15 and x = 3.8, 2. the value of a when b = 13, x = 8.5 and y = 35, 3. the values of b when a = 23, x = 15.6 and y = 56. 20. The resistance of a wire, R ohms (Ω), is directly proportional to its length, l m, and inversely proportional to the square of its radius, r m. a. Express R in terms of l, r and a constant k. b. Make r the subject of the formula in (a). c. i. Given that a wire with a length of 2 m and a radius of 0.5 cm has a resistance of 40 Ω, use your answer in (a) to find the value of k. ii. Hence, use your answer in (b) to find the radius of a wire with a length of 4 m and a resistance of 60 Ω. ******ebook converter DEMO Watermarks*******

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1. The value of a fraction remains unchanged if both its numerator and denominator are multiplied or divided by the same non-zero number or expression, i.e.

and

,

where b, c ≠ 0. 2. When we multiply

, we have:

where b, d ≠ 0. When we divide

, we have:

where b, c, d ≠ 0.

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1. Simplify each of the following. a. b. c. d. e. f. g. h. 2. Simplify each of the following. a. b. c. d. ******ebook converter DEMO Watermarks*******

e. f. g. h. 3. Express each of the following as a fraction in its simplest form. a. b. c. d. e. f. g. h. 4. In each of the following cases, make the letter in the bracket the subject of the formula. a. b. c. d.

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e. f. 5. Given that

, find

a. the value of k when a = 3, b = 6 and c = 20, b. the value of c when a = 4, b = 7 and k = 11. 6. Given that a2 + a2b = 320, find a. the value of b when a = 8, b. the values of a when b = 2 1 5. 7. Given that

, find

a. the value of V when = 3.142, R = 12, h = 14 and r = 9, b. the value of h when = 3.142, R = 8, V = 3800 and r = 6, c. the values of R when = 3.142, V = 3500, h = 17 and r = 8.5, d. the value of r when = 3.142, R = 11, V = 4600 and h = 6.9. 8. Solve each of the following equations. a. b. 9. In an electrical circuit, when two resistors are connected in parallel, the effective resistance, R ohms (Ω), of the two resistors is given by the formula , where R1 and R2 are the resistances of the individual resistors in ohms. i. Make R the subject of the formula

.

ii. Hence, find the effective resistance of the two resistors if R1 = 2 and R2 = 3. ******ebook converter DEMO Watermarks*******

10. The equation of a circle is given by the formula (x – a)2 + (y – b)2 = r2, where (a, b) are the coordinates of the center of the circle and r is the radius. i. Make y the subject of the formula (x – a)2 + (y – b)2 = r2. ii. Hence, find the y-coordinate(s) of the point(s) on the circle when a = 2, b = 3, r = 5 and x = 5.

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1. Simplify . Hint: You need to apply one of the laws of indices. 2. Given that , find the value of n and of k. Hint: You need to apply one of the laws of indices. 3. If a, b and c are non-zero constants such that , show that .

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and

Chapter Three Graphs of Linear Equations A moving walkway is similar to a conveyor belt that allows people to travel a fixed distance. Some moving walkways are steeper than others.

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LEARNING OBJECTIVES At the end of this chapter, you should be able to: illustrate the rectangular coordinate system and its uses, illustrate linear equations in two variables, illustrate the slope of a line, find the slope of a line given two points, an equation and a graph.

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3.1 Cartesian Coordinates

Battleship Game (Two Players)

Search on the Internet for free online battleship games. Sink your opponent’s battleships before your opponent sinks all your ships.

Fig. 3.1 1. On the grid labeled ‘Self’, arrange the following five ships: An aircraft carrier A battleship A submarine A cruiser A destroyer The ships must be placed horizontally or vertically; none of the ships can be placed diagonally. Note that players are not allowed to see each other’s ******ebook converter DEMO Watermarks*******

grids. 2. Take turns to try to sink your opponent’s ships. Call out a location on the grid, e.g. D7, F4 and G10, to hit your opponent’s ship. If a ship is found at the location called out, the other player says ‘hit’. Otherwise, the player says ‘missed’. On the ‘Opponent’ grid, record each location you have called out by shading it for a ‘hit’ and drawing a cross for a ‘miss’. Similarly, on the ‘Self’ grid, record the locations that your opponent has called out. 3. A ship is sunk when all the spaces it occupies have been called out. The player whose ship is sunk says ‘My ship has been sunk.’. 4. A player wins the game when all his opponent’s ships have been sunk. In the battleship game, we use the labels of the columns and the rows to call out locations on the grids. We shall use the same idea to locate a student in a classroom where the students are seated at desks that are arranged neatly in rows and columns as shown in Fig. 3.2.

Fig. 3.2 To locate a student, the teacher can associate the student with the column and the row his or her seat is at. The teacher can write a pair of numbers against ******ebook converter DEMO Watermarks*******

the name of a student in the class list as follows: A(2, 4)

B(4, 6)

C(3, 3)

From the pair of numbers (2, 4), we will be able to know that student A is seated at column 2 and in row 4. The pairs of numbers (4, 6) and (3, 3) tell us that student B is seated at column 4 and in row 6, and student C is seated at column _____ and in row _____ respectively. Are you able to write down the pairs of numbers which correspond to students D, E and F?

Ordered Pairs Discuss each of the following questions with your classmates. 1. Is a single number sufficient to describe the exact position of a student in the classroom seating plan? Can the location of a seat in a cinema be represented by a single number? 2. Is the order in which the two numbers are written important, i.e. do (5, 3) and (3, 5) indicate the same position? The pairs of numbers (2, 4), (4, 6), (3, 3) and so on are examples of ordered pairs. Do you know why they are called ordered pairs?

1. Using a horizontal scale of 1 to 10 and a vertical scale of A to J, design a map that includes the locations of your house, a bus stop and a shopping mall in your neighborhood. 2. Using a horizontal scale of 1 to 12 and a vertical scale of A to K, design a ground floor map for a shopping mall. 3. Use suitable horizontal and vertical scales to design the seating plan of a ******ebook converter DEMO Watermarks*******

100-seat cinema. Now let us display the same classroom plan in Fig. 3.2 with horizontal and vertical lines drawn through the centers of the boxes, showing the positions of the students. The horizontal lines and vertical lines are numbered as shown in Fig. 3.3.

Fig. 3.3 In the battleship game, each location is called out by the column labels followed by the row labels. Similarly, the first number in each ordered pair is with reference to the horizontal scale while the second number is with reference to the vertical scale. To further simplify it, we can use a point (indicated by a cross) to show the position of each student. This gives us an idea of locating a point in a plan.

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Fig. 3.4 Fig. 3.4 shows a rectangular or Cartesian plane which consists of two number lines intersecting at right angles at the point O, known as the origin. The horizontal and vertical axes are called the x-axis and the y-axis respectively. The position of any point in the plane can be determined by its distance from each of the axes. In Fig. 3.4, point A is 2 units to the right of the y-axis and 3 units above the x-axis. Thus its position is given by the ordered pair (2, 3). Similarly, the ordered pair (–3, –1) determines point B and (–3, 3) represents point C. Name the ordered pair that determines point D. What is the origin, O, represented by? In general, each point P in the plane is located by an ordered pair (x, y). We call x the x-coordinate (or abscissa) of P and y the y-coordinate (or ordinate) of P, i.e. P has coordinates (x, y). ******ebook converter DEMO Watermarks*******

On a sheet of graph paper, use a scale of 1 cm to represent 1 unit to draw the x-axis for values of x from –2 to 3 and the y-axis for values of y from –2 to 3. Plot the points A(2, 2), B(–2, 3), C(–1, –2), D(3, –1) and E(1, 0).

Exercise 3A Questions 1-6

Story Time coordinates were invented by René Descartes Cartesian (1596 – 1650) when he tried to describe the path of a fly crawling along crisscrossed beams on the ceiling while he lay on his bed. Due to his poor health, he had developed a lifetime habit of sleeping until 11 a.m. every morning. He broke this habit only when Queen Christina of Sweden persuaded him to go to Stockholm to teach her how to draw tangents at 5 a.m.! After a few months of walking in the cold climate to the palace at 5 a.m., he died of pneumonia. Even up till his death, Descartes displayed a strong sense of diligence.

BASIC LEVEL 1. Write down the coordinates of each point shown in the figure.

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2. Write down the coordinates of the vertices of the following figure.

3. On a sheet of graph paper, use a scale of 1 cm to represent 1 unit to draw the x-axis for values of x from –3 to 6 and the y-axis for values of y from – 2 to 5. Plot the points A(2, 5), B(1, 2), C(–2, –1), D(6, –2), E(3, –2) and F(–1, 2). INTERMEDIATE LEVEL 4. Plot each set of the given points on a sheet of graph paper. Join the points in order with straight lines and identify each geometrical shape obtained. 1. (6, 4), (–6, 4), (–6, –4), (6, –4) ******ebook converter DEMO Watermarks*******

2. (0, 5), (–6, 0), (0, –5), (6, 0) 3. (0, 0), (0, 8), (5, 4) 4. (1, 0), (0, 3), (–1, 4), (–5, –2) 5. (5, 2), (–1, 3), (–1, –3), (5, –2) 5. The vertices of a right-angled triangle are A(1, 0), B(7, 0) and C(1, 8). Plot the points A, B and C on a sheet of graph paper. Hence, find the area of ∆ABC. 6. Plot each of the following points on a sheet of graph paper. (3, –5), (2, –3), (1, –1), (0, 1), (–1, 3), (–2, 5), (–3, 7) Do you notice that the points lie in a special pattern? Describe the pattern. 7. 1. Plot the points A(−2, −3), B(3, −1) and C(1, 4) on a sheet of graph paper. 2. Mark D, the fourth vertex of the square ABCD. 3. Draw the square and give the coordinates of D. ADVANCED LEVEL 8. Two of the vertices of a triangle ABC are A(1, 1) and B(5, 5). The area of ∆ABC is 12 units2 and the y-coordinate of the point C is 1. By plotting the points A and B on a sheet of graph paper, determine the possible xcoordinates of C.

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3.2 Graphs of Linear Equations in Two Variables Consider the linear equation y = 2x. We shall look at all pairs of values x and y that satisfy the equation. When x = 1, y = 2 × 1 = 2;      x = 2, y = 2 × 2 = 4;      x = 2.5, y = 2 × 2.5 = 5;      x = 3, y = 2 × 3 = 6;      x = 3.1, y = 2 × 3.1 = 6.2, etc. Since we are not able to list all the values of x and y that satisfy the linear equation y = 2x, we can use a graph to display the equation as illustrated in Worked Example 1. (Drawing the Graph of a Linear Equation in Two Variables)

Worked Example

1

i. On a sheet of graph paper, using a scale of 2 cm to represent 1 unit on the x-axis and 1 cm to represent 1 unit on the y-axis, draw the graph of the equation y = 2x for values of x from 0 to 4. ii. The point (3, p) lies on the graph in (i). Find the value of p.

Solution: i. We first set up a table of values for x and y. These pairs of values for x and ******ebook converter DEMO Watermarks*******

y satisfy the equation y = 2x.

Using the scale of 2 cm to represent 1 unit on the x-axis and 1 cm to represent 1 unit on the y-axis, the three pairs of values are plotted as points in the Cartesian plane and a straight line is drawn to pass through these three points.

ii. From the graph in (i), when x = 3, p = y = 6

We only need to plot 3 points to obtain the graph of a linear equation. In fact, a straight line can be determined by plotting 2 points. We use the 3rd point to check for mistakes in the graph.

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Ensure that the graph is only drawn for values of x from 0 to 4. The graph must be labeled with y = 2x.

1. i. On a sheet of graph paper, using a scale of 2 cm to represent 1 unit on the x-axis and 1 cm to represent 1 unit on the y-axis, draw the graph of the equation y = 2x + 1 for values of x from 0 to 4. ii. The point (q, 6) lies on the graph in (i). Find the value of q. 2. On a sheet of graph paper, using a scale of 2 cm to represent 1 unit on the x-axis and 1 cm to represent 1 unit on the y-axis, draw the graphs of the equations y = 3x and y = 2 – 2x for values of x from –2 to 2.

Exercise 3B Questions 1-7

Linear Equation in Two Variables Refer to Worked Example 1. 1. The coordinates of the points A and B are (1, 2) and (3, 7) respectively. Do the coordinates of each of the points satisfy the linear equation y = 2x? Explain your answers. 2. Using the graph drawn, state the coordinates of two points that satisfy the linear equation y = 2x. 3. Kristel says that ‘the coordinates of every point on the line satisfy the linear equation y = 2x.’ Discuss with your classmates whether she is right. ******ebook converter DEMO Watermarks*******

The line drawn in Worked Example 1 is said to be the graph of the linear equation y = 2x because the graph is a straight line. 1. What can you say about the coordinates of the points that lie on the line y = 2x? 2. Tricia says that the graphs of the equations y = x + 3 and y = –2x – 1 are linear. Do you agree with her? Explain your answer.

BASIC LEVEL 1. a. On a sheet of graph paper, using a scale of 1 cm to represent 1 unit on the x-axis and 1 cm to represent 2 units on the y-axis, draw the graph of each of the following equations for values of x from 0 to 4. i. y = 2x + 8 ii. y = 2x + 2 iii. y = 2x – 3 iv. y = 2x – 6 b. What do you notice about the lines you have drawn in (a)? 2. a. On a sheet of graph paper, using a scale of 1 cm to represent 1 unit on the x-axis and 1 cm to represent 2 units on the y-axis, draw the graph of each of the following equations for values of x from –4 to 4. i. y = 3x + 7 ******ebook converter DEMO Watermarks*******

ii. y = 3x + 5 iii. y = 3x – 3 iv. y = 3x – 6 b. What do you notice about the lines you have drawn in (a)? 3. a. On a sheet of graph paper, using a scale of 1 cm to represent 1 unit on the x-axis and 1 cm to represent 2 units on the y-axis, draw the graph of each of the following equations for values of x from –4 to 4. i. y = –2x + 5 ii. y = –2x + 3 iii. y = –2x – 4 iv. y = –2x – 7 b. What do you notice about the lines you have drawn in (a)? INTERMEDIATE LEVEL 4. a. On a sheet of graph paper, using a scale of 1 cm to represent 1 unit on the x-axis and 1 cm to represent 2 units on the y-axis, draw the graph of each of the following equations for values of x from –4 to 4. i. y = –4x + 8 ii. y = –4x + 2 iii. y = –4x – 3 iv. y = –4x – 6 b. Write down another set of four linear equations whose graphs are parallel to each other. 5. i. On a sheet of graph paper, using a scale of 2 cm to represent 1 unit on ******ebook converter DEMO Watermarks*******

the x-axis and 1 cm to represent 1 unit on the y-axis, draw the graph of the equation y = −3x for the values of x from −2 to 2. ii. The point (−1, r) lies on the graph in (i). Find the value of r. 6. i. On a sheet of graph paper, using a scale of 2 cm to represent 1 unit on the x-axis and 1 cm to represent 1 unit on the y-axis, draw the graph of the equation y = 6 – 3x for values of x from –3 to 3. ii. The points (a, 0), (–2, b) and (c, 1.5) lie on the graph in (i). Find the values of a, b and c. 7. On a sheet of graph paper, using a scale of 2 cm to represent 1 unit on the x-axis and 1 cm to represent 1 unit on the y-axis, draw the graphs of the equations y = 2x + 4 and y = 2 – 3x for values of x from –2 to 2.

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3.3 Slope of a Straight Line The equation of a straight line is in the form y = mx + c, where m and c are constants. What happens to the line if we change the value of m and of c?

Equation of a Straight Line In this investigation, we shall explore how the graph of a straight line in the form y = mx + c changes when either m or c varies. Go to http://www.shinglee.com.sg/StudentResources/ spreadsheet ‘Equation of a Straight Line’.

and

open

the

1. Change the value of c from –3 to 3 in steps of 1 by clicking on the scroll bar. What happens to the line? State the coordinates of the point where the line cuts the y-axis. 2. Change the value of m from 0 to 5 in steps of 1. What happens to the line? 3. Change the value of m from 0 to –5 in steps of –1. What happens to the line? 4. What is the difference between a line with a positive value for m and a line with a negative value for m?

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Fig. 3.5

Slope of a Straight Line Look at the roads around you. Some roads are steeper than others. The steeper a road is, the harder it is to walk up the road. Slope is a measure of steepness. From the investigation on the previous page, The equation of a straight line is in the form of y = mx + c, where the constant m is the slope of the line and the constant c is the yintercept.

Note: The y-intercept refers to the y-coordinate of the point of intersection of the line with the y-axis. From the investigation, we have discovered that the slope of a line can be either positive or negative. ******ebook converter DEMO Watermarks*******

Fig. 3.6(a) shows a line that goes __________ from the left to the right and its slope is positive. Fig. 3.6(b) shows a line that goes __________ from the left to the right and its slope is negative.

Fig. 3.6 In real life, can the slope be negative? From the investigation, we can also see that as the absolute value of the slope m increases, the steepness of the line increases. For example, a line with a slope of 3 is steeper than a line with a slope of 2 since 3 > 2; but a line with a slope of –3 is also __________ than a line with a slope of –2 although –3 < – 2. How do we find the slope of a line? The slope of a straight line is the measure of the ratio of the vertical change (or rise) to the horizontal change (or run), i.e.

In Fig. 3.6(a), the line goes downward from the right to the left but its slope is positive. Thus it is important to specify whether the line is going upward or ******ebook converter DEMO Watermarks*******

downward from the left to the right.

Worked Example

(Finding Slopes of Straight Lines) Find the slope of each of the following lines.

2 a.

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b.

Solution: a.

Take two points A and B on the line and draw dotted lines to form the ******ebook converter DEMO Watermarks*******

right-angled triangle ABC. Vertical change (or rise) Horizontal change (or run) Since the line goes upward from the left to the right, its slope is positive.

b.

Take two points P and Q where the line cuts the y-axis and x-axis respectively. Let O be the origin (0, 0). ******ebook converter DEMO Watermarks*******

Vertical change (or rise) OP = 6 Horizontal change (or run) OQ = 2 Since the line goes downward from the left to the right, its slope is negative.

Exercise 3C Questions 6(a)-(h)

Find the slope of each of the following lines. a.

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Slopes of Straight Lines Work in pairs. Consider the points D(1, 3) and E(2.5, 6) on the line in Worked Example 2(a). 1. Find the slope of DE. 2. Is slope of DE = slope of AB? 3. Hence, we can choose any two points on a line to find its slope because the slope of a straight line is ____________. After learning how to find the slope of a straight line, we need to have a sense of the magnitude of the steepness of a line.

Slopes in the Real World 1. How steep is a road with a slope of 1? Fig. 3.7 shows a line with a slope of 1. Measure the angle of inclination between the line and the horizontal dotted line.

Fig. 3.7 2. How steep is a road with a slope of 2? Make an accurate drawing of a line with a slope of 2, and indicate the vertical change and horizontal change ******ebook converter DEMO Watermarks*******

clearly. Measure the angle of inclination. 3. Repeat Step 2 for a road with a slope of 1 2. 4. Do you consider a road with a slope of 1 steep or gentle? Discuss with your classmates if there are many roads in the Philippines that have a slope of 1. 5. Do you consider a road with a slope of 1 2 steep or gentle? Discuss with your classmates whether the slopes of most roads in the Philippines are greater than or less than 1 2.

The steepest street in the world is Baldwin Street in Dunedin, New Zealand, with a slope of about 0.38.

Finding the Slope of a Straight Line 1. In Fig. 3.8(a) and (b), A and B are two points on the line.

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Fig. 3.8 i. Find the slope of each line segment. ii. Choose two other points that lie on each of the line segments and calculate the slope of each line segment. Compare your answers with those obtained in (i). What do you notice? Explain your answer. 2. Given any two points A(x1, y1) and B(x2, y2), how would you find the slope of the line passing through A and B? 3. Using your answer in Question 2, find the slope of the line passing through each of the following pairs of points. a. (−1, 4) and (3, 7) b. (−4, −3) and (2, −11) c. (6, 3) and (–4, 3) d. (2, –1) and (2, 8) Compare your answers with those obtained by your classmates. ******ebook converter DEMO Watermarks*******

From the class discussion, we observe that if A(x1, y1) and B(x2, y2) are two points on a line, then

Fig. 3.9

Instead of writing the slope of AB as

, we can also write it as

. Is

? Explain your answer.

Horizontal Line

Slope of a Horizontal Line In this investigation, we shall find out what the slope of a horizontal line is. Fig. 3.10 shows a horizontal line.

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Fig. 3.10 1. There are four points on the line. The coordinates of A and C are given. Write down the coordinates of B and of D. 2. In the line segment AC, rise = ________ and run = ________. 3. In the line segment BD, rise = ________ and run = ________. 4. What can you conclude about the slope of a horizontal line?

In this case, the equation of the horizontal line is y = 2. Notice that the ycoordinates of all the points on the horizontal line are equal to 2.

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Vertical Line

Slope of a Vertical Line In this investigation, we shall find out what the slope of a vertical line is. Fig. 3.11 shows a vertical line.

Fig. 3.11 1. There are four points on the line. The coordinates of P and of R are given. Write down the coordinates of Q and of S. 2. In the line segment PR, rise = ________ and run = ________. 3. In the line segment QS, rise = ________ and run = ________. 4. What can you conclude about the slope of a vertical line? ******ebook converter DEMO Watermarks*******

In this case, the equation of the vertical line is x = 3. Notice that the xcoordinates of all the points on the vertical line are equal to 3.

Exercise 3C Questions 1, 7, 13

Slope of a Straight Line 1. Using a suitable geometry software, draw a line segment with the endpoints as A(–2, 1) and B(0, 5). 2. Find the slope of the line segment AB and record it in Table 3.1. Describe the slope of the line segment AB using one of the following terms: positive, negative, zero or undefined. 3. Write down the value of y2 – y1 and of x2 – x1 in Table 3.1. 4. Repeat Steps 1–3 for each of the following pairs of points. a. C(7, 5) and D(4, 8) b. E(–2, 6) and F(–4, 3) c. G(1, 1) and H(3, 1) d. I(–4, 3) and J(–4, 6) Table 3.1 1. 1. When y2 – y1 > 0 and x2 – x1 < 0, what do you notice about the sign of the slope? 2. When y2 – y1 < 0 and x2 – x1 > 0, what do you notice about the sign of ******ebook converter DEMO Watermarks*******

the slope? 3. When the signs of y2 – y1 and x2 – x1 are the same, what do you notice about the sign of the slope? 4. When y2 – y1 = 0, what do you notice about the slope of the line? 5. When x2 – x1 = 0, what do you notice about the slope of the line? From the investigation, we observe that the slope of a straight line can be positive, negative, zero or undefined, if y2 – y1 and x2 – x1 have the same signs, the slope of the straight line is positive,

if y2 – y1 and x2 – x1 have opposite signs, the slope of the straight line is negative,

if y2 – y1 = 0 or y2 = y1, the slope of a horizontal line is zero,

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if x2 – x1 = 0 or x2 = x1, the slope of a vertical line is undefined.

When the slope of the line is positive, as the value of x increases (from point A to point B), the value of y also increases.

Worked Example

3

(Finding Slope given Two Points) Find the slope of the line passing through each of the following pairs of points. a. A(2, 3) and B(7, 5) b. P(–2, 8) and Q(1, –1)

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Solution: a.

b.

Exercise 3C Questions 2(a)-(f), 3, 12

Find the slope of the line passing through each of the following pairs of points. a. C(3, 1) and D(6, 3) b. H(5, –7) and K(0, –2) c. M(–4, 1) and N(16, 1) ******ebook converter DEMO Watermarks*******

d. P(8, 0) and Q(–2, –4) e. U( 1 2, –3) and V(–3, 4)

Worked Example

4

(Using the Slope to Determine the Coordinates of a Point on the Line) 1. If the slope of the line joining the points (k, 5) and (2, k) is –2, find the value of k. 2. The points A, B and C have coordinates (−6t, 6), (8, −t) and (11, 3 − t) respectively. If the slope of AB is the same as the slope of BC, find the value of t.

Solution: a.

b.

a. If the slope of the line joining the points (4, –9) and (–3, h) is –3, find the ******ebook converter DEMO Watermarks*******

value of h. b. If the slope of the line joining the points (k, 3) and (7, −k) is −1, find the value of k. c. The points M and N have coordinates (p, p) and (−6, 2) respectively. If the line MN has the slope 1 5, find the value of p. d. The points D, E and F have coordinates (2w, 7), (3, w + 4) and (9, w + 8) respectively. If the slope of DE is the same as the slope of EF, find the value of w. e. The points G(−1, r + 1), H(r + 6, 12) and I(r + 7, 17) are collinear, i.e. they lie on a straight line. Find the value of r.

Exercise 3C Questions 4-5, 8-11

BASIC LEVEL 1.

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Write down the slope of each of the given lines. 2. Find the slope of the line passing through each of the following pairs of points. a. A(0, 0) and B(–2, 1) b. C(2, –3) and D(1, 7) c. E(–2, 4) and F(–5, 8) d. G(–4, 7) and H(1, –8) e. I(–2, –5) and J(2, 6) f. K(–7, 9) and L(6, 9) 3. The points A(0, 1), B(7, 1), C(6, 0), D(0, 5) and E(6, 4) are shown in the diagram.

Find the slope of each of the line segments AB, AE, DC and DE. 4. If the slope of the line joining the points (–3, –7) and (4, p) is 5 3, find the value of p. 5. The coordinates of A and B are (3k, 8) and (k, –3) respectively. Given that the slope of the line segment AB is 3, find the value of k. INTERMEDIATE LEVEL 6. Given that the equation of the line representing each of the following linear graphs is in the form y = mx + c, find the slope m and state the y-intercept ******ebook converter DEMO Watermarks*******

c. a.

b.

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c.

d.

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e.

f.

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g.

h.

7. The

figure

shows

five

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line

segments.

Find the slope of each of the line segments. 8. The slope of the line joining the points (9, a) and (2a, 1) is 3. Find the value of a. 9. The points P, Q and R have coordinates (6, –11), (k, –9) and (2k, –3) respectively. If the slope of PQ is equal to the slope of PR, find the value of k. 10. The points P(2, –3), Q(3, –2) and R(8, z) are collinear, i.e. they lie on a straight line. Find the value of z. 11. The line joining the points A(2, t) and B(7, 2t + 7) has a slope of 2. Find the value of t. ADVANCED LEVEL 12. The coordinates of the vertices of a square ABCD are A(0, 6), B(2, 1), C(7, 3) and D(5, 8). i. Find the slope of all four sides of ABCD. ii. What do you observe about the slopes of the opposite sides of a square? ******ebook converter DEMO Watermarks*******

13.

In the figure, Line 1 is parallel to the x-axis and Line 3 is parallel to the yaxis. Line 2 is parallel to Line 5 and Line 4 is parallel to Line 6. If the slopes of Line 5 and Line 6 are –3 and 1 2 respectively, write down the slopes of Line 1, Line 2, Line 3 and Line 4.

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1. A Cartesian plane consists of two axes – the x-axis and the y-axis – intersecting at right angles at the origin O(0, 0). 2. The coordinates of a point P in the Cartesian plane are (x, y), where x is the x-coordinate and y is the y-coordinate of the point. 3. The equation of a straight line is in the form y = mx + c, where the constant m is the slope of the line and the constant c is the y-intercept. 4. The slope of a straight line is the measure of the ratio of the vertical change (or rise) to the horizontal change (or run), i.e.

5. If A(x1, y1) and B(x2, y2) are two points on a line, then slope of

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1. Plot each set of the given points on a sheet of graph paper. Join the points in order with straight lines and identify each geometrical shape obtained. a. (–2, 2), (–2, 6), (4, 6), (4, 2) b. (2, –2), (6, 2), (2, 6), (–2, 2) c. (2, –4), (8, 4), (6, 8), (–2, 4) d. (0, 7), (2, 7), (2, 5), (–4, 1) 2. The figure shows a circle.

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a. Write down the coordinates of each of the points shown in the figure. b. State the point on the circle that has i. the same x-coordinate as E, ii. the same y-coordinate as J. 3. The equation of a line is

. Find the value of y when

i. x = 12, ii.

,

iii.

.

4. The equation of a line is y = 250 – 20x. Find the value of x when ******ebook converter DEMO Watermarks*******

i. y = 150, ii. y = 450, iii. y = –1150. 5. i. On a sheet of graph paper, using a scale of 2 cm to represent 1 unit on the x-axis and 1 cm to represent 1 unit on the y-axis, draw the graph of the linear equation y = 2 1/2 x + 3 for values of x from –3 to 3. ii. The points (–2, a) and (b, 3) lie on the graph in (i). Find the value of a and of b. 6. Given that the equation of the line representing each of the following linear graphs is in the form y = mx + c, find the slope m and state the y-intercept c. a.

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b.

7. A straight line has a slope of 2 and passes through the point (0, –3). i. Write down the equation of the straight line. ii. Given that the line also passes through the point (4, k), find the value of k. 8. The equation of a straight line is 6x + 2y = 7. i. Find the slope of the line. Another line with equation y = mx + c has the same slope as 6x + 2y = 7 and passes through the point (3, 5). ii. Find the value of c. 9. The coordinates of the points A and B are (1, 5) and (2, –3) respectively. Find the equation of the line passing through the origin and having the same slope as AB. ******ebook converter DEMO Watermarks*******

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The figure shows a distance-time graph. Write a scenario that describes the graph. Come up with questions that you can ask based on the graph and provide the solutions.

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Chapter Four Linear Graphs and Simultaneous Linear Equations During festive seasons, people usually buy gifts for their friends and relatives. When shopping for gifts, if we narrow down our gift choices to two items and we have a fixed budget, how do we determine the number of each item to buy?

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LEARNING OBJECTIVES At the end of this chapter, you should be able to: write the linear equation ax + by = c in the form y = mx + b and vice versa, graph a linear equation given (a) any two points; (b) the x – and y – intercepts; (c) the slope and a point on the line, describe the graph of a linear equation in terms of its intercepts and slope, find the equation of a line given (a) two points; (b) the slope and a point; (c) the slope and its intercepts, solve problems involving linear equations in two variables, illustrate a system of linear equations in two variables, graph a system of linear equations in two variables, categorize when a given system of linear equations in two variables has graphs that are parallel, intersecting and coinciding, solve a system of linear equations in two variables by (a) graphing; (b) substitution; (c) elimination, solve problems involving systems of linear equations in two variables.

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4.1 Horizontal and Vertical Lines In Chapter 3, we have learned that the equation of a straight line is in the form y = mx + c, where the constant m is the slope of the line and the constant c is the y-intercept. Consider the equation of the straight line y = 2x + 1. For each value of x, there is a corresponding value of y. When x = 0, y = 2(0) + 1 = 1. When x = 2, y = 2(2) + 1 = 5. If we plot these points on a sheet of graph paper, we obtain a straight line. Since the points (0, 1) and (2, 5) lie on the line, the coordinates of these points satisfy the equation y = 2x + 1. Does the point (4, 10) lie on the line with equation y = 2x + 1? We replace x by 4 and y by 10 in the equation y = 2x + 1.

Since the coordinates of (4, 10) do not satisfy the equation y = 2x + 1, the point does not lie on the line. What are the equations of a horizontal line and a vertical line?

Equation of a Horizontal Line ******ebook converter DEMO Watermarks*******

Fig. 4.1 shows a horizontal line.

Fig. 4.1 1. What is the slope of the horizontal line? 2. There are 4 points on the line. The coordinates of A and of C are given. Write down the coordinates of B and of D. 3. What do you notice about the y-coordinates of all the four points on the horizontal line? 4. What do you think the equation of the horizontal line is? From the investigation, since the slope m of a horizontal line is 0, the equation of a horizontal line is

The y-coordinates of all the points on a horizontal line are equal to c(a constant).

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Exercise 4A Question 1

a. Write down the equation of each of the given horizontal lines. b. In the graph, draw each of the lines with the following equations. i. y = 2 ii. y = 0 Describe the lines. ******ebook converter DEMO Watermarks*******

Equation of a Vertical Line Fig. 4.2 shows a vertical line.

Fig. 4.2 1. What is the slope of the vertical line? 2. There are 4 points on the line. The coordinates of P and of R are given. Write down the coordinates of Q and of S. 3. What do you notice about the x-coordinates of all the four points on the vertical line? 4. What do you think the equation of the vertical line is? ******ebook converter DEMO Watermarks*******

From the investigation, since the slope m of a vertical line is undefined, we cannot write the equation of a vertical line in the form y = mx + c. As the xcoordinates of all the points on a vertical line are equal to the same constant value a, the equation of a vertical line is

Exercise 4A Question 2

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a. Write down the equation of each of the given vertical lines. b. In the graph, draw each of the lines with the following equations. i. x = –3.5 ii. x = 0 Describe the lines.

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4.2 Graphs of Linear Equations in the form ax + by = k and y = mx + c Equation of a Straight Line y = mx + c In Chapter 3, we have learned that the equation of a straight line is in the form y = mx + c, where the constant m is the slope of the line and the constant c is the y-intercept.

Fig. 4.3 In Fig. 4.3, the straight line passes through the points A(0, c) and P(x, y). If the slope of the line is m, then

In general, ******ebook converter DEMO Watermarks*******

for a straight line passing through the point (0, c) and with slope m, the equation is y = mx + c. The equation y = mx + c is known as the slope-intercept form of the equation of a straight line. In this equation, m gives the slope of the straight line, c gives the intercept on the y-axis and (0, c) is the point where the line cuts the y-axis.

Worked Example

1

(Finding the Equation of a Straight Line given the Coordinates of 2 Points and Drawing the Line) Find the equation of the straight line passing through each of the following pairs of points and draw the line. a. A(1, 2) and B(3, 7) b. C(2, 3) and D(7, 3) c. E(5, 1) and F(5, 6)

We can also substitute (3, 7) into the equation of AB to find the value of c.

Solution: a.

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b. C(2, 3) and D(7, 3) have the same y-coordinate of value 3. ∴ CD is a horizontal line with equation y = 3.

c. E(5, 1) and F(5, 6) have the same x-coordinate of value 5. ∴ EF is a vertical line with equation x = 5.

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Exercise 4A Questions 5(a)-(h), 10, 17

Find the equation of the straight line passing through each of the following pairs of points and draw the line. a. A(–2, 1) and B(5, 3) b. C(6, 4) and D(–4, 4) c. E(–3, 5) and F(–3, 8) d. G(0, 1) and H(6, 2) e. I(– 1 2, −1) and J(0, −2)

Worked

(Finding the Equation of a Straight Line given the Slope and a Point and Drawing the Line)

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Example

2

Find the equation of the straight line with slope 2 and passing through (−2, 1). Draw the line.

Solution: Equation of the line is in the form y = 2x + c Since (−2, 1) lies on the line, the coordinates (−2, 1) must satisfy the equation,

∴ Equation of the line is y = 2x + 5

Exercise 4A Questions 3, 4, 6, 7, 9(a)-(d), 11, 13-16, 20

Find the equation of each of the following straight lines. Draw the line. ******ebook converter DEMO Watermarks*******

a. Slope −2 and passing through (3, 1) b. Slope 3 and passing through (1, 5) c. Slope 0 and passing through (−2, 3) d. Slope

and passing through (0, −4)

e. Slope a and passing through (a, 0), where a > 1

Worked Example

3

(Finding the Equation of a Straight Line given its Intercepts and Drawing the Line) Find the equation of the straight line with xintercept 3 and y-intercept −2. Draw the line.

Solution: Equation of the line is in the form y = mx − 2 Slope of the line is positive. ∴ Equation of the line is

Alternatively, equation of the line can be found using the intercept-intercept form,

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where a gives the intercept on the x-axis and b gives the intercept on the yaxis.

∴ Equation of the line is

Exercise 4A Questions 8, 18

Find the equation of the straight line with each of the following pairs of xintercept and y-intercept. Draw the line. a. 1, ½ b. –2, 4 c. –1, –3 d. 8, –4 e. –⅓, 2

Comparing Graphs of Linear Equations in the ******ebook converter DEMO Watermarks*******

form ax + by = k and y = mx + c We have learned how to draw graphs of linear equations in the form y = mx + c, where m and c are constants. In this section, we shall take a look at the graphs of linear equations in the form ax + by = k, where a, b and k are constants.

Graphs of ax + by = k 1. Consider the equation 2x + y = 3. i. Using a graphing software, draw the graph of 2x + y = 3. ii. Do the points A(2, –1) and B(–2, 5) lie on the graph in (i)? Do the coordinates of each of the points satisfy the equation 2x + y = 3? Explain your answers. iii. The point (1, p) lies on the graph in (i). Determine the value of p. iv. The point (q, –7) lies on the graph in (i). Determine the value of q. v. On the same axes in (i), draw the graph of y = −2x + 3. What do you notice? Hence, show algebraically that y = −2x + 3 can be obtained from 2x + y = 3. 2. Consider the equation 3x – 4y = 6. i. Using a graphing software, draw the graph of 3x – 4y = 6. ii. The point (2, r) lies on the graph in (i). Determine the value of r. iii. The point (s, –1.5) lies on the graph in (i). Determine the value of s. iv. State the coordinates of two other points that satisfy the equation 3x – 4y = 6. v. On the same axes in (i), draw the graph of Hence, show algebraically that 6.

. What do you notice?

can be obtained from 3x − 4y =

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Worked Example

4

(Drawing the Graph of ax + by = k) The variables x and y are connected by the equation 2x – 3y = 2. Some values of x and the corresponding values of y are given in the table.

a. Calculate the value of p. b. On a sheet of graph paper, using a scale of 1 cm to represent 1 unit on the x-axis and 2 cm to represent 1 unit on the y-axis, draw the graph of 2x – 3y = 2 for –2 ≤ x ≤ 4. c. The point (1, q) lies on the graph in (b). Find the value of q. d. i. On the same axes in (b), draw the graph of y = 1. ii. State the x-coordinate of the point on the graph of 2x – 3y = 2 that has a y-coordinate of 1. e. Express 2x – 3y = 2 in the form y = mx + c. On the same axes in (b), draw the line. What can you say about the graphs of the linear equations?

–2 ≤ x ≤ 4 represents values of x that are more than or equal to –2 but less than or equal to 4.

Solution: a. When x = –0.5, y = p,

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b.

c. From the graph in (b), When x = 1, q=y=0 d. (ii) x-coordinate of point = 2.5 e.

The graphs of 2x − 3y = 2 and

are the same.

The variables x and y are connected by the equation 3x + y = 1. Some values of x and the corresponding values of y are given in the table. ******ebook converter DEMO Watermarks*******

a. Find the value of p. b. On a sheet of graph paper, using a scale of 4 cm to represent 1 unit on the x-axis and 1 cm to represent 1 unit on the y-axis, draw the graph of 3x + y = 1 for –2 ≤ x ≤ 2. c. The point (–1, q) lies on the graph in (b). Find the value of q. d. 1. On the same axes in (b), draw the graph of y = –0.5. 2. State the x-coordinate of the point on the graph of 3x + y = 1 that has a y-coordinate of –0.5. e. Express 3x + y = 1 in the form y = mx + c. On the same axes in (b), draw the line. What can you say about the graphs of the linear equations?

Exercise 4A Questions 12, 19, 21

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BASIC LEVEL 1.

a. Write down the equation of each of the given horizontal lines. b. In the graph, draw each of the lines with the following equations. i. y = –3 2. Describe the lines.

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2.

a. Write down the equation of each of the given vertical lines. b. In the graph, draw each of the lines with the following equations. i. x = 1 ii. Describe the lines. 3. Given that y = –x + c passes through the point (1, 2), find the value of c. 4. The point (–3, 3) lies on the line y = 4x + k. Find the value of k. 5. Find the equation of the straight line passing through each of the following pairs of points and draw the line. a. A(0, 0) and B(1, –1) b. C(1, 3) and D(2, 5) c. E(2, 4) and F(–2, 3) d. G(–6, –5) and H(4, 4) ******ebook converter DEMO Watermarks*******

e. I(–2, –4) and J(1, –7) f. K(–7, –5) and L(–1, –1) g. M(8, 0) and N(–9, 0) h. O(0, 0) and P(0, 7) 6. Find the equation of each of the straight lines, given the slope and the coordinates of a point that lies on it. Draw the line. a. b. 3, (1, 1) c. –3, (2, –5) d. e. 0, (5, 4) f. a, (0, a) 7. Write down the equation of the straight line which passes through the origin and with slope 2. 8. Find the equation of the straight line with each of the following pairs of xintercept and y-intercept. Draw the line. a. 1, 1 b. 3, 1 c. –1, 4 d. –2, –3 e. f. 9. In each of the following diagrams, find the slope and the y-intercept of the line where possible. State the equation of each line.

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a.

b.

c.

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d.

INTERMEDIATE LEVEL 10. The diagram shows ΔABC with vertices A(1, 1), B(1, 3) and C(5, 5).

i. Find the area of ΔABC. ii. Find the slope of the line passing through B and C. iii. Find the equation of the line passing through A and C. 11. The lines 2x – 5 = ky and (k + 1)x = 6y – 3 have the same slope. Find the possible values of k. 12. Given the line x 3 + 2 y = 1, i. make y the subject of the formula x 3 + 2 y = 1, ii. find the slope of the line, ******ebook converter DEMO Watermarks*******

iii. find the coordinates of the point at which the line cuts the x-axis. 13. i. Find the equation of the straight line which passes through the point (–3, 5) and with slope − 2 3. ii. Given that the line in (i) also passes through the point (p, 3), find the value of p. 14. Find the equation of the straight line passing through the point (3, –2) and having the same slope as the line 2y = 5x + 7. 15. i. Find the equation of the straight line which passes through the point (3, 1) and with slope 3. ii. Hence, find the coordinates of the point of intersection of the line in (i) with the line y = x. 16. The line l has equation 5x + 6y + 30 = 0. Given that P is the point (3, –1), find i. the coordinates of the point where l crosses the x-axis, ii. the coordinates of the point of intersection of l with the line x = 2, iii. the equation of the line passing through P and having the same slope as l, iv. the equation of the line passing through P and having a slope of 0. 17. A straight line l passes through the points A(0, 3) and B(3, 12). a. Find i. the slope of the line l, ii. the equation of the line l. b. The line x = 3 is the line of symmetry of ΔABC. Find the coordinates of C. 18. ******ebook converter DEMO Watermarks*******

i. Find the equation of the straight line which has x-intercept 3 and yintercept −6. ii. Hence, find the coordinates of the point of intersection of the line in (i) with the line x = 2. ADVANCED LEVEL 19. The variables x and y are connected by the equation –x + 2y = 4. Some values of x and the corresponding values of y are given in the table.

a. Find the value of p and of q. b. On a sheet of graph paper, using a scale of 1 cm to represent 1 unit on the x-axis and 2 cm to represent 1 unit on the y-axis, draw the graph of –x + 2y = 4 for –5 ≤ x ≤ 5. c. The point (r, 0.5) lies on the graph in (b). Find the value of r. d. i. On the same axes in (b), draw the graph of x = 3. ii. State the y-coordinate of the point on the graph of –x + 2y = 4 that has an x-coordinate of 3. e. Express −x + 2y = 4 in the form y = mx + c. On the same axes in (b), draw the line. What can you say about the graphs of the linear equations? 20. If the line mx = ny + 2 has the same slope as the x-axis, find the value of m. State the condition for the line to be parallel to the y-axis instead. 21. Consider the equation –2x + y = –3. a. Copy and complete the table.

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b. On a sheet of graph paper, using a scale of 4 cm to represent 1 unit on the x-axis and 2 cm to represent 1 unit on the y-axis, draw the graph of – 2x + y = –3 for –1 ≤ x ≤ 2. c. Express –2x + y = –3 in the form y = mx + c. On the same axes in (b), draw the line. What can you say about the graphs of the linear equations? d. i. On the same axes in (b), draw the graph of y = –1. ii. Find the area of the trapezium bounded by the lines –2x + y = –3, y = –1, and the x- and y-axes.

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4.3 Solving Simultaneous Linear Equations Using Graphical Method We have learned how to draw the graphs of etc. What is the relationship between the graphs of two linear equations when we draw them on the same axes? What is the connection between the coordinates of the point of intersection of the two graphs and the pair of values of x and y that satisfies both the equations?

Solving Simultaneous Linear Equations Graphically 1. Consider the linear equations 2x + 3y = 5 and 3x – y = 2. i. Using a graphing software, draw the graphs of 2x + 3y = 5 and 3x – y = 2 on the same axes. ii. What are the coordinates of the point of intersection of the two graphs? iii. Five pairs of values of x and y are given in Table 4.1.

Table 4.1 ******ebook converter DEMO Watermarks*******

Determine the pair of values of x and y that satisfies both the equations 2x + 3y = 5 and 3x – y = 2. What do you notice? 2. Consider the linear equations 3x – 4y = 10 and 5x + 7y = 3. i. Using a graphing software, draw the graphs of 3x – 4y = 10 and 5x + 7y = 3 on the same axes. ii. What are the coordinates of the point of intersection of the two graphs? iii. Hence, state the pair of values of x and y that satisfies both the equations 3x – 4y = 10 and 5x + 7y = 3. What can we conclude about the coordinates of the point of intersection of the two graphs and the pair of values of x and y that satisfies both the equations? Explain your answer. In the investigation, the graphs of 2x + 3y = 5 and 3x – y = 2 intersect at the point (1, 1). x = 1 and y = 1 satisfies the two linear equations simultaneously. We say that x = 1 and y = 1 is the solution of the simultaneous linear equations 2x + 3y = 5 and 3x – y = 2. Can we say the same for the linear equations 3x – 4y = 10 and 5x + 7y = 3?

Choice of Appropriate Scales for Graphs Before we proceed to draw a graph, we have to choose a suitable scale. The following guidelines may be useful: Use a convenient scale for both the x-axis and the y-axis. For example, we may use 1 cm to represent 1 unit, 2 units, 4 units, 5 units or 10 units. Avoid using awkward scales such as 1 cm to represent 3 units or 1 cm to represent 4.3 units. The scale used for the x-axis need not be the same as the scale used in the yaxis. Choose a suitable scale so that the graph will occupy more than half the size of the graph paper. Look at the largest and the smallest value of x and estimate the scale to be ******ebook converter DEMO Watermarks*******

used. Repeat the process for the values of y.

Choice of Appropriate Scales for Graphs and Accuracy of Graphs Work in pairs. 1. Using a suitable scale, draw the graph of y = 3x – 1. Compare your graph with that of your classmate. Do the graphs look different? 2. Use your graph in Question 1 to find i. the value of y when x = 1.3, ii. the value of x when y = −2.8. 3. How do you check for the accuracy of your answers in Question 2? 4. If your answers in Question 2 are inaccurate, how can you improve your graph?

Worked Example

5

(Solving Simultaneous Linear Equations Using Graphical Method) Using the graphical method, solve the simultaneous equations

Solution: See Table.

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The graphs intersect at the point (6, –4). ∴ The solution is x = 6 and y = –4.

We only need to plot 3 points to obtain the graph of a linear equation. In fact, a straight line can be determined by plotting 2 points. We use the 3rd point to check for mistakes in the graph.

Exercise 4B Questions 1(a)-(f), 2(a)-(d), 3

1. Using the graphical method, solve the simultaneous equations ******ebook converter DEMO Watermarks*******

2. Using the graphical method, solve the simultaneous equations

Coincident Lines and Parallel Lines Work in pairs. 1. a. Using a graphing software, on separate axes, draw the graphs of each of the following pairs of simultaneous equations. i. x + y = 1 3x + 3y = 3 ii. 2x + 3y = –1 20x + 30y = –10 iii. x – 2y = 5 5x – 10y = 25 b. What do you notice about the graphs of each pair of simultaneous equations? c. Does each pair of simultaneous equations have any solutions? If yes, what are the solutions? 2. a. Using a graphing software, on separate axes, draw the graphs of each of the following pairs of simultaneous equations. i. x + y = 1 3x + 3y = 15 ii. 2x + 3y = –1 20x + 30y = –40 iii. x – 2y = 5 5x – 10y = 30 b. What do you notice about the graphs of each pair of simultaneous ******ebook converter DEMO Watermarks*******

equations? c. Does each pair of simultaneous equations have any solutions? If yes, what are the solutions? From the class discussion, we notice that the graphs of each pair of simultaneous equations in Question 1 are identical, i.e. the two lines coincide. Since every point on each line is a point of intersection of the graphs, the graphs have an infinite number of points of intersection. Hence, the simultaneous equations have an infinite number of solutions. The graphs of each pair of simultaneous equations in Question 2 are parallel lines. Since the graphs do not intersect, they have no point of intersection. Hence, the simultaneous equations have no solution.

Exercise 4B Questions 4(a)-(d), 5(a)-(b)

What type of simultaneous equations have 1. infinitely many solutions? 2. no solution?

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BASIC LEVEL 1. Using the graphical method, solve each of the following pairs of simultaneous equations. a. b. c. d. e. f.

INTERMEDIATE LEVEL 2. Using the graphical method, solve each of the following pairs of simultaneous equations. a. b. c. d.

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3. a. Consider the equation y = 2x + 9. i. Copy and complete the table.

ii. On a sheet of graph paper, using a scale of 1 cm to represent 1 unit on the x-axis and 1 cm to represent 4 units on the y-axis, draw the graph of y = 2x + 9 for –8 ≤ x ≤ 4. b. Consider the equation

.

i. Copy and complete the table.

ii. On the same axes in (a)(ii), draw the graph of 4.

for –8 ≤ x ≤

c. Hence, solve the simultaneous equations 2x – y = –9 and x – 4y = –8. 4. Using the graphical method, solve each of the following pairs of simultaneous equations. a. b. c. d. ADVANCED LEVEL ******ebook converter DEMO Watermarks*******

5. Using the graphical method, solve each of the following pairs of simultaneous equations. a. b.

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4.4 Solving Simultaneous Linear Equations Using Algebraic Methods In Grade 7, we have learned how to solve linear equations in one variable such as 3x – 4 = 11 and 4x – 10 = 5x + 7. To solve a linear equation in one variable x means to find the value of x so that the values on both sides of the equation are equal, i.e. x satisfies the equation.

What are the solutions to a linear equation in two variables, e.g. 2x + y = 13? Do you obtain the same solutions as your classmates? In Section 4.3, we have learned that from the graphs of two linear equations, the coordinates of the point(s) of intersection give the solution(s) to the pair of simultaneous linear equations. In this section, we shall take a look at two algebraic methods that can be used to solve a pair of simultaneous equations: the elimination method and the substitution method.

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Solving Simultaneous Linear Equations Using Elimination Method Let us use the elimination method to solve the following pair of equations. We shall label the equations as equation (1) and equation (2).

The elimination method is usually used when the absolute values of the coefficients of one variable in both the equations are the same. For example, the absolute values of the coefficients of y in equations (1) and (2) are the same. What happens when we add equation (2) to equation (1)?

Notice that the terms in y are eliminated. We are left with a linear equation in one variable x.

∴ The solution of the simultaneous equations is x = 5 and y = 3. Check: Substitute x = 5 and y = 3 into (1) and (2):

Since x = 5 and y = 3 satisfies both the equations, it is the solution of the simultaneous equations.

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It is a good practice to check your solution by substituting the values of the unknowns which you have found into the original equations.

Worked Example

6

(Solving Simultaneous Linear Equations Using Elimination Method) Using the elimination method, solve the simultaneous equations

The coefficient of x in both the equations is 3. Hence, when we subtract equation (2) from equation (1), the terms in x are eliminated.

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Solution:

To eliminate a variable, the absolute values of the coefficients of the variable in both the equations must be the same.

1. Using the elimination method, solve each of the following pairs of simultaneous equations. a. x – y = 3 4x + y = 17 b. 7x + 2y = 19 7x + 8y = 13 c. 13x + 9y = 4 17x – 9y = 26 d. 4x – 5y = 17 x – 5y = 8 ******ebook converter DEMO Watermarks*******

2. Using the elimination method, solve the simultaneous equations

Exercise 4C Questions 1(a)-(l), 5(a), 6(a)-(b) It is sometimes necessary to manipulate one of the equations before we can eliminate a variable by addition or subtraction.

Worked Example

7

(Solving Simultaneous Linear Equations Using Elimination Method) Using the elimination method, solve the simultaneous equations

In this case, it is easier to eliminate y first. We multiply equation (2) by 2 so that the absolute values of the coefficients of y in both the equations are the same.

Solution:

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Using the elimination method, solve each of the following pairs of simultaneous equations. 1. 2. It is sometimes necessary to manipulate both the equations before we can eliminate a variable.

Exercise 4C Questions 2(a)-(f), 5(b), 6(c)-(d)

Worked Example

8

(Solving Simultaneous Linear Equations Using Elimination Method) Using the elimination method, solve the simultaneous equations

Solution:

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The LCM of 6 and 4 is 12. We multiply equation (1) by 2 and equation (2) by 3 so that the absolute values of the coefficients of y in both the equations are the same.

In Worked Example 8, is it easier to eliminate x first? Explain your answer by showing how x can be eliminated.

Exercise 4C Questions 3(a)-(f), 5(c)-(d), 6(e)-(f)

Using the elimination method, solve each of the following pairs of simultaneous equations. a. b.

Worked Example

9

(Solving Simultaneous Fractional Equations Using Elimination Method) Using the elimination method, solve the simultaneous equations

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Solution: Method 1:

Method 2:

Exercise 4C Questions 7(a)-(d) ******ebook converter DEMO Watermarks*******

Using the elimination method, solve the simultaneous equations

Solving Simultaneous Linear Equations Using Substitution Method Now, we shall take a look at how we can solve a pair of simultaneous equations using the substitution method. In this method, we first rearrange one equation to express one variable in terms of the other variable. Next, we substitute this expression into the other equation to obtain an equation in only one variable.

Worked Example

10

(Solving Simultaneous Linear Equations Using Substitution Method) Using the substitution method, solve the simultaneous equations

Solution:

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It is easier to obtain the value of y by substituting the value of x into equation (3) instead of equation (1) or (2).

If we make x the subject of equation (1) or (2) in Worked Example 10, will we get the same solution? Which way is easier?

Using the substitution method, solve the simultaneous equations

Exercise 4C Questions 4(a)-(n)

Worked Example

11

(Solving Simultaneous Linear Equations Using Substitution Method) Using the substitution method, solve the simultaneous equations

Solution:

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Using the substitution method, solve the simultaneous equations

Exercise 4C Questions 8(a)-(f), 12-14

Imee was asked to solve the simultaneous equations

using the substitution method. She did it this way: ******ebook converter DEMO Watermarks*******

What was wrong? Explain your answer.

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Worked Example

12

(Solving Simultaneous Fractional Equations Using Substitution Method) Using the substitution method, solve the simultaneous equations

Solution:

Consider

, where b, d ≠ 0.

Multiply by bd on both sides,

Using the substitution method, solve each of the following pairs of ******ebook converter DEMO Watermarks*******

simultaneous equations. a.

b.

Exercise 4C Questions 9(a)-(d), 10(a)-(f), 11(a)-(d)

BASIC LEVEL 1. Using the elimination method, solve each of the following pairs of simultaneous equations. a. x + y = 16 x–y=0 b. x – y = 5 x + y = 19 c. 11x + 4y = 12 9x – 4y = 8 d. 4y + x = 11 3y – x = 3 e. 3x + y = 5 x+y=3 f. 2x + 3y = 5 2x + 7y = 9 ******ebook converter DEMO Watermarks*******

g. 7x – 3y = 15 11x – 3y = 21 h. 3y – 2x = 9 2y – 2x = 7 i. 3a – 2b = 5 2b – 5a = 9 j. 5c – 2d = 9 3c + 2d = 7 k. 3f + 4h = 1 5f – 4h = 7 l. 6j – k = 23 3k + 6j = 11 2. Using the elimination method, solve each of the following pairs of simultaneous equations. a. 7x – 2y = 17 3x + 4y = 17 b. 16x + 5y = 39 4x – 3y = 31 c. x + 2y = 3 3x + 5y = 7 d. 3x + y = –5 7x + 3y = 1 e. 7x – 3y = 13 2x – y = 3 f. 9x – 5y = 2 3x – 4y = 10 3. Using the elimination method, solve each of the following pairs of simultaneous equations. a. 7x – 3y = 18 6x + 7y = 25 b. 4x + 3y = –5 ******ebook converter DEMO Watermarks*******

3x – 2y = 43 c. 2x + 3y = 8 5x + 2y = 9 d. 5x + 4y = 11 3x + 5y = 4 e. 4x – 3y = –1 5x – 2y = 4 f. 5x – 4y = 23 2x – 7y = 11 4. Using the substitution method, solve each of the following pairs of simultaneous equations. a. x + y = 7 x–y=5 b. 3x – y = 0 2x + y = 5 c. 2x – 7y = 5 3x + y = –4 d. 5x – y = 5 3x + 2y = 29 e. 5x + 3y = 11 4x – y = 2 f. 3x + 5y = 10 x – 2y = 7 g. x + y = 9 5x – 2y = 4 h. 5x + 2y = 3 x – 4y = –6 i. x – y = 1 2x – y = 8 j. 3x + 2y = 8 2y – 5x = 8 ******ebook converter DEMO Watermarks*******

k. 2x + y = 5 2x + 3y = 1 l. 2x – 3y = 7 5x – 7y = 18 m. x + 3y = 2 5x – 2y = 27 n. 4x – 3y = 8 6x + y = 1 INTERMEDIATE LEVEL 5. Using the elimination method, solve each of the following pairs of simultaneous equations. a. x + y = 0.5 x–y=1 b. 2x + 0.4y = 8 5x – 1.2y = 9 c. 10x – 3y = 24.5 3x – 5y = 13.5 d. 6x + 5y = 10.5 5x – 3y = –2 6. Using the elimination method, solve each of the following pairs of simultaneous equations. a. 4x – y – 7 = 0 4x + 3y – 11 = 0 b. 7x + 2y – 33 = 0 3y – 7x – 17 = 0 c. 5x – 3y – 2 = 0 x + 5y – 6 = 0 d. 5x – 3y – 13 = 0 7x – 6y – 20 = 0 e. 7x + 3y – 8 = 0 ******ebook converter DEMO Watermarks*******

3x – 4y – 14 = 0 f. 3x + 5y + 8 = 0 4x + 13y – 2 = 0 7. Using the elimination method, solve each of the following pairs of simultaneous equations. a.

b.

c.

d.

8. Using the substitution method, solve each of the following pairs of simultaneous equations. a. 2x + 5y = 12 4x + 3y = –4 b. 4x – 3y = 25 6x + 5y = 9 c. 3x + 7y = 2 6x – 5y = 4 d. 9x + 2y = 5 7x – 3y = 13 e. 2y – 5x = 25 4x + 3y = 3 f. 3x – 5y = 7 4x – 3y = 3 ******ebook converter DEMO Watermarks*******

9. Using the substitution method, solve each of the following pairs of simultaneous equations. a.

b.

c.

d.

10. Using the substitution method, solve each of the following pairs of simultaneous equations. a.

b.

c. d. e.

f.

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ADVANCED LEVEL 11. Using either the elimination or the substitution method, solve each of the following pairs of simultaneous equations. 1. 2.

3.

4. 12. If x = 3 and y = –1 is the solution of the simultaneous equations 3px + qy = 11, –qx + 5y = p, find the value of p and of q. 13. If x = –11 and y = 5 is the solution of the simultaneous equations px + 5y = q, qx + 7y = p, find the value of p and of q. 14. A computer animation shows a cat moving in a straight line. Its height, h meters, above the ground, is given by 8s – 3h = –9, where s is the time in seconds after it starts moving. In the same animation, a mouse starts to move at the same time as the cat and its movement is given by –29s + 10h = 16. Find the height above the ground and the time when the cat meets the mouse.

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4.5 Applications of Simultaneous Equations in Real-World Contexts In this section, we will learn how to apply the concept of simultaneous equations to solve mathematical and real-life problems. A familiar problem which we have learned how to solve in primary school is to formulate a pair of simultaneous equations to solve the problem. Consider the following problem: 7 cups of coffee and 4 pieces of toast cost ₱350. 5 cups of coffee and 4 pieces of toast cost ₱290. Find the cost of each item.

In primary school, we have learned how to solve the problem by using the following representation (or drawing a diagram). Method 1:

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∴ Cost of 2 cups of coffee ∴ Cost of 1 cup of coffee = ₱30 5 × ₱30 + cost of 4 pieces of toast = ₱290 Cost of 4 pieces of toast ∴ Cost of 1 piece of toast = ₱35 We shall now take a look at how we can solve the problem by using the algebraic method. Method 2: Let the cost of 1 cup of coffee be ₱x and the cost of 1 piece of toast be ₱y.

Substitute x = 1 into

∴ Cost of 1 cup of coffee = ₱30   Cost of 1 piece of toast = ₱35

Worked Example

13

(Finding Two Numbers Given Sum and Difference) The sum of two numbers is 67 and their difference is 3. Find the two numbers.

Solution: ******ebook converter DEMO Watermarks*******

Let the smaller number be x and the greater number be y.

Can you solve Worked Example 13 by using only one variable x?

1. The sum of two numbers is 36 and their difference is 9. Find the two numbers. 2. One third of the sum of two angles is 60˚ and one quarter of their difference is 28˚. Find the two angles. 3. The figure shows a rectangle with its length and width as indicated. Find the perimeter of the rectangle.

Exercise 4D Questions 1-2, 5-10, 19 (Finding a Fraction) If 1 is added to the numerator and 2 to the ******ebook converter DEMO Watermarks*******

Worked Example

14

denominator of a fraction, the value obtained is . If 2 is subtracted from its numerator and 1 from its denominator, the resulting value is . Find the fraction.

Solution: Let the numerator of the fraction be x and its denominator be y, i.e. let the fraction be .

If 1 is added to the numerator and to the denominator of a fraction, the value obtained is . If 5 is subtracted from its numerator and from its denominator, the resulting value is . Find the fraction.

Exercise 4D Question 11

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Worked Example

15

(Finding ages) The sum of the ages of Eric and his mother is 60. Two years ago, Eric’s mother was three times as old as Eric. Calculate 1. Eric’s present age, 2. the age of Eric’s mother when he was born.

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Solution: i. Let the present age of Eric's mother be x years and that of Eric be y years. Then two years ago, Eric’s mother was (x – 2) years old and Eric was (y – 2) years old.

∴ Eric’s present age = 16 years ii. Substitute y = 16 into (3): x = 3(16) – 4 = 44 ∴ Age of Eric’s mother when he was born = 44 – 16 = 28 years

1. In five years’ time, Kate’s father will be three times as old as Kate. Four years ago, her father was six times as old as her. Find their present ages. 2. The ratio of the ages of Carlo and Daniel is 3 : 5. In 6 years’ time, the ratio of their ages will be 3 : 4. How old will Carlo and Daniel be in 6 years’ time? 3. To visit the Manila Zoo, 11 adults and 5 children have to pay ₱270 whereas 14 adults and 9 children have to pay ₱370. Find the total amount a family of 2 adults and 3 children has to pay to visit the zoo.

Exercise 4D Questions 3-4, 12-18, 21-22 ******ebook converter DEMO Watermarks*******

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Worked Example

16

(Finding a Two-Digit Number) The sum of the digits of a two-digit number is 8. When the digits of the number are reversed and the number is subtracted from the original number, the result obtained is 18. Find the original number.

Solution: Let the tens digit of the original number be x and its ones digit be y. Then the original number is 10x + y, and the number obtained when the digits of the original number are reversed is 10y + x.

Search on the Internet for ‘Psychic Mind Reader’. This is an interactive applet that claims to read your mind. Follow the instructions on the applet. Do you believe that the applet can read your mind? If not, try to show, by using algebra, how this fascinating trick works! Explain why the symbols for the numbers 90 and 99 are different from the symbol that corresponds to your final number.

A two-digit number is such that the sum of its digits is 11. When the digits of ******ebook converter DEMO Watermarks*******

the number are reversed and the number is subtracted from the original number, the result obtained is 9. Find the original number.

Exercise 4D Question 20

In each of the following questions, formulate a pair of linear equations in two variables to solve the problem. BASIC LEVEL 1. The sum of two numbers is 138 and their difference is 88. Find the two numbers. 2. The difference between two numbers is 10 and their sum is four times the smaller number. Find the two numbers. 3. A belt and a wallet cost ₱1380 belts and 4 wallets cost ₱7005. Find the cost of each item. 4. 8 kg of potatoes and 5 kg of carrots cost ₱923 whereas 2 kg of potatoes and 3 kg of carrots cost ₱369. Find the cost of 1 kg of each item. INTERMEDIATE LEVEL 5. Two numbers are such that if 7 is added to the first number, a number twice the second number is obtained. If 20 is added to the second number, the number obtained is four times the first number. Find the two numbers. 6. The sum of two numbers is 48. If the smaller number is one fifth of the larger number, find the two numbers. 7. One fifth of the sum of two angles is 24˚ and half their difference is 14˚. Find the two angles. 8. The figure shows an equilateral triangle with its sides as indicated. Find the ******ebook converter DEMO Watermarks*******

length of each side of the triangle.

9. The figure shows a rectangle with its length and width as indicated. Given that the perimeter of the rectangle is 120 cm, find the area of the rectangle.

10. The figure shows a rhombus with its sides as indicated. Find the perimeter of the figure.

11. If 1 is subtracted from the numerator and from the denominator of a fraction, the value obtained is ½. If 1 is added to its numerator and to its denominator, the resulting value is ⅔. Find the fraction. 12. In 2013, the sum of the ages of two bears, Lulu and Nana, was 11 years. In 2022, Lulu will be three times as old as Nana was in 2013. Find their ages in 2014. 13. 6 adults and 4 senior citizens have to pay ₱140 while 13 adults and 7 senior citizens have to pay ₱290 to visit an exhibition. Find the total amount 2 adults and a senior citizen have to pay to visit the exhibition. ******ebook converter DEMO Watermarks*******

14. Megan intends to buy either Gift A, which costs ₱330, or Gift B, which costs ₱260, as Christmas gifts for each of her parents, 2 siblings, 13 relatives and 10 friends. Given that she intends to spend ₱7510, find the number of each gift she should buy. 15. There are some chickens and goats on a farm. Given that the animals have a total of 50 heads and 140 legs, how many more chickens than goats are there? 16. ₱2600 is divided between Sam and Miguel such that one quarter of Sam’s share is equal to one sixth of Miguel’s share. How much does each of them receive? 17. Carlo deposited a total of ₱823 000 in Bank A and Bank B at the beginning of 2013. Bank A and Bank B pay simple interest at rates of 0.6% and 0.65% per annum respectively. He withdrew all his money from the two banks at the end of 2013. If the amount of interest he earned from each bank is the same, find the amount of money he deposited in each bank. 18. The sum of the ages of Mr Reyes and his son Angelo is 61 years. The difference in their ages is 29 years. 1. How old is Angelo now? 2. How old will Mr Reyes be when Angelo is 21 years old? ADVANCED LEVEL 19. Two numbers are such that when the larger number is divided by the smaller number, both the quotient and the remainder are equal to 2. If five times the smaller number is divided by the larger number, both the quotient and the remainder are also equal to 2. Find the two numbers. 20. A two-digit number is such that the sum of its digits is of the number. When the digits of the number are reversed and the number is subtracted from the original number, the result obtained is 45. Find the original number. 21. Chris has ₱330. If he buys 8 pears and 5 mangoes, he will be short of ₱40. If he buys 5 pears and 4 mangoes, he will receive ₱55 in change. Find the price of 1 pear and 1 mango. ******ebook converter DEMO Watermarks*******

22. Gemma’s mother buys some shares of Company A on Day 0. On Day 7, the share price of Company A is ₱152. If she sells all her shares of Company A and buys 2000 shares of Company B on Day 7, she would receive ₱244 000. On Day 12, the share price of Company A is ₱158 and the share price of Company B is ₱16 less than that on Day 7. If she sells all her shares of Company A and buys 5000 shares of Company B on day 12, she would have to pay ₱198 000. Find i. the number of shares of Company A Gemma’s mother has, ii. the share price of Company B on Day 12.

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1. Equation of a horizontal line The equation of a straight line that is parallel to the x-axis and which passes through the point (a, b) is y = b. It has a slope of 0. Equation of a vertical line The equation of a straight line that is parallel to the yaxis and which passes through the point (a, b) is x = a. Its slope is undefined. 2.

Equation of an oblique line The equation of a straight line passing through the point (0, c) and with slope m is y = mx + c.

3. If x = a and y = b satisfy each of the two simultaneous equations in two ******ebook converter DEMO Watermarks*******

variables, then x = a and y = b is known as a solution of the two equations. 4. A pair of simultaneous linear equations in two variables can be solved by the graphical method, the elimination method, the substitution method. 5. The solution of a pair of simultaneous linear equations is given by the coordinates of the point of intersection of the graphs of the two equations. 6. A pair of simultaneous linear equations has an infinite number of solutions if the graphs of the two equations are identical. 7. A pair of simultaneous linear equations has no solution if the graphs of the two equations are parallel. 8. For mathematical and real-life problems that involve simultaneous equations, we formulate a pair of linear equations in two variables before solving for the unknowns in the problems.

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1. Consider the equation 2x + y = 2. a. Copy and complete the table.

b. On a sheet of graph paper, using a scale of 2 cm to represent 1 unit on the x-axis and 1 cm to represent 1 unit on the y-axis, draw the graph of 2x + y = 2 for –4 ≤ x ≤ 4. c. The point (p, –2) lies on the graph in (b). Find the value of p. d. i. On the same axes in (b), draw the graph of x = –0.5. ii. State the coordinates of the point of intersection of the graphs of 2x + y = 2 and x = –0.5. 2. The coordinates of the points A and B are (0, 6) and (8, 0) respectively. i. Find the equation of the line passing through A and B. Given that the line y = x + 1 cuts the line AB at the point M, find ii. the coordinates of M, iii. the equation of the line which passes through M and is parallel to the xaxis, iv. the equation of the line which passes through M and is parallel to the yaxis. ******ebook converter DEMO Watermarks*******

3. A spring is suspended freely. When a mass of 20 g is attached to the spring, it has a length of 12 cm. When a mass of 50 g is attached to the spring, it has a length of 15 cm. The graph below shows how the length, y cm, of the spring, varies with the mass, x g, attached to it.

i. Find an expression for y in terms of x. ii. State what the value of the y-intercept represents. 4. a. The variables x and y are connected by the equation 5x – 3y = 2. Some values of x and the corresponding values of y are given in the table.

i. Find the value of p and of q. ii. On a sheet of graph paper, using a scale of 1 cm to represent 1 unit on both axes, draw the graph of 5x – 3y = 2 for –5 ≤ x ≤ 7. b. Consider the equation 3x + 4y = 7. i. Copy and complete the table. See Table. ii. On the same axes in (a)(ii), draw the graph of 3x + 4y = 7 for –5 ≤ x ≤ 7. c. Hence, solve the simultaneous equations 5x – 3y = 2 and 3x + 4y = 7. ******ebook converter DEMO Watermarks*******

5. Solve each of the following pairs of simultaneous equations. a. 7x + 2y = 10 5x + 2y = 6 b. 9x + 4y = 28 4y – 11x = –12 c. 2x – 5y = 22 2x – 3y = 14 d. 6x – y = 16 3x + 2y = –12 e. 4x + 3y = 0 5y + 53 = 11x f. 5x – 4y = 4 2x – y = 2.5 6. Two numbers are such that if 11 is added to the first number, a number twice the second number is obtained. If 20 is added to the second number, the number obtained is twice the first number. Find the two numbers. 7. The figure shows a parallelogram with its sides as indicated. Find the perimeter of the parallelogram.

8. If 1 is subtracted from the numerator and 2 is added to the denominator of a fraction, the value obtained is ½. If 3 is added to its numerator and 2 is subtracted from its denominator, the resulting value is 1¼. Find the fraction. 9. A two-digit number is such that the sum of its digits is 12 and the ones digit is twice its tens digit. Find the number. 10. In four years’ time, Daniel’s mother will be three times as old as Daniel. Six years ago, his mother was seven times as old as him. Find ******ebook converter DEMO Watermarks*******

i. Daniel’s present age, ii. the age of Daniel’s mother when he was born. 11. If Ann gives ₱20 to Kristel, Kristel will have twice as much as Ann. If Kristel gives ₱50 to Ann, Ann will have twice as much as Kristel. How much does each of them have? 12. A vendor buys 36 smartphones and tablet computers for ₱919 500. Given that a smartphone costs ₱29 500 and a tablet computer costs ₱20 000, find the number of each item the vendor buys. 13. 5 cups of ice-cream milk tea and 4 cups of citron tea cost ₱880 whereas 7 cups of ice-cream milk tea and 6 cups of citron tea cost ₱1270. Find the difference between the cost of 1 cup of ice-cream milk tea and 1 cup of citron tea. 14. Angelo mixes coffee powder that costs ₱80 per kg with coffee powder that costs ₱110 per kg. Given that he sold 20 kg of the mixture at ₱101 per kg such that he does not make any profit or incur any loss, find the mass of each type of coffee powder that he uses for the mixture. 15. A mobile company charges a fixed rate of ₱x per minute for the first 120 minutes of talk time and another rate of ₱y per minute for each additional minute of talk time. Chris paid ₱851 and ₱1026 for 175 minutes and 210 minutes of talk time on two different occasions respectively. Find the amount he has to pay if he uses 140 minutes of talk time. 16. In a Mathematics test, the average score obtained by Class 2A is 72 and the average score obtained by Class 2B is 75. The average score obtained by the two classes is 73.48. Given that there is a total of 75 students in the two classes, find the number of students in each class.

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1. Consider the simultaneous equations px – y = 6, 8x – 2y = q. Determine the conditions that p and q must satisfy if the simultaneous equations have i. an infinite number of solutions, ii. no solution, iii. a unique solution, i.e. only one solution. 2. Solve the simultaneous equations

3. Two positive numbers are such that the sum of 11 times the square of the first number and 13 times the cube of the second number is 395. If 218 is subtracted from 26 times the cube of the second number, the number obtained is 121 times the square of the first number. Find the two numbers. 4. Some spiders, dragonflies and houseflies are kept in three separate enclosures. They have a total of 20 heads, 136 legs and 19 pairs of wings. Given that a spider has 8 legs and 0 pairs of wings, a dragonfly has 6 legs and 2 pairs of wings, and a housefly has 6 legs and 1 pair of wings, find i. the number of spiders, ii. the number of dragonflies, iii. the number of houseflies. 5. A rooster costs ₱150 and a hen costs ₱90. Chicks are sold at 3 for ₱30. A ******ebook converter DEMO Watermarks*******

farmer bought 100 birds of these three types for ₱3000. How many of each type of bird did he buy? Hint: There are three possible sets of answers.

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A1 Revision Exercise 1. Factorize each of the following expressions completely. a. 15x − 20 b. 22 + 11y c. −64a − 16ax d. 9by − 3y + 12cy 2. Factorize each of the following expressions completely. a. 10 − 15x + 20x2 b. 5a2x − 3a3x2 + 6a2x2 c. 6a2 + 8a3 − 10a5 d. 12x3y − 9x2y2 + 6xy3 3. Factorize each of the following expressions completely. a. 4f2 – 10f + 6 b. 1 – 12hk + 36h2k2 c. 5m2n – 15mn2 – 25mn d. 2px + 3qy – 2py – 3qx e. 25x2 − 20x + 4 f. x2 − 11x + 28 4. Factorize each of the following expressions completely. a. 2c2d2 + 5cd – 12 b. 25h2k2 + 10hk + 1 ******ebook converter DEMO Watermarks*******

c. 16 – 4(m + 2)2 d. 3pr – ps + 6qr – 2qs 5. Without using a calculator, evaluate each of the following. a. 4622 − 4522 b. 9032 – 972 6. Express each of the following as a fraction in its simplest form. a. b. 7. i. Make x the subject of the formula

.

ii. Hence, find the value of x when a = 1 and y = 5. 8. Simplify each of the following. a. b. 9. The centripetal force, F Newtons, acting on an object of mass m kg, moving at a tangential velocity v m s–1 along a path with radius r m, is given by the formula . i. Given that v ≥ 0, make v the subject of the formula

.

ii. Hence, find the tangential velocity of a motorcycle of mass 225 kg moving along a circular track with radius 5 m, if the centripetal force acting on it is 4500 Newtons.

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A2 Revision Exercise 1. i. The vertices of a triangle are A(0, –4), B(4, –2) and C(2, 2). Plot the points A, B and C on a sheet of graph paper. ii. Plot the point D such that ABCD is a square. Hence, write down the coordinates of D. iii. Find the slopes of AB, BC and CD. What do you notice about the slopes of AB and CD? Explain its significance. 2. Find the equation of the line joining the points A(5, 7) and B(8, 12). 3. Find the equation of the line passing through the point (2, –5) and parallel to the line 5x + 7y = 46. 4. Find the coordinates of the point at which the line

cuts the y-axis.

5. Solve the simultaneous equations 8x + 3y = 14, 2x + y = 4. 6. The coordinates of the point of intersection of the lines px + y = 3 and x + 2y = q are (2, –3). Find the value of p and of q. 7. There are x chickens and y rabbits on a farm. Given that the animals have a total of 70 heads and 196 legs, formulate a pair of simultaneous equations involving x and y. By solving the simultaneous equations, find the number of chickens and rabbits on the farm. 8. 5 bottles of milk and 2 cans of apple juice cost ₱136 while 7 bottles of milk and 6 cans of apple juice cost ₱280. Find the cost of one bottle of milk and one can of apple juice.

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Chapter Five Linear Inequalities in Two Variables Many companies are devoted to using systems of inequalities to ensure that they are making the most profit. This can be achieved by producing the right combination of items in terms of quantity or spending the least amount of money in the production of certain items. Linear programming is the mathematical process of analyzing a system of inequalities to make the best decisions given the constraints of the situation.

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LEARNING OBJECTIVES At the end of this chapter, you should be able to: illustrate linear inequalities in two variables, differentiate linear inequalities in two variables from linear equations in two variables, graph linear inequalities in two variables, solve problems involving linear inequalities in two variables, solve a system of linear inequalities in two variables, solve problems involving systems of linear inequalities in two variables.

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5.1 Linear Inequalities in Two Variables Recap (Properties of Inequalities) In Grade 7, we have learned some properties of inequalities. We can add or subtract a positive number from both sides of an inequality without having to reverse the inequality sign, i.e. if x ≥ y and a > 0, then x + a ≥ y + a and x – a ≥ y – a.

This is also true for a negative number b. if x ≥ y and b < 0, then x + b ≥ y + b and x – b ≥ y – b.

We can multiply or divide both sides of an inequality by a positive number without having to reverse the inequality sign, i.e. if x ≥ y and c > 0, then cx ≥ cy and .

However, if we multiply or divide both sides of an inequality by a negative number, we will have to reverse the inequality sign, i.e.

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if x ≥ y and d < 0, then dx ≤ dy and

.

For any three numbers x, y and z, if x > y and y > z, then x > z.

This is known as the transitive property of inequalities.

Recap (Linear Equations in Two Variables) We have learned linear equations in two variables and how to draw graphs of linear equations in the form y = mx + c, where x and y are the variables. In general, for a straight line passing through the point (0, c) and with gradient m, the equation is y = mx + c.

In this section, we shall learn about linear inequalities in two variables.

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Linear Inequalities in Two Variables Case 1 (On the line) 1. The graph of the linear equation x + 2y = 4 is shown in Fig. 5.1(a). Some points are marked on the line.

Fig. 5.1(a) Find the values of x + 2y by substituting the coordinates of each point on the line. Record the values in Table 5.1. What do you notice about the values of x + 2y? ******ebook converter DEMO Watermarks*******

Case 2 (Below the line) 1. The graph of the linear equation x + 2y = 4 is shown in Fig. 5.1(b). Some points are marked below the line.

Fig. 5.1(b) Find the values of x + 2y by substituting the coordinates of each point below the line. Record the values in Table 5.1. What do you notice about the values of x + 2y? Case 3 (Above the line) 1. The graph of the linear equation x + 2y = 4 is shown in Fig. 5.1(c). Some points are marked above the line.

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Fig. 5.1(c) Find the values of x + 2y by substituting the coordinates of each point above the line. Record the values in Table 5.1. What do you notice about the values of x + 2y? Table 5.1 From the investigation, we observe the following: 1. In Case 1, the values of x + 2y are always equal to 4; 2. In Case 2, the values of x + 2y are always less than 4; 3. In Case 3, the values of x + 2y are always greater than 4. We are familiar with Case 1 as the points on the line represent a linear equation in two variables x + 2y = 4. In Case 2, since the values of x + 2y are always less than 4 for points below the line of linear equation x + 2y = 4, we can write the inequality x + 2y < 4 to ******ebook converter DEMO Watermarks*******

represent the region below a straight line graph. In Case 3, since the values of x + 2y are always greater than 4 for points above the line of linear equation x + 2y = 4, we can write the inequality x + 2y > 4 to represent the region above a straight line graph. We can say that the points in Fig. 5.1(b) satisfy the inequality x + 2y < 4 and the points in Fig. 5.1(c) satisfy the inequality x + 2y > 4.

Worked Example

1

(Drawing Graphs of Linear Inequalities in Two Variables) Show, unshaded, the region satisfied by the following inequalities: x ≥ 0, y ≥ 1, x + y ≤ 5, y < x + 3.

Solution: The graphs of x = 0, y = 1, x + y = 5 and y = x + 3 are drawn.

For x ≥ 0, we shade the region to the left of the y-axis. For y ≥ 1, we shade the region below the line y = 1. For x + y ≤ 5, the region above x + y = 5 is shaded. ******ebook converter DEMO Watermarks*******

For y < x + 3, the region above the line y = x + 3 is shaded. Since y < x + 3, we use a dotted line for y = x + 3.

We use a solid line for inequalities with ≥ or ≤ sign, and a dotted line for inequalities with > or < sign.

The required region is represented by the unshaded part. Check whether the unshaded region is the correct region by using a point in the unshaded region, e.g. (1, 2), to verify the inequalities.

Show, unshaded, the region satisfied by the inequalities x + 2y ≤ 8 and x > y.

Exercise 5A Questions 1(a)-(f)

Worked Example

2

(Writing Linear Inequalities in Two Variables from Graphs) Write down the inequalities which define the unshaded region.

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Solution: Equation of l1:

The unshaded region lies below l1. Hence y ≤ x + 4 defines a part of the unshaded region. Equation of l2:

The unshaded region lies above l2. Hence ******ebook converter DEMO Watermarks*******

defines a part of the

unshaded region. Equation of l3:

The unshaded region lies below l3. Hence unshaded region.

defines a part of the

Equation of l4:

The unshaded region lies above l4. Hence unshaded region.

defines a part of the

∴ The unshaded region is defined by the four inequalities: and

.

Write down the inequalities which define the unshaded region.

Exercise 5A Questions 2(a)-(f), 3(a)-(f)

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BASIC LEVEL 1. Show, unshaded, the regions satisfied by the following inequalities: 1. x ≥ 0, y ≥ 2x 2. x > 2, y ≥ x + 1 3. x + y < 4, y ≥ x − 1 4. x > 0, y > 2, y ≤ 6 − x 5. x > 0, 2x + y ≤ 10, y ≥ 1 6. ) y < x + 3, x ≤ 5, y ≥ −1 INTERMEDIATE LEVEL 1. In each of the following cases, write down the inequalities which define the unshaded region.

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a.

b.

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c.

d.

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e.

f.

ADVANCED LEVEL 1. In each of the following cases, write down the inequalities which define the unshaded region.

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a.

b.

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c.

d.

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e.

f.

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5.2 Application of Systems of Linear Inequalities in Two Variables in Real-World Contexts In this section, we will learn how to apply systems of linear inequalities in two variables to solve mathematical and real-life problems.

Worked Example

3

(Solving Real-life Problems involving Systems of Linear Inequalities in Two Variables) A farmer plans to divide his land into not more than 36 plots to plant either a banana tree or coconut tree on each plot. He decides that he will plant at least 20 banana trees and that there will be at least twice as many banana trees as coconut trees. 1. Taking x to represent the number of banana trees and y to represent the number of coconut trees, write down three inequalities, other than x ≥ 0 and y ≥ 0, which satisfy the above conditions. 2. On a sheet of graph paper, show, unshaded, the region satisfied by the inequalities in (a). 3. A plot of land used to plant a banana tree has an area of 16 m2 and a plot of land used to plant a coconut tree has an area of 4 m2. Use your graph to estimate the maximum possible land area that the farmer has.

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Solution: a. The three inequalities are x + y ≤ 36, x ≥ 20 and x ≥ 2y. b. Draw the lines x + y = 36, x = 20 and x = 2y. Shade the regions not required by the inequalities: and y ≥ 0 i. Above x + y = 36 ii. Left of x = 20 iii. Above x = 2y iv. Below the x-axis v. Left of the y-axis

1. Let the land area that the farmer has be A m2. ******ebook converter DEMO Watermarks*******

A is given by 16x + 4y and must be satisfied by the unshaded region. If the farmer plants 36 banana trees, i.e. x = 36 and y = 0, he has the maximum possible land area. Maximum possible value of A = 16(36) + 4(0) = 576 m2

When shading the region not required by an inequality involving x and y, write the inequality such that y is on the LHS of the inequality.

The maximum possible land area is obtained by substituting the coordinates of one of the vertices of the unshaded region on the graph. Since the plot of land used to plant a banana tree has a greater area than that used to plant a coconut tree, the maximum possible land area occurs when as many banana trees as possible are planted. That is, x = 36 and y = 0.

A shopkeeper stocks two brands of drinks called Coola and Shiok. He is ordering fresh supplies and finds that he has room for up to 1000 cans. He proposes to order at least twice as many cans of Shiok as of Coola. He wishes to have at least 100 cans of Coola and not more than 800 cans of Shiok. 1. Taking x to represent the number of cans of Coola and y to represent the number of cans of Shiok that he orders, write down four inequalities involving x and/ or y, other than x ≥ 0 and y ≥ 0, which satisfy the above conditions. 2. The point (x, y) represents x cans of Coola and y cans of Shiok ordered. Using a scale of 1 cm to represent 100 cans on each axis, construct and indicate clearly, by shading the unwanted regions, the region in which (x, ******ebook converter DEMO Watermarks*******

y) must lie. 3. The profit made by selling a can of Coola is 6 and that of a can of Shiok is 5. Use your graph to estimate the number of cans of each brand that the shopkeeper should order to give the maximum profit.

Exercise 5B Questions 1-7

BASIC LEVEL 1. Kate plans to divide her ribbon into not more than 30 pieces. Each piece of ribbon is to be used to wrap either a box of cookies or a packet of candies. She will wrap at least 8 boxes of cookies and there will be at least twice as many packets of candies as boxes of cookies. a. Taking x to represent the number of boxes of cookies and y to represent the number of packets of candies that she wraps, write down three inequalities, other than x ≥ 0 and y ≥ 0, which satisfy the above conditions. b. On a sheet of graph paper, show, unshaded, the region satisfied by the inequalities in (a). c. A piece of ribbon used to wrap a box of cookies is 30 cm long and a piece of ribbon used to wrap a packet of candies is 15 cm long. Use your graph to estimate the maximum possible length of ribbon that Kate has. 2. A chef plans to divide his dough into not more than 40 portions. Each portion of dough is to be used to make either a pizza or bread. He will make at least a dozen pizzas and there will be at least twice as many pizzas as loaves of bread. a. Taking x to represent the number of pizzas and y to represent the number of loaves of bread that he makes, write down three inequalities, other ******ebook converter DEMO Watermarks*******

than x ≥ 0 and y ≥ 0, which satisfy the above conditions. b. On a sheet of graph paper, show, unshaded, the region satisfied by the inequalities in (a). c. The dough used to make a pizza weighs 6 g and the dough used to make a loaf of bread weighs 8 g. Use your graph to estimate the maximum possible weight of dough that the chef has. 3. Mrs Reyes plans to divide her orange syrup into not more than 25 bottles. Each big bottle has a capacity of 1 liter and each small bottle has a capacity of 0.5 liter. She decides to pour the syrup into at least 5 big bottles and there will be at least half as many small bottles as big bottles. a. Taking x to be the number of big bottles and y to be the number of small bottles of syrup that she prepares, write down three inequalities, other than x ≥ 0 and y ≥ 0, which satisfy the above conditions. b. On a sheet of graph paper, show, unshaded, the region satisfied by the inequalities in (a). c. Use your graph to estimate the maximum possible amount of syrup that Mrs Reyes has. INTERMEDIATE LEVEL 4. A supermarket manager stocks two brands of detergent called Power Clean and Disappear. His stock is running low and finds that he has room for up to 200 bottles. He proposes to order at least twice as many bottles of Power Clean as of Disappear. He wishes to have at least 50 bottles of Disappear and not more than 140 bottles of Power Clean. a. Taking x to represent the number of bottles of Power Clean and y to represent the number of bottles of Disappear that he orders, write down four inequalities involving x and/or y, other than x ≥ 0 and y ≥ 0, which satisfy the above conditions. b. The point (x, y) represents x bottles of Power Clean and y bottles of Disappear ordered. Using a scale of 2 cm to represent 50 bottles on each axis, construct and indicate clearly, by shading the unwanted regions, the region in which (x, y) must lie. ******ebook converter DEMO Watermarks*******

c. The profit of a bottle of Power Clean is 100 and the profit of a bottle of Disappear is 80. Use your graph to estimate the number of bottles of each brand that the supermarket manager should order to give the maximum profit. 5. A manufacturer of baking flour supplies it in packets of two sizes: “Economy” weighs 3 kg and costs 60, and “Giant” weighs 6 kg and costs 100. A restaurant owner is buying his monthly supply of baking flour but does not wish to spend more than 10 000. He is buying at least 300 kg of baking flour. He estimates that he will require at least 40 but not more than 80 “Economy” packets. a. Taking x to represent the number of “Economy” packets and y to represent the number of “Giant” packets that he buys, write down three inequalities, other than x ≥ 0 and y ≥ 0, which satisfy the above conditions. b. The point (x, y) represents x “Economy” packets of baking flour and y “Giant” packets of baking flour bought. Using a scale of 1 cm to represent 10 packets on each axis, construct and indicate clearly, by shading the unwanted regions, the region in which (x, y) must lie. c. Use your graph to estimate the number of “Economy” packets and “Giant” packets he should buy to minimize his expenses on flour. 6. Two types of ship, Gigantic and Jumbo, are available to move 300 men and 20 000 kg of equipment. Each Gigantic ship can carry 40 men and 3000 kg of equipment. Each Jumbo ship can carry 50 men and 2000 kg of equipment. a. If x Gigantic ships and y Jumbo ships are used, write down the inequalities, other than x ≥ 0 and y ≥ 0, which x and y must satisfy. b. The point (x, y) represents the number of Gigantic ships, x, and the number of Jumbo ships, y. Using a suitable scale on each axis, construct and indicate clearly, by shading the unwanted regions, the region in which (x, y) must lie. c. Use your graph to estimate the least number of ships that can move 300 men and 20 000 kg of equipment. ******ebook converter DEMO Watermarks*******

ADVANCED LEVEL 7. A dealer produces two blends of tea, Fragrant and Instant, by mixing two varieties of tea leaves, A and B. In Fragrant blend, and in Instant blend,

.

Given that he produces x kilograms of Fragrant and y kilograms of Instant, copy and complete the following table. See Table. a. The dealer has at most 3200 kg of variety A and at least 3000 kg of variety B. Write down two inequalities involving x and y which satisfy these conditions and show that they simplify to 2x + y ≤ 8000 and x + 3y ≤ 15 000. b. He wishes to produce less Fragrant than Instant and has sufficient containers for only 2300 kg of Fragrant and 5000 kg of Instant. Write down three inequalities which satisfy these conditions. c. The point (x, y) represents x kilograms of Fragrant and y kilograms of Instant. Using a scale of 2 cm to represent 1000 kg on each axis, construct and indicate clearly, by shading the unwanted regions, the region in which (x, y) must lie. d. The dealer makes the same profit per kilogram on Fragrant as on Instant. Use your graph to estimate the weight of each blend that he should produce to maximize the profit.

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1. A linear inequality in two variables can be illustrated on a graph by drawing the linear equation in two variables and shading the unwanted region. 2. The region below y = mx + c represents the inequality y < mx + c, and the region above y = mx + c represents the inequality y > mx + c. 3. A system of linear inequalities in two variables can be solved by the graphical method. The solutions to the system of linear inequalities are represented by the unshaded region.

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1. Show, unshaded, the regions satisfied by the following inequalities: a. y ≥ 0, y ≤ 2x b. x > −1, y ≥ 2x + 1 c. x + 2y < 2, y ≥ x + 1 d. x > 1, y ≤ 2, y ≤ x − 6 e. x > 0, 2x + 2y ≤ 9, y ≥ 2 f. ) y < x + 1, y − 2x + 3 ≥ 0, y ≥ 1 2. In each of the following cases, write down the inequalities which define the unshaded region. a.

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b.

c.

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d.

e.

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f.

g.

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h.

i.

3. A children’s book is to be published in both hardback and paperback edition. A bookstore is ordering more than 20 hardback copies, 40 or more paperback copies, at least 70 but not more than 80 copies altogether. (a) Using x to represent the number of hardback copies and y to represent the number of paperback copies, write down the inequalities to represent these conditions. ******ebook converter DEMO Watermarks*******

a. The point (x, y) represents x hardback copies and y paperback copies. Using a scale of 1 cm to represent 10 books on each axis, construct and indicate clearly, by shading the unwanted regions, the region in which (x, y) must lie. b. Given that the profit of each hardback copy is ₱300 and the profit of each paperback copy is ₱200, calculate the number of each edition that the bookstore must buy to give the maximum profit. 4. In an hour, Machine A can package 2000 canned drinks and 4000 cup noodles, and Machine B can package 5000 canned drinks and 3000 cup noodles. A company has to package 100 000 canned drinks and 100 000 cup noodles in an hour. a. If x Machine A and y Machine B are used, write down the inequalities, other than x ≥ 0 and y ≥ 0, which x and y must satisfy. b. The point (x, y) represents the number of Machine A, x, and the number of Machine B, y. Using a suitable scale on each axis, construct and indicate clearly, by shading the unwanted regions, the region in which (x, y) must lie. c. Use your graph to estimate the least number of machines that can package the required quantity of canned drinks and cup noodles. 5. Mr Gonzales who intended to keep chickens and ducks on his farm asked each of his four children how many chickens and/or ducks he should keep. a. Daniel suggested that he should keep more than 10 ducks. b. Kristel suggested that the number of chickens should be at least 20 but not more than 50. c. Imee suggested that the total number of chickens and ducks should be less than 70. d. Eric suggested that the number of chickens should be greater than or equal to the number of ducks. e. Taking x to be the number of chickens and y to be the number of ducks, write down the inequalities which represent these conditions. f. The point (x, y) represents x chickens and y ducks that Mr Gonzales kept. Using a scale of 2 cm to represent 10 chickens on the x-axis and a ******ebook converter DEMO Watermarks*******

scale of 2 cm to represent 10 ducks on the y-axis, and indicate clearly by shading the unwanted regions, the region in which (x, y) must lie. g. Assuming Mr Gonzales took all his children’s suggestions. When he sold the animals, he made a profit of 60 on each chicken and ₱120 on each duck. Find the minimum number of ducks he kept on his farm to ensure a profit of at least ₱4800. 6. Brand A of potato chips contains 240 calories per kilogram and 200 units of vitamins per kilogram. Brand B of potato chips contains 160 calories per kilogram and 80 units of vitamins per kilogram. It is desired to have at least 10 kg mixture of brands A and B that contains not more than 2400 calories and at least 1600 units of vitamins. a. If x kilograms of Brand A and y kilograms of Brand B are mixed, write down the inequalities, other than x ≥ 0 and y ≥ 0, which x and y must satisfy. b. The point (x, y) represents the weight of Brand A, x kilograms, and the weight of Brand B, y kilograms. Using a suitable scale on each axis, construct and indicate clearly, by shading the unwanted regions, the region in which (x, y) must lie. c. Use your graph to estimate the maximum weight of the mixture of potato chips that contains the desired amount of calories and vitamins.

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A banker has ₱1000 000 to invest in three different funds. The government bond fund has a 5% return, the local bank's fund has a 7% return, and a highrisk account has an expected 10% return. To minimize risk, the banker decides not to invest more than ₱100 000 in the high-risk account. For regulation reasons, he needs to invest at least three times as much in the government bond fund as in the bank’s fund. How should the money be invested to maximize the expected returns?

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Chapter Six Functions and Linear Graphs Consider a fruit juicer. Every time we put an orange into it, orange juice is produced. If we put starfruit into it, we get starfruit juice. Under no circumstance will we get orange juice by putting in some other types of fruit. This is a convenient analogy for our next topic – functions. In mathematics, we can define an operation on a set of numbers so that every time we apply the operation on a given number x, say, we will always get a result y. Such an operation is known as a function.

LEARNING OBJECTIVES ******ebook converter DEMO Watermarks*******

At the end of this chapter, you should be able to: illustrate a relation and a function, verify if a given relation is a function, determine dependent and independent variables, find the domain and range of a function, illustrate a linear function, graph a linear function’s (a) domain; (b) range; (c) table of values; (d) intercepts; and (e) slope, solve problems involving linear functions.

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6.1 Relations Let A = {2, 4, 5} and B = {6, 8, 10}. Consider the relation “is a factor of”. This relation connects the elements in set A to the elements in set B since 2 is a factor of 6, 4 is a factor of 8 and 5 is a factor of 10 and so on. We can display the relation “is a factor of” from set A to set B using an arrow diagram as shown in Fig. 6.1.

Fig. 6.1

The arrow is always from the domain to the codomain. Set A is called the domain and set B the codomain of this relation. The arrows represent the relation “is a factor”. Hence 2 “is a factor of” 6, 4 “is a factor of” 8, etc. These can be conveniently represented by pairs of numbers (2, 6), (4, 8), etc., which are known as ordered pairs. Thus the set of ordered pairs {(4, 8), (2, 6), (2, 8), (2, 10), (5, 10)} represents the relation “is a factor” from set A to set B. In each ordered pair, the first component must be an element in the domain and the second component an element in the codomain. The second component of an ordered pair is called the image of the corresponding first component. ******ebook converter DEMO Watermarks*******

Note that an element of the domain may be related to one or more elements in the codomain and vice versa.

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6.2 Functions A function performs one or more operations on the inputs, i.e. the values it takes in, to produce outputs, i.e. the results. The operations performed on the inputs are known as the rule of the function. In this section, we will take a look at functions.

Search on the Internet for an interactive ‘Function Machine’ which you can key in an input and the machine will give you an output. Then you can guess the equation of the function.

Function Machine In this investigation, we shall explore the concept of a function by looking at how a function machine works. Fig. 6.2 shows a function machine whose function is to ‘add 3’ to any input x to produce an output y. For example, if you input x = 2, the output will be y = 2 + 3 = 5.

We can represent a function using words.

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Fig. 6.2

We can represent a function using an equation. 1. Write down an equation that shows the relationship between the output y and the input x. y = __________ 2. Write down the output y for each of the following inputs x. a. a) Input x = 4 → Output y = _____ b. b) Input x = −7 → Output y = _____ 3. Write down the input x for each of the following outputs y. a. a) Input x = _____ → Output y = 9 b. b) Input x = _____ → Output y = 0 4. The above data can be represented by Table 6.1. Complete the table.

Table 6.1

We can represent a function using a table.

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5. In Fig. 6.3, the point (2, 5) is shown. Plot the rest of the points based on Table 6.1 and draw a straight line that passes through all the points.

Fig. 6.3

We can represent a function using a graph. ******ebook converter DEMO Watermarks*******

Select any point on the straight line. Do the coordinates satisfy the equation which you have written down in Question 1? 6. Based on Table 6.1 and Fig. 6.3, state the number of output(s) y for each input x. From the above, we can see that a function is such that every input produces only one output. The input x and the output y of a function can be written as an ordered pair (x, y). A function can be represented using words, an equation, a table or a graph. y = x + 3 is called the equation of the function. Fig. 6.4 shows another function machine whose function is to ‘multiply –2’ to any input x before ‘subtracting 1’ from the result to produce an output y. For example, if you input x = 3, the output will be y = 3 × (–2) – 1 = –7.

Fig. 6.4 7. Representation of a function using an equation Write down the equation of the function. y = __________ 8. Representation of a function using a table Complete Table 6.2 to show the corresponding output values for the input values. Table 6.2 9. Representation of a function using a graph In Fig. 6.5, the point (3, –7) is shown. Plot the rest of the points based on Table 6.2 and draw a straight line that passes through all the points. ******ebook converter DEMO Watermarks*******

Fig. 6.5 10. Based on Table 6.2 and Fig. 6.5, state the number of output(s) y for each input x. From the investigation, in general, A function is a relationship between two variables x and y such that every input x produces exactly one output y. The input x is called an independent variable while the output y is called a dependent variable.

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1. y2 = x is not the equation of a function because there are two values of y for every positive value of x, e.g. if the input x = 9, then the output y = ± _____, there is no value for the output y if the input x is negative. 2. Is it possible for a function to have two input values x with the same output value y? Hint: Consider the equation of the function y = x2.

1. The equation of a function is y = 2x – 3. Find 1. the value of y when x = 4, 2. the value of x when y = –5. 2. The equation of a function is

. Find

1. the value of y when x = 0, 2. the value of x when y = −⅔.

Exercise 6A Questions 1-2, 6

Relations and Functions Fig. 6.6 shows the arrow diagram of a relation.

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Fig. 6.6 We notice that only one arrow leaves each element in the domain. Thus every element in the domain of the relation has a unique (exactly one) image in the codomain. The relations, whose arrow diagrams are shown in Fig. 6.7 and Fig. 6.8, also satisfy the property that every element in the domain has a unique image in the codomain.

Fig. 6.7

Fig. 6.8

Relations in Fig. 6.6, Fig. 6.7 and Fig. 6.8 are examples of a special kind of relation that we call functions. In particular, Fig. 6.6 is a one-to-one correspondence which we call a one-to-one function. A function is a relation in which every element in the domain has a unique image in the codomain. ******ebook converter DEMO Watermarks*******

What are the differences among the relations shown in Fig. 6.6, Fig. 6.7 and Fig. 6.8?

Worked Example

(Verifying if a Relation is a Function) State, with reason, whether each of the following arrow diagrams defines a function.

1 a.

b.

Solution: 1. The relation is a function since every element in the domain A has a unique image in the codomain B. 2. The relation is not a function since the element a in the domain A has three images, Math, Science and Geography, in the codomain B.

Exercise 6A Questions 3(a)-(d) ******ebook converter DEMO Watermarks*******

State, with reason, whether each of the following arrow diagrams defines a function. a.

b.

Notation of a Function Consider the function f whose arrow diagram is displayed in Fig. 6.9. The domain of the function is the set X = {1, 2, 3, 4} and its codomain is the set Y = {1, 2, 3, 4, 5, 6, 7, 8}.

Fig. 6.9 We often use a lower case letter such as f to name a function. The notation f : X → Y is used to indicate that the function f has domain X and codomain Y. ******ebook converter DEMO Watermarks*******

We read this as “a function f from X to Y”. We can also write X →f Y. This reminds us of the diagram in Fig. 6.10.

Fig. 6.10 For a function f : X → Y, each element x in the domain X has a unique image y in the codomain Y. We often say y is a function of x and write it as y = f(x). The function may be written as f : x ↦ f(x), linking an element of the domain to its image f(x) in the codomain. Note that the vertical stroke on the arrow distinguishes it from f : X → Y. f(x) is also called the value of the function f at x.

f(x) is read as “f of x”. Consider the example in Fig. 6.6. Suppose f represents the function, then f(1) = b, f(2) = c and f(3) = a. In Fig. 6.7, if g represents the function, then g(Peter) = g(Jim) = g(John) = A, g(Adam) = g(Tom) = B and g(Thomas) = C. In Fig. 6.8, if h represents the function, then h(A) = 40, h(B) = 70, h(C) = 50 and h(D) = 60. For the function f in Fig. 6.9, we have f(1) = 4 = 1 + 3, f(2) = 5 = 2 + 3, f(3) = 6 = 3 + 3 and f(4) = 7 = 4 + 3 ******ebook converter DEMO Watermarks*******

Thus, in general, f(x) = x + 3 i.e. to each x in the domain, the function f assigns the image f(x) by adding 3 to x. The function f can be completely described as follows: f : x ↦ x + 3, x = 1, 2, 3 or 4 or x ∈ {1,2, 3, 4}

Worked Example

2

(Determining Dependent and Independent Variables) Given the function f : x ↦ 3x + 2, where x is real, find the value of a. f(2), b. f(a), where a is a constant, c. x for which f(x) = 17.

Solution: a. f(2) = 3(2) + 2 = 8 b. f(a) = 3a + 2 c. When f(x) = 17, we have

Given the function g : x ↦ 5x − 4, where x is real, find the value of 1. g(3), 2. g(a + 2), ******ebook converter DEMO Watermarks*******

3. x for which g(x) = 1.

Exercise 6A Questions 4, 7-12, 15-17

Worked Example

3

(Drawing and Using Arrow Diagrams for Functions) Illustrate, by means of an arrow diagram, the function f : x ↦ 2x + 1 where the domain is the set of integers.

1. 2. The arrow diagram represents part of the function f : x ↦ ax2 + bx, where x is an integer. Find the value of a and of b.

Solution: a. Two parallel lines are drawn to represent the domain and the codomain. Positive and negative numbers are marked on the line. Each element in the domain is linked to its image by means of an arrow as shown below.

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Since there are infinite elements in the domain, it is impossible to draw all the arrows. Therefore, only a few points are taken. b. f : x ↦ ax2 + bx f(x) = ax2 + bx From the arrow diagram,

Substitute a = 2 into ∴ a = 2 and b = −5

a. Illustrate, by means of an arrow diagram, the function f : x ↦ 2x − 1 where the domain is the set of integers. b. The arrow diagram represents part of the function f : x ↦ ax3 + bx, where x is an integer. Find the value of a and of b.

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Exercise 6A Questions 5, 13-14

BASIC LEVEL 1. The equation of a function is y = 4x + 5. Find the value of y when i. x = 3, ii. x = −2. 2. The equation of a function is y = 25 − 3x. Find the value of x when i. y = 34, ii. y = −5. 3. Each of the following relations has the set of integers {2, 4, 6, 8} as its domain. State whether each of the following arrow diagrams defines a function. If the answer is no, state the reason.

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a.

b.

c.

d.

4. A function f is defined by f : x ↦ 6x − 4 for all real values of x. What are the images of 2, −4, 1 3 and – 1 2 under f? 5. Draw arrow diagrams with two parallel lines to show the following functions. Let the domain be the set of integers and draw six arrows for each function. a. f : x ↦ x + 2 b. f : x ↦ 2x2 − 2 INTERMEDIATE LEVEL 6. The equation of a function is

. Find

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a. the value of y when i. x = −3, ii. x = 1 1 2. b. the value of x when a. y = 1, b. y = – 6. 7. Given the function f : x ↦ 5 − 2x, evaluate each of the following: i. f(1) ii. f(−2) iii. f(0) iv. f(3) + f(−3) 8. Given the function g(x) = 7x + 4, find the value of each of the following: a. g(2) b. g(−3) c. d. g(0) + g(−1) e. 9. Given the functions f : x ↦ 5x − 9 and g : x ↦ 2 − 6x, find the value of x for which a. f(x) = 16 b. g(x) = 14 c. g(x) = x d. f(x) = 2x e. f(x) = g(x) f. 2f(x) = 3g(x) ******ebook converter DEMO Watermarks*******

10. Given the function g(x) = mx + c and that g(1) = 5 and g(5) = −4, find the values of m and c. Hence evaluate g(3) and g(−4). 11. If g(x) = x2 + 5, find an expression for each of the following: a. g(a) b. g(a + 1) c. g(a + 1) − g(a − 1) d. g(a2) e. g(a3) f. g(a3 + 1) 12. If h(x) = −5x + 4, a. express h(2a) − h(a) in terms of a, b. find the value of a for which h(a) = 0, c. express h(a2) + h(a) in terms of a. 13. The diagram represents part of the function f : x ↦ ax2 + b, where x is a real number.

a. Find the value of a and of b. b. An arrow is to be drawn starting at x = 2. At what point should this arrow end? 14. The arrow diagram shows part of the function f : x ↦ ax + b, where x is any ******ebook converter DEMO Watermarks*******

real number.

a. Calculate the value of a and of b. b. An arrow is to be drawn starting at x = 2. At what point should this arrow end? ADVANCED LEVEL 15. Given the function f(x) = 4x + 9, evaluate f(1), f(2) and f(3). Is it true that a. f(1) + f(2) = f(1 + 2)? b. f(3) − f(2) = f(3 − 2)? c. f(1) × f(2) = f(1 × 2)? d. f(2) ÷ f(1) = f(2 ÷ 1)? 16. Given the functions

and

, evaluate

.

a. Is it true that f(2) + f(3) = f(2 + 3)? Show your working clearly. b. Is it true that g(4) − g(2) = g(4 − 2)? Show your working clearly. c. Find the value of x for which f(x) = g(x). d. Find the expressions of f(2a) and g(3a) in terms of a. e. Find the value of k for which f(2k) = g(6k). f. Find the value of k for which f(k) + g(k) = 5. 17. The function f is defined by 1. Find the value of a for which

for all real values of x except x =

a. f(2) = 5, b. f(3) = a, ******ebook converter DEMO Watermarks*******

c. f(a) = a.

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6.3 Linear Functions

Ordered Pairs Satisfying a Linear Function Consider the linear function y = 2x, we shall look at all pairs of numbers x and y that satisfy the linear function y = 2x. When x = 1, y = 2(1) = 2     x = 2, y = 2(2) = 4     x = 3, y = 2(3) = 6     x = 4, y = 2(4) = 8 etc Thus the pair of numbers x = 1, y = 2 satisfy the linear function y = 2x. Similarly, x = 3, y = 6 also satisfy the linear function y = 2x etc. Obviously, we cannot list all the values of x and y that satisfy the linear function y = 2x. Can we find a way to display the linear function y = 2x? The table below shows five pairs of values of x and y that satisfy the linear function y = 2x. See Table.. The five pairs of numbers (x, y) in the table on the previous page are plotted as points in the Cartesian plane as shown in Fig. 6.11 below. The scale for both the x- and y-axes is 1 cm to 1 unit.

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Fig. 6.11 1. a. Do you notice any pattern formed by the points? b. How does the value of y change as the value of x increases by 1? c. Do all the values of y change accordingly as the values of x increase? In the following graph, Fig. 6.12, a straight line is drawn to pass through the five points.

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Fig. 6.12 The line is said to be a graph of the linear function y = 2x. Do you agree that every pair of numbers (x, y) satisfying y = 2x appear as a point somewhere on the line and that every point on the line has coordinates that satisfy the function y = 2x? The graph of the function is the set of all points whose coordinates satisfy the function. Let us consider another function y = 2x + 1, we shall look at some pairs of numbers x and y that satisfy the function y = 2x + 1. When x = –2, y = 2(–2) + 1 = –3      x = –1, y = 2(–1) + 1 = –1      x = 0, y = 2(0) + 1 = 1      x = 1, y = 2(1) + 1 = 3      x = 2, y = 2(2) + 1 = 5 ******ebook converter DEMO Watermarks*******

Thus the pairs of numbers x = –2, y = –3, and x = –1, y = –1 etc satisfy the function y = 2x + 1. The table below shows five pairs of values of x and y that satisfy the function y = 2x + 1. See Table.. The five pairs of numbers (x, y) appearing in the table on the previous page are plotted as points in the Cartesian plane as shown in Fig. 6.13 below where we use 1 cm to represent 1 unit on both the x and y-axes.

Fig. 6.13 2. ******ebook converter DEMO Watermarks*******

a. Do you notice any pattern formed by the points? b. How does the value of y change as the value of x increases by 1? c. Do all the values of y change accordingly as the values of x increase? In the following graph, Fig. 6.14, a straight line is drawn to pass through the five points.

Fig. 6.14 The line is the graph of the linear function y = 2x + 1.

In the above investigation, we have plotted five points and a line is then ******ebook converter DEMO Watermarks*******

drawn to join the five points. a. What is the minimum number of points we need to draw a linear function? b. How do we ensure that we have calculated the coordinates of the points satisfying the functions accurately? c. How many points should we plot to draw a line accurately?

1. A linear function is given by y = 4 − 2x. Complete the table below and draw the graph of the function. See Table.. 2. A linear function is given by f(x) = x + 5. a. State its x- and y-intercepts. b. Draw the graph of the function. 3. Given a linear function a. state the slope of its graph, b. draw the graph of the function.

Exercise 6B Questions 1-3, 6-8

Range of a Function Let f : X → Y be a function. The set of values of f(x) is called the range of f. The range of the function f in Fig. 6.9 is {4, 5, 6, 7}. The range may or may not consist of all of the elements of the codomain. The range {4, 5, 6, 7} consists of only some of the elements of Y = {1, 2, 3, 4, 5, 6, 7, 8}. Very often, we are interested in the range and not the codomain of a function. Hence it is often inadequate to define a function f by stating its domain and ******ebook converter DEMO Watermarks*******

the rule which determines the unique image f(x) of each x in the domain. Fig. 6.15, Fig. 6.16 and Fig. 6.17 show the graphs of the functions f, g and h respectively.

Fig. 6.15

Fig. 6.16

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Fig. 6.17 Notice that the graph of f consists of 7 isolated points (−3, 9), (−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4) and (3, 9). These 7 points are joined by a smooth curve to obtain the graph of g. By extending the graph of g upwards indefinitely, the graph of h results. The domain of f is {−3, −2, −1, 0, 1, 2, 3} and the range of f is the set {0, 1, 4, 9}. From the graph of g, the domain of g is −3 ≤ x ≤ 3 and the range of g is 0 ≤ g(x) ≤ 9 and from the graph of h, the domain is the set of real numbers and the range of h is h(x) ≥ 0. In general, the range of a function can be obtained by projecting the graph of the function on the vertical axis as shown in Fig. 6.18.

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Fig. 6.18

Worked Example

4

(Finding the Range of a Function) A function f is defined by f : x ↦ 2x − 1. Find the range of f for the domain −1 < x ≤ 3.

Solution: f(x) = 2x − 1 f(−1) = 2(−1) − 1 = −3 f(3) = 2(3) − 1 = 5 The graph of f for the domain −1 < x ≤ 3 is shown below.

From the graph, the range of f is −3 < f(x) ≤ 5. Alternatively, −1 < x ≤ 3 −2 < 2x ≤ 6 (multiply throughout by 2) −3 < 2x − 1 ≤ 5 (subtract 1 from each term) −3 < f(x) ≤ 5

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This method is useful for linear functions y = ax + b.

Notice that in the graph in Worked Example 4, the empty circle is used to represent the point (−1, −3). Can you explain why this is done?

A function g is defined by g : x ↦ x − 4. Find the range of g for the domain −1 < x ≤ 3.

Exercise 6B Questions 4(a)-(d), 9(a)-(d)

Worked Example

5

(Finding the Domain of a Function) A function f is defined by f : x ↦ −3x − 1. Find the domain of f for the range −3 ≤ f(x) < 2.

Solution: f(x) = −3x − 1 When

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When

The graph of f for the range −3 ≤ f(x) < 2 is shown below.

From the graph, the domain of f is

A function g is given by ≤ g(x) ≤ 1.

.

. Find the domain of g for the range −1

Exercise 6B Questions 5(a)-(d), 10(a)-(d)

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BASIC LEVEL 1. A linear function f is given by f(x) = 2x + 5. Complete the table below and draw the graph of the function. See Table.. 2. A linear function is defined by y = −x − 5. a. State its x- and y-intercepts. b. Draw the graph of the function. 3. A linear function is given by the equation 6y = 3x − 2. a. State the slope of the graph of the function. b. Draw the graph of the function. 4. Sketch the graphs of the following functions for the given domain on separate diagrams and state the range in each case. a. f : x ↦ x, −1 ≤ x ≤ 1 b. f : x ↦ x − 6, 2 < x < 4 c. f : x ↦ 2x − 3, −1 ≤ x ≤ 3 d. f : x ↦ x + 2, x ≥ −1 5. Find the domain of each of the following functions for the range −2 ≤ f(x) ≤ 5. a. f : x ↦ 2x − 3 b. f : x ↦ x + 6 c. f : x ↦ −2x + 1 d. f : x ↦ 2 − x

INTERMEDIATE LEVEL 6. A linear function g is given by g(x) = ax + b. See Table.. ******ebook converter DEMO Watermarks*******

a. Find the value of a and of b. b. Complete the table above and draw the graph of the function. 7. A linear function is defined by y = ax + b. Given that its x-intercept is and y-intercept is −4, a. find the value of a and of b, b. draw the graph of the function. 8. Given that the slope and y-intercept of the graph of a linear function f are −1 and respectively, a. state the equation of the function, b. draw the graph of the function. 9. The graph of each of the following functions is shown. State the range of each function for the given domain. a. g : x ↦ x2 + 1, −1 ≤ x ≤ 3

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b. g : x ↦ (x + 2)2 − 1, −4 ≤ x ≤ 1

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c. g : x ↦ −x2 + 4, −3 ≤ x ≤ 1

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d. g : x ↦ x2 − 4x + 3, 0 ≤ x ≤ 5

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ADVANCED LEVEL 10. For each of the following functions, state the largest possible domain corresponding to the given range. The graph of each function is provided. a. f : x ↦ x2 − 4 for the range −4 ≤ f(x) ≤ −3

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b. f : x ↦ x2 − 2x + 1 for the range 0 ≤ f(x) ≤ 1

c. f : x ↦ (x − 2)(x − 6) for the range −4 ≤ f(x) ≤ 0

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1. A relation connects elements in set A (domain) to elements in set B (codomain) according to the definition of the relation. 2. A function is a relationship between two variables x and y such that every input x produces exactly one output y. 3. Every pair of values (x, y) that satisfies the equation of a function appears as a point on the graph of the function. Conversely, every point on the graph of the function has coordinates that satisfy the equation of the function. 4. A function is a relation in which every element in the domain is connected to a unique (exactly one) element (image) in the codomain. 5. The range of a function f is the set of values of f(x) (images under f) for the given domain.

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1. The equation of a function is y = 20 − 6x. Find the value of y when i. x = −5, ii. x = −2, iii. x = 1, iv. x = 4. 2. The equation of a function is y = 3x + 4. Find the value of x when i. y = 34, ii. y = −5, iii. y = 19, iv.

.

3. Each of the following relations has the set of integers {2, 4, 6, 8} as its domain. State whether each of the following arrow diagrams defines a function. If the answer is no, state the reason. a.

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b.

4. A function f is defined by f : x ↦ 3x − 2 for all real values of x. What are the images of 1, −4 and under f? 5. Draw arrow diagrams with two parallel lines to show the following functions. Let the domain be the set of integers and draw six arrows for each function. a. f : x ↦ 2 − x b. f : x ↦ x2 − 2 6. Given the function f : x ↦ x2 − 2x + 3, evaluate each of the following: a. f(1) b. f(−2) c. f(0) d. f(1) + f(−1) 7. Given the functions f : x ↦ 3x − 9 and g : x ↦ 4x, find the value of x for which a. f(x) = 15 b. g(x) = 12 c. g(x) = x d. f(x) = 2x e. f(x) = g(x) f. 2f(x) = −3g(x) 8. If f(x) = 7x + 6, a. express f(2a) + f(a) in terms of a, ******ebook converter DEMO Watermarks*******

b. express f(2a) − f(a) in terms of a, c. find the value of a for which f(a) = 0. 9. The diagram represents part of the function f : x ↦ ax3 + b, where x is a real number.

a. Find the value of a and of b. b. An arrow is to be drawn starting at x = 2. At what point should this arrow end? 10. A linear function f is given by f(x) = 3 − draw the graph of the function. See Table..

. Complete the table below and

11. A linear function is defined by g(x) = 8x − 16. a. State its x- and y-intercepts. b. Draw the graph of the function. 12. A linear function is given by the equation 2y − 3x = 8. a. State the slope of the graph of the function. b. Draw the graph of the function. 13. Sketch the graphs of the following functions for the given domain on separate diagrams and state the range in each case. a. f : x ↦ 1 − x, −2 ≤ x ≤ 2 ******ebook converter DEMO Watermarks*******

b. f : x ↦ 5x − 6, 2 < x < 4 c. f : x ↦ x 3 + 2, −1 ≤ x < 3 d. f : x ↦ −x − 2, x ≥ 1 14. Find the domain of each of the following functions for the range 0 ≤ f(x) ≤ 5. a. f : x ↦ −4x + 5 b. f : x ↦ 4x − 6 c. f : x ↦ 2x + 3 d. f : x ↦ −x − 2 15. A function g is given by g(x) = ax2 + b. See Table. a. Find the value of a and of b. b. Complete the table above. 16. The graph of each of the following functions is shown. State the range of each function for the given domain. a. g : x ↦ x2 − 4x + 1, 0 ≤ x ≤ 4

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b. g : x ↦ −2x2 − x, −1 ≤ x ≤ −0.5

c.

, all real values of x

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A

function

f

is

defined

such

that

and

where a and b are natural numbers. 1. Show that

.

2. Hence find the value of f(2).

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Chapter Seven Geometry Logic Statements The development of mathematical proof is primarily the product of ancient Greek mathematics. The idea of proving a statement is true is said to have begun in the 5th century BC when Greek philosophers developed a way of convincing each other of the truth of particular mathematical statements. They had to agree on definitions of certain basic ideas and axioms, which were statements about the starting points.

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LEARNING OBJECTIVES At the end of this chapter, you should be able to: determine the relationship between the hypothesis and the conclusion of an ‘if-then’ statement, transform a statement into an equivalent ‘if-then’ statement, determine the inverse, converse and contrapositive of an ‘ifthen’ statement, illustrate the equivalences of: (a) the statement and its contrapositive; and (b) the converse and inverse of a statement, use inductive or deductive reasoning in an argument, write a proof (both direct and indirect).

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7.1 Statements Simple Statements Statements are sentences that are either true or false. For example: "The Philippines is in Asia" is a true statement; “Snakes are mammals” is a false statement since snakes are reptiles; “This pasta is delicious” is not a statement since whether a food is delicious is subjective among different people. These statements are called simple statements as they cannot be broken down into simpler ones and they convey one idea only.

Worked Example

1

(Determining a Statement) Determine if each of the following sentences is a true statement, false statement or not a statement. a. 8 − 2 = 5 b. 4 × 7 = 28 c. x + 1 = 8

Solution: a. 8 − 2 = 5 is a false statement since 8 − 2 = 6. b. 4 × 7 = 28 is a true statement. c. x + 1 = 8 is not a statement since the value of x is not known.

While x + 1 = 8 is not a statement, the following is a statement: x + 1 = 8 for ******ebook converter DEMO Watermarks*******

a real number x.

Determine if each of the following sentences is a true statement, false statement or not a statement. a. A quadrilateral has 4 sides. b. 0 is not an integer. c. a + b = b + a for all real numbers a and b.

Exercise 7A Questions 1(a)-(f)

Compound Statements Compound statements can be formed in two ways: Negation The negation of a true statement is a false statement and the negation of a false statement is a true statement. Sometimes, it is necessary to change a statement to its opposite meaning. Using connectives Compound statements are usually formed by two simple statements using the following connectives: a. and b. or c. if … then d. if and only if We shall learn about the ‘if-then’ statements and ‘if and only if’ statements in the next section.

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Worked Example

2

(Writing a Negation Statement) Write a negation statement for each of the following and state whether the result is true or false. a. Potato is low in sodium. b. There are seven planets in the solar system. c. 20% of 80 is 16. d. All prime numbers are odd numbers.

Solution: a. Potato is high in sodium. (false statement) b. The number of planets in the solar system is not seven. (true statement) c. 20% of 80 is not 16. (false statement) d. Not all prime numbers are odd numbers. (true statement)

We must show some caution in writing negation statements involving all, none and some. Writing “All prime numbers are even numbers” will give a false statement.

Write a negation statement for each of the following and state whether the result is true or false. a. The young of a cow is an ox. b. Water freezes to become ice. c. Each interior angle of an equilateral triangle is 180°. d. All integers are rational numbers. ******ebook converter DEMO Watermarks*******

Exercise 7A Questions 2(a)-(d)

Worked Example

3

(Determining if an ‘And’ Statement or ‘Or’ Statement is True) a. Determine whether the following ‘and’ statements are true. i. 20 − 5 = 15 and 10 × 3 = 30 ii. 20 − 5 = 10 and 10 × 3 = 30 iii. 20 − 5 = 10 and 10 × 3 = 40 b. Determine whether the following ‘or’ statements are true. i. A triangle has 3 interior angles or the sum of interior angles of a triangle is 180°. ii. A triangle has 4 interior angles or the sum of interior angles of a triangle is 180°. iii. A triangle has 4 interior angles or the sum of interior angles of a triangle is 360°.

Solution: a. Only statement (i) is true. Each of the other statements is false since at least one of its simple statements is false. b. Only statement (iii) is false. Each of the other statements is true since at least one of its simple statements is true.

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Determine whether the following statements are true. a. 10% discount of $50 is $5 and half of 16 is 9. b. 12 + 3 > 10 and 30 − 16 < 20 c. An acute angle is less than 180° or an obtuse angle is greater than 180°. d. A square has 4 equal sides or 4 right angles.

Exercise 7A Questions 3(a)-(c), 4(a)-(c), 5(a)-(e)

BASIC LEVEL 1. Determine if each of the following sentences is a true statement, false statement or not a statement. a. a − b = b − a for all real numbers a and b. b. x2 + y2 for all real numbers x and y. c. (a + b)2 = a2 + 2ab + b2 d. x − 4 < 3 where x is a real number. e. ab = ba f. 22 ≠ (−2)2 INTERMEDIATE LEVEL 2. Write a negation statement for each of the following and state whether the result is true or false. ******ebook converter DEMO Watermarks*******

a. There are 26 letters in the alphabet. b. The slope of a straight line is constant. c. 12 and 18 have three common factors. d. 1 is a perfect square. 3. Determine whether the following ‘and’ statements are true. a. 15 + 5 > 16 and 30 ÷ 4 = 10 b. 20 + 16 ÷ 2 = 28 and 20 ÷ 4 − 2 = 3 c.

and

4. Determine whether the following ‘or’ statements are true. a. Adjacent angles on a straight line add up to 180° or vertically opposite angles are equal. b. Adjacent angles on a straight line are equal or vertically opposite angles add up to 180°. c. Adjacent angles on a straight line add up to 90° or vertically opposite angles are equal. ADVANCED LEVEL 5. Determine whether the following statements are true. a. A rectangle is a square. b. A rectangle is neither a quadrilateral nor a square. c. A rectangle is either not a quadrilateral or a square. d. A rectangle is a quadrilateral but not a square. e. It is not true that a rectangle is a quadrilateral and a square.

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7.2 ‘If-then’ Statements and ‘If and only if’ Statements ‘If-then’ Statements ‘If-then’ statements are commonly used in the proving of Geometry. For example, “If two sides are opposite sides of a rectangle, then they have the same length”; “If a figure has 3 sides, then it is a triangle”. Such a statement is called an implication. Let a be “two sides are opposite sides of a rectangle” and b be “the two sides have the same length”. We can say that a implies b or a ⇒ b, where ⇒ is the implication symbol. The clause following the word “if” of an ‘if-then’ statement is called the hypothesis of the statement and the clause following “then” is called the conclusion.

It is important to get the hypothesis and conclusion right. It is wrong to say “If you qualify for college, then you get good grades”. For example, If you get good grades, then you qualify for college. Hypothesis: You get good grades. ******ebook converter DEMO Watermarks*******

Conclusion: You qualify for college. If two circles have equal radii, then they have the same area. Hypothesis: Two circles have equal radii. Conclusion: The circles have the same area.

Worked Example

4

(Writing an Equivalent ‘If-then’ Statement) Write each of the following statements in the ‘if-then’ form. a. The diagonals of a rectangle have the same length. b. ∠a and ∠b are vertically opposite angles and they are equal. c. The product of an even number and another number is an even number.

Solution: a. If the diagonals are those of a rectangle, then they have the same length. b. If ∠a and ∠b are vertically opposite angles, then they are equal. c. If one of two numbers is an even number, then the product of the numbers is an even number.

Write each of the following statements in the ‘if-then’ form. a. The angles of a square are right angles. b. Two squares of the same area have sides of equal lengths. c. A rational number is a number that can be expressed as the fraction of two integers, with the denominator not equal to zero.

Exercise 7B Questions 1(a)-(f) ******ebook converter DEMO Watermarks*******

‘If and only if’ Statements ‘If and only if’ statements or bi-implication statements are commonly used. Recall that “If a figure has 3 sides, then it is a triangle”. It is also true that “If a figure is a triangle, then it has 3 sides”. Let a be “A figure has 3 sides” and b be “A figure is a triangle”. The resulting ‘if and only if’ statement is “A figure has 3 sides if and only if it is a triangle”. We can say that a is equivalent to b or a ⇔ b, where ⇔ is the biimplication symbol. Let c be “A figure is a square” and d be “A figure has 4 sides”. While it is true that “If a figure is a square, then it has 4 sides”, it is false that “If a figure has 4 sides, then it is a square”. In this case, c is not equivalent to d.

Worked Example

5

(Determining True ‘If-then’ and ‘If and only if’ Statements) Write true ‘if-then’ or ‘if and only if’ statement for each of the following.

a

b

(a)

Two angles of a triangle are equal.

Two angles are base angles of an isosceles triangle.

(b)

A number is divisible by 8.

A number is divisible by 2.

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(c)

B is the midpoint of AC.

AB = BC

2, 4 and 6 are divisible by 2, but they are not divisible by 8. Therefore, a number is divisible by 2 does not imply that it is divisible by 8.

AB = BC does not mean that point B lies on the line AC. It would be true that “If A, B and C lie on a straight line and AB = BC, then B is the midpoint of AC”.

Solution: a. Both a ⇒ b and b ⇒ a are true. So a ⇔ b is true. Statement: Two angles of a triangle are equal if and only if they are base angles of an isosceles triangle. b. a ⇒ b is true but b ⇒ a is false. Statement: If a number is divisible by 8, then it is divisible by 2. c. a ⇒ b is true but b ⇒ a is false. Statement: If B is the midpoint of AC, then AB = BC.

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Write true ‘if-then’ or ‘if and only if’ statement for each of the following. a

b

(a)

Three angles of a triangle are equal.

A triangle is equilateral.

(b)

A whole number is divisible by 3.

A whole number is divisible by 9.

(c)

x and y are odd numbers.

xy is an odd number.

Exercise 7B Questions 2(a)-(f), 3(a)-(d)

BASIC LEVEL 1. Write each of the following statements in the ‘if-then’ form. a. The diagonals of a parallelogram bisect each other. b. ∠a and ∠b are adjacent angles on a straight line and the sum of ∠a and ∠b is 180°. c. The product of two positive numbers is a positive number. ******ebook converter DEMO Watermarks*******

d. An obtuse angle is greater than 90°. e. The sum of two even numbers is an even number. f. An equilateral triangle has 3 equal sides. INTERMEDIATE LEVEL 2. Write true ‘if-then’ or ‘if and only if’ statement for each of the following. a

b

(a)

A quadrilateral has 4 right angles.

A quadrilateral is a square.

(b)

A number is a multiple of 5.

A number is a multiple of 10.

(c)

ΔABC is a rightangled triangle.

∠ABC is a right angle.

(d)

x cannot be expressed as the fraction of two integers, with the denominator not equal to zero.

x is an irrational number.

A fraction is in

The numerator and denominator of a fraction

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(f)

its lowest terms.

have no common factors except 1.

ΔABC is isosceles.

ΔABC is equilateral.

ADVANCED LEVEL 3. Explain if the following statements are true. a. If x is an even number and y is an odd number, then xy + y2 is an odd number. b. xy + x + y − 1 is an odd number if and only if x and y are even numbers. c. x2 + 7x + 12 = 0 if and only if x = −3 or x = −4. d. If x2 + 2x − 3 = 0, then x = 1 and x = −3.

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7.3 Converse, Inverse and Contrapositive of 'If-then' Statements Converse of a Statement When the hypothesis and the conclusion in an ‘if-then’ statement are reversed, the resulting statement is called the converse of the ‘if-then’ statement. When an ‘if-then’ statement and its converse are true, the ‘if and only if’ statement is true. For example, both “If a figure has 3 sides, then it is a triangle” and its converse “If a figure is a triangle, then it has 3 sides” are true. Therefore, “A figure has 3 sides if and only if it is a triangle” is true.

Worked Example

6

(Writing the Converse of a Statement) Write the converse of the statement “The opposite sides of a rectangle are parallel”. Explain if the converse is true.

Solution: First, write the statement in ‘if-then’ form. Statement: If two sides of a rectangle are opposite sides, then they are parallel. Converse: If two sides of a rectangle are parallel, then they are opposite sides ******ebook converter DEMO Watermarks*******

of the rectangle. The converse is true. Only the opposite sides of a rectangle are parallel (the adjacent sides are perpendicular).

Write the converse of the statement “If a quadrilateral is a rectangle, then it has two pairs of parallel sides”. Explain if the converse is true.

Exercise 7C Questions 1(a)-(f), 2(a)-(f), 3(a)-(d)

Inverse of a Statement When the hypothesis and the conclusion in an implication are negated, the resulting statement is called the inverse of the ‘if-then’ statement. For example, the inverse of the statement “If a figure has 3 sides, then it is a triangle” is “If a figure does not have 3 sides, then it is not a triangle”.

Worked Example

7

(Writing the Inverse of a Statement) Write the inverse of the statement “If x = 2, then x2 = 4”. Explain if the inverse is true.

Solution: Statement: If x = 2, then x2 = 4. Inverse: If x ≠ 2, then x2 ≠ 4. The inverse is false since x = −2 also gives x2 = 4.

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Write the inverse of the statement “If x < 5, then x < 6”. Explain if the inverse is true.

Exercise 7C Questions 1(a)-(f), 2(a)-(f), 3(a)-(d)

Contrapositive of a Statement When the hypothesis and the conclusion in an implication are reversed and negated, the resulting statement is called the contrapositive of the ‘if-then’ statement. For example, we reverse the hypothesis and the conclusion of the statement “If a figure has 3 sides, then it is a triangle” to form the converse of the statement, i.e. “If a figure is a triangle, then it has 3 sides”. Then we negate the hypothesis and conclusion to form the contrapositive of the statement, i.e. “If a figure is not a triangle, then it does not have 3 sides”.

Worked Example

8

(Writing the Contrapositive of a Statement) Write the contrapositive of the statement “If 4x = 24, then x = 6”. Explain if the contrapositive is true.

Solution: Statement: If 4x = 24, then x = 6. Converse: If x = 6, then 4x = 24. Contrapositive: If x ≠ 6, then 4x ≠ 24. The contrapositive is true since 6 is the only number when multiplied by 4 gives 24.

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Write the contrapositive of the statement “If x = y, then x + z = y + z”. Explain if the contrapositive is true.

Exercise 7C Questions 1(a)-(f), 2(a)-(f), 3(a)-(d)

BASIC LEVEL 1. Write the converse, inverse and contrapositive for each of the following statements. Explain if the resulting statement is true. a. If a polygon has 5 sides, then it is a pentagon. b. If two lines are parallel, then they do not meet. c. The diagonals of a square bisect at right angles. d. If ∠ABC is 110°, then ΔABC is an obtuse-angled triangle. e. If x = 3, then x2 = 9. f. If x > 15, then x > 14. INTERMEDIATE LEVEL 2. Write the converse, inverse and contrapositive for each of the following statements. Explain if the resulting statement is true. a. A regular hexagon has 6 equal sides. b. A rectangle is a parallelogram. c. If Q is the midpoint of PR, then P, Q and R are collinear. ******ebook converter DEMO Watermarks*******

d. If x3 = −27, then x = −3. e. If x = 2, then x2 + 12 = x4. f. If a > b and c > 0, then ac > bc. ADVANCED LEVEL 3. Write the converse, inverse and contrapositive for each of the following statements. Explain if the resulting statement is true. a. If x is an even number and y is an odd number, then x2 − y2 is an odd number. b. (x + y)2 is an even number if and only if x and y are even numbers. c. x2 + y2, where x and y are variables, is an algebraic expression that cannot be factorized. d. ax2 + bx + c is a quadratic expression, where x is the variable, a, b and c are constants and a ≠ 0.

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7.4 Proving Statements Mathematical Inductive and Deductive Reasoning

Inductive Reasoning versus Deductive Reasoning Work in groups of 4-5. Mr Ramos gave Daniel and Riza the following problem to solve.

Daniel solved the question by substituting different pairs of values of x and y into the expressions on the LHS and RHS of the equation. If the equation is true, he would get the same value for the two expressions. After substituting five pairs of values of x and y, he drew a table of values as shown below: ******ebook converter DEMO Watermarks*******

See Table.. He concluded that x2 − y2 = (x + y)(x − y) since all five pairs of x and y values showed that LHS of the equation equals RHS of the equation. On the other hand, Riza solved the question using the following steps:

She concluded that x2 − y2 = (x + y)(x − y) for all values of x and y. Discuss with your classmates the following questions: a. What is the difference between Daniel’s method and Riza’s method of solving the question? b. What assumption did Daniel make in his method? c. Write down the reasoning for each of Riza’s steps. d. Is Daniel wrong in using his method? e. Whose method do you prefer? Why? Daniel used inductive reasoning in his approach of solving the question while Riza used deductive reasoning. Inductive reasoning makes broad generalizations from specific observations whereas deductive reasoning makes a specific and logical conclusion from a general statement or hypothesis.

(Determining Inductive Reasoning or Deductive Reasoning) ******ebook converter DEMO Watermarks*******

Worked

Example

9

Reasoning) Determine whether each of the following problems is solved by inductive or deductive reasoning. Explain why. l1 and l2 are straight lines. a. Sam proved that ∠a and ∠b will always be equal by the following steps: ∠a + ∠c = 180° (adj. ∠s on a str. line) ∠b + ∠c = 180° (adj. ∠s on a str. line) ∠a + ∠c = ∠b + ∠c ∴ ∠a = ∠b b. 5, 12, 19, 26, … Gemma conjectured that the next number in the sequence is 33 by observing that 7 is added to each term to get the next term. Therefore, the next term in the sequence is 26 + 7 = 33.

Solution: a. Sam used mathematical rules and logic to arrive at a valid conclusion. Hence, he solved the problem by deductive reasoning. b. Gemma used inductive reasoning to arrive at her answer. She used specific examples (the given terms in the sequence) to make a general rule to add 7 to each term of the sequence. However, if the numbers of the sequence represent the dates of a month, then the next number could be 2 (31 days in the month), 3 (30 days in the month), 5 (February) or 4 (February in a leap year).

Gemma’s answer may not be right although it is a reasonable answer. She recognized a number pattern and generalized her observations. ******ebook converter DEMO Watermarks*******

Determine whether each of the following problems is solved by inductive or deductive reasoning. Explain why. a. When the class was asked to find the nth term of the sequence 3 + 5 + 7 + 9 + …, Antonio gave the expression 2n + 1. b. Two angles of a triangle are 40° and 60°. Amelia found the third angle using “angle sum of a triangle”.

Exercise 7D Question 1

Writing a Proof A proof is a deductive argument for a mathematical statement. In the argument, we use deductive reasoning to prove that the statement is always true. Mathematical rules and laws are used in logical steps to demonstrate that the statement is always true, rather than enumerating many confirmatory cases. Recall that in the previous section, Daniel used inductive reasoning to solve Mr Ramos’ problem. In his approach, he extended a set of numerical values to show that the equation, which is the given statement, is true for all values of x and y. In Riza’s approach which uses deductive reasoning, she started with the given statement and used logical steps based on mathematical rules and laws to arrive at the conclusion that the original statement is true for all values of x and y. What Riza did was to write a proof to show that the given statement is always true. In this section, we shall learn to write proofs with the reasoning for each step.

Worked Example

(Writing an Algebraic Proof) Prove that if w > x and y > z, then w + y > x + z. Give a ******ebook converter DEMO Watermarks*******

reason for each step.

10 Solution: Statement

Reason

w>x w+y>x+y y>z

Given statement Add y to both sides Given statement

x+y>x+z

Add x to both sides

w+y>x+z

Transitive law for inequality

Prove that if x > y and z < 0, then xz < yz. Give reasoning for each step.

Exercise 7D Questions 2(a)-(d), 8(a)-(d), 14

Worked Example

(Writing a Geometric Proof)

11

Given that both ∠ABD and ∠EBF are right angles, prove that ******ebook converter DEMO Watermarks*******

∠ABF = ∠CBE. Give a reason for each step.

Solution: Statement

Reason

∠ABF + ∠FBD = ∠ABD

Sum of complementary angles

∠ABF + ∠ABE = ∠EBF

Sum of complementary angles

∠ABF + ∠FBD = ∠ABF + ∠ABE

∠ABD = ∠EBF = 90°

∠FBD = ∠ABE

By comparison

∠ABC + ∠ABD = 180°

Adj. ∠s on a str. line

∠ABC = 180° − 90°

Subtract 90° from both sides

= 90° ∠ABE + ∠CBE = ∠ABC

Sum of complementary angles

∠FBD + ∠ABF = ∠ABD

Commutative law of addition (1st statement)

∠ABF = ∠CBE

∠FBD = ∠ABE, ∠ABD = ∠ABC, by comparison

In the figure, ABE and BCD are straight lines. ∠ACD = ∠CBE. Prove that ΔABC is isosceles. Give reasoning for each step.

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Exercise 7D Questions 3-6, 9-12, 15

Direct Proof and Indirect Proof The proofs that we have learned in the previous section are called direct proofs. A direct proof in geometry is the most common form of proof used. In a direct proof, the conclusion to be proved is shown to be true directly by deductive reasoning using mathematical facts and theories pertaining to the specific situation. The indirect proof is a less common type of proof in geometry, though equally effective. In an indirect proof, we prove a conclusion in a roundabout way. The direct proof and indirect proof are two different ways of getting the same result, where the direct proof is more straightforward than the indirect proof. We follow these steps when using an indirect proof to prove a statement. 1. Assume the negation of the statement to be true. 2. Use deductive reasoning to do the proving and eventually run into a contradiction: two statements that cannot both be true. 3. The contradiction in step 2 shows that the assumption of the negation of the statement is false. Hence, the statement to be proved must be true. An indirect proof is also called a proof by contradiction, because we are making use of the contradiction to the negation of a statement in order to prove that the statement is true.

Worked Example

12

(Writing an Indirect Proof) Write an indirect proof that a triangle has at most one obtuse angle. Give a reason for each step.

Solution: ******ebook converter DEMO Watermarks*******

Let us assume ΔABC to be a triangle that has two obtuse angles. Statement

Reason

∠A and ∠B are obtuse.

Assumption that the negation of the statement is true

∠A + ∠B > 180°

∠A > 90°, ∠B > 90°

∠A + ∠B + ∠C = 180°

∠ sum of ΔABC

∠C = 180° − ∠A − ∠B

Subtract ∠A and ∠B from both sides

180° − ∠A − ∠B > 0

∠C > 0

∠A + ∠B < 180°

Add ∠A and ∠B to both sides

Contradiction

∠A + ∠B cannot be more than and less than 180°

A triangle has at most one obtuse angle.

The negation is false

ΔABC is an isosceles triangle with AB = AC. Write an indirect proof that ∠ABC = ∠ACB. Give a reason for each step.

Exercise 7D Questions 7, 13

BASIC LEVEL

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1. Mrs Gonzales asked the class to find the sum of interior angles of a quadrilateral. Determine whether the student/group of students used inductive or deductive reasoning to solve the problem. Explain your answer. a. A group of 5 students each drew a quadrilateral and measured the four interior angles of his/her quadrilateral. The angles were added and all of them obtained the sum of 360°. Hence, they concluded that the sum of interior angles of a quadrilateral is 360°. b. Tricia reasoned that since a quadrilateral is made up of two triangles and the sum of interior angles of a triangle is 180°, the sum of interior angles of a quadrilateral is 180° × 2 = 360°. 2. Prove the following statements. Give a reason for each step. a. If p > q, q > r, r > s, then p > s. b. If a + b = c + d, then a2 + ab = ac + ad. c. If 3x + 8 = −7, then x = −5. d. If x = 4, then

.

3. The figure shows ΔABE and ΔDCE. AEC and DEB are straight lines. Prove that if ∠BAE = ∠CDE, then ∠ABE = ∠DCE. Give a reason for each step.

4. In the figure, ∠AGF = ∠EGF, ∠BGC = ∠DGC, and FGC is a straight line. Prove that ∠AGB = ∠EGD. Give a reason for each step.

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5. Prove that x = 12 in the figure. Give a reason for each step.

6. In the figure, BE is the bisector of ∠ABC and BCD is a straight line. Prove that ∠ABE = 28°. Give a reason for each step.

7.

ΔABC is a right-angled triangle as shown. Write an indirect proof that the acute angles of a right-angled triangle are complementary. Give a reason for each step. ******ebook converter DEMO Watermarks*******

INTERMEDIATE LEVEL 8. Prove the following statements. Give a reason for each step. a. (a + b)(a2 − ab + b2) = a3 + b3 b. If a or b is odd, then ab + a2 + b2 is odd. c. 4(6x − 5y) − 3(2y − 8x) = 2(24x −13y) d. If 2 − 5x > 27, then x < −5. 9. The figure shows ΔABC, where ∠ABC is a right angle. If AC = CD, prove that ∠BAC = 90° − 2x°. Give a reason for each step.

10. The figure shows ΔABE and ΔBCD where ∠BCD = x°, ∠BDC = y° and ∠BAE = z°. ABC is a straight line. Prove that ∠AED = x° + y° + z°. Give a reason for each step.

11. In the figure, AOB, COD and EOF are straight lines. Prove that if ∠BOD is a right angle, then a = 8. Give a reason for each step.

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12. In the figure, ACE, BCD and DEF are straight lines. Prove that x = 47. Give a reason for each step.

13.

Lines l1 and l2 are cut by a transversal t such that ∠a = ∠b. Write an indirect proof that ∠c = ∠d. Give a reason for each step.

ADVANCED LEVEL ******ebook converter DEMO Watermarks*******

14. Prove that (x2 − 3)3 + 3(x2 − 3) − (x2 − 3)2 − 3 = (x + 2)(x − 2)(x4 − 6x2 + 12). Give a reason for each step. 15. The figure shows ΔADE where AB = AC, BD = BE and AF = DF. ABD, DFE, ACF and BCE are straight lines. Prove that ∠AEC + ∠CAE = 72°. Give a reason for each step.

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1. Statements are sentences that are either true or false. 2. An ‘if-then’ statement is an implication in which the clause following the word “if” is called the hypothesis of the statement and the clause following “then” is called the conclusion. 3. An ‘if and only if’ statement is a bi-implication in which the ‘if-then’ statement is true and reversing the hypothesis and conclusion also results in a true statement. 4. When the hypothesis and the conclusion in an ‘if-then’ statement are reversed, the resulting statement is called the converse of the ‘if-then’ statement. 5. When the hypothesis and the conclusion in an implication are negated, the resulting statement is called the inverse of the ‘if-then’ statement. 6. When the hypothesis and the conclusion in an implication are reversed and negated, the resulting statement is called the contrapositive of the ‘if-then’ statement. 7. Inductive reasoning makes broad generalizations from specific observations whereas deductive reasoning makes a specific and logical conclusion from a general statement or hypothesis. 8. A proof is a deductive argument that demonstrates the truth of a mathematical statement. It can be a direct proof or indirect proof.

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1. Determine if each of the following sentences is a true statement, false statement or not a statement. a. (a + b) + c = a + (b + c) for all real numbers a, b and c. b. 3 × 8 < 8 × 3 c. x2 − 2xy + y2 for all real numbers x and y. d. x + 6 = 2 − x e. 9 is a prime number and −4 is a negative integer. f. A right angle is 90° or the sum of supplementary angles is 90°. 2. Write each of the following statements in the ‘if-then’ form. a. An isosceles triangle has 2 equal angles. b. The opposite angles of a parallelogram are equal. c. ∠a and ∠b are vertically opposite angles and they are equal. d. A prime number is a whole number that has exactly 2 different factors, 1 and itself. e. The cross section is a quadrilateral and the sum of angles of the cross section is 360°. f. x = 5 and x2 = 25.

3. Write true ‘if-then’ or ‘if and only if’ statement for each of the following. ******ebook converter DEMO Watermarks*******

a

b

(a)

ABCD is a rhombus.

ABCD is a parallelogram.

(b)

A chocolate bar has a uniform cross section.

A chocolate bar is a prism.

(c)

∠ABC is an obtuse angle.

ΔABC is an obtuse-angled triangle.

(d)

We multiply or divide both sides of an inequality by a negative number.

We reverse the inequality sign.

(e)

A number is a multiple of 6.

A number is a multiple of 2.

(f)

A quadrilateral has 2 pairs of equal angles.

A quadrilateral is a rhombus.

4. Write the converse, inverse and contrapositive for each of the following statements. Explain if the resulting statement is true. a. A regular octagon has 8 equal angles. b. If ΔABC is a right-angled triangle, then it is not an equilateral triangle. c. If a whole number is not prime, then it must be composite. d. If a(a + 1) = 6, then a = 2. e. If x = −1 and y = 5, then x2 + xy = −4. ******ebook converter DEMO Watermarks*******

f. If ab > 0, then a > 0 and b > 0 or a < 0 and b < 0. 5. Prove the following statements. Give a reason for each step. a. If v2 = u2 + 2gs, then b. If p > q and q > r, then pr − p2 − qr + pq < 0. c. If 5x − 9 = 2x − 3, then x = 2. d. If x = −4 and y = 5, then 6. Prove that if ∠AOC = ∠BOD, then ∠AOB = ∠COD. Give a reason for each step.

7. In the figure, AB = AC and BOC is a straight line. Prove that if OA is the angle bisector of ∠BAC, then OA is perpendicular to BC. Give a reason for each step.

8. In the figure, ΔABD is an isosceles triangle and ΔBCD is an equilateral triangle. ABC is a straight line. Show that ∠ADC is a right angle. Give a reason for each step.

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9. In the figure, AB // PQ, BP // RQ, ∠ABT = (12x + 22)° and ∠PQT = 11x°. Prove that x = 4. Give a reason for each step.

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An empty tank is placed below two taps. One tap takes x min to fill the tank and the other tap takes y min to fill the tank. Prove that, starting with an empty tank, it takes min for both taps together to fill the tank. Explain how you arrive at the answer.

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B1 Revision Exercise 1. Show, unshaded, the regions satisfied by the following inequalities: a. x ≤ 1, y > 0, y < 2x + 3 b. x < 0, 2y ≤ x + 8, y + 2x > 0 2. In each of the following cases, write down the inequalities which define the unshaded region. a.

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b.

3. The set of points with coordinates (x, y) satisfies the three inequalities: y + x ≤ 8, 4y + 3x ≥ 24 and 3y ≤ 2x. i. Using 2 cm to represent 1 unit on each axis, construct accurately on graph paper, and clearly indicate by shading the unwanted regions, the region in which the set of points (x, y) must lie. ii. Using your graph, find the greatest value of (x − y) for points in the region. 4. On a sheet of graph paper, using a scale of 1 cm to represent 1 unit on both axes, draw the graphs of the functions and y = – x + 6 for values of x from – 4 to 6. 5. a. On a sheet of graph paper, using a scale of 1 cm to represent 1 unit on both axes, draw the graph of each of the following functions. i. y = x + 2 ii. y = x – 3 iii. y = 2 ******ebook converter DEMO Watermarks*******

iv. y = –3 b. Find the area enclosed by the four lines.

B2 Revision Exercise 1. The function f is defined by f : x ↦ 3x − k. Given that f(−2) = 14, find the value of k. 2. a. Sketch the graph of f : x ↦ 2x + 5 for the domain −1 ≤ x ≤ 4. b. Write down the range of f for the graph in (a). 3. a. Sketch the graph of f : x ↦ −3x − 2. b. Find the domain of f for the range −5 < f(x) ≤ 1. 4. Given the function h(x) = mx + c and that h(6) = −1 and h(−4) = −6, find the values of m and c. Hence evaluate h(3) and h(4). 5. Write the converse, inverse and contrapositive for each of the following statements. Explain if the resulting statement is true. a. The perpendicular bisector of a line segment passes through the midpoint of the line segment. b. If

, then a = 0.

6. Prove that if

, then x = −2. Give a reason for each step.

7. The figure shows ΔABD formed by two isosceles triangles, ΔABC and ΔACD. BCD is a straight line and ∠ABD = x°. Prove that ∠ADB = 2x°.

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8. It is given that

. Show the mathematical statement is

true using a. inductive reasoning, b. deductive reasoning.

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Chapter Eight Axiomatic Structure of a Mathematical System Euclidean geometry is a mathematical system attributed to the Alexandrian Greek mathematician Euclid, which was described in his book on geometry: the Elements. Euclid's Elements organized the geometry then known into a systematic presentation known as the axiomatic method that is still used in many texts. This axiomatic method has since been adopted not only throughout mathematics but in many other fields as well. New axiom and postulate systems were developed by various mathematicians.!

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LEARNING OBJECTIVES At the end of this chapter, you should be able to: describe a mathematical system, illustrate the need for an axiomatic structure of a mathematical system in general, and in Geometry in particular: (a) defined terms; (b) undefined terms; (c) postulates; and (d) theorems.

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8.1 The Axiomatic System In mathematics, an axiomatic system is any set of statements that are known to be true and some or all of these statements can be used to logically derive certain results. This type of mathematical reasoning consists of the following parts: undefined terms, defined terms, postulates (axioms) and theorems.

Undefined Terms and Defined Terms The time we started to learn mathematics, we also started to learn its vocabulary. We have learned about the definitions of terms such as integer, quadrilateral, polynomial, etc. The reasons that we need defined terms are: We need to be precise and concise in what we say or write; We need to understand each other and make sure that we mean the same thing when we say or write a particular word. A definition has two main parts: the term to be defined and the explanation of the term. Why is there a need for undefined terms? Undefined terms are terms that do not require a definition as these terms are used as a base to define other terms. They serve as building blocks of other mathematical terms, such as definitions, axioms and theorems. How do we know the meaning of a term if we do not define it? By describing some of its characteristics, we may be able to have a good idea of its meaning.

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The real number is considered as an undefined term. However, many have attempted to define it by describing its properties as far as possible. One such example is as follows: For any real numbers x and y, we say that x is less than y (denoted by x < y) if x lies to the left of y on the real number line; x is greater than y (denoted by x > y) if x lies to the right of y on the real number line. Can you think of other ways to define the 'real number'? Present your answers to the class.

Postulates Postulates (or axioms) are statements which are assumed to be true without proof. Thus an axiom is a mathematical statement that is believed to be clearly true that it needs not be proved. In general, we use 'postulate' in geometry and 'axiom' in other sections of mathematics. These statements are beginning assumptions from which logical consequences follow. Listed below are some of the axioms often used in Algebra and Geometry. The letters represent real numbers. Axioms for Real Numbers Commutative Axiom

a+b=b+a ab = ba

Associative Axiom

(a + b) + c = a + (b + c) (ab)c = a(bc)

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Distributive Axiom

a(b + c) = ab + ac

Reflexive Axiom

a=a

Symmetric Axiom

If a = b, then b = a.

Transitive Axiom

If a = b and b = c, then a = c. If a > b and b > c, then a > c.

Addition Axiom

If a = b, then a + c = b + c. If a > b, then a + c > b + c.

Multiplication Axiom

If a = b, then ac = bc. If a > b and c > 0, then ac > bc.

Existence of Additive Inverse

For every real number a, there exists a real number −a such that a + (−a) = (−a) + a = 0.

Existence of Multiplicative Inverse

For every non-zero real number a, there exists a real number such that .

Existence of Additive Identity

For every real number a, a + 0 = a.

Existence of Multiplicative For every real number a, a × 1 = 1 × a = a. Identity ******ebook converter DEMO Watermarks*******

Trichotomy Axiom

For every real numbers a and b, exactly one of the following is true: a = b, a > b or a < b.

Note that most of the axioms are true for all real numbers a, b and c. The exception is the Existence of Multiplicative Inverse where a is defined as a non-zero real number.

Theorems Recall that the axiomatic structure of a mathematical system starts with a set of definitions and axioms that are clearly stated. From these definitions and axioms, a new result is derived through reasoning. The axiomatic structure of a mathematical system follows this sequence:

In an axiomatic mathematical system, the set of axioms is the point from which theorems will be derived. Therefore, it is very important that the axioms on which theorems are based be true. A theorem is any mathematical statement that can be shown to be true using ******ebook converter DEMO Watermarks*******

accepted logical mathematical arguments. For example, defining a and b as real numbers and a > b and using the Addition Axiom, i.e. add −a − b to both sides of the inequality, we get −b > −a. Thus we can prove the theorem: If a and b are real numbers and a > b, then −a < −b.

A proof is a deductive argument that demonstrates the truth of a mathematical statement.

Worked Example

1

(Axioms for Real Numbers) Prove that if x and y are real numbers such that x + y = x, then y = 0. State any axiom used.

Solution: Statement x+y=x x+y−x=x−x y+x−x=0 y=0

Reason Given statement Addition axiom Commutative axiom of addition

Solve the equation 5x = 2x + 9. State any axiom used.

Exercise 8A Questions 1(a)-(f), 2-6 ******ebook converter DEMO Watermarks*******

The Evolution of the Axiomatic System Work in groups of 3-4. The foundation of the axiomatic structure of a mathematical system was laid by the ancient Greek mathematicians Thales, Pythagoras and Euclid. However, it was not until the nineteenth century that mathematical theory formalized with the work of Boole and Frege. By the early twentieth century, modern mathematics further developed with the work of Greg Cantor and G. H. Hardy. Today, mathematicians continue to develop new ideas of mathematics and at the same time extend and generalize existing areas of mathematics. Find out more about the development of the axiomatic system. You could read and discover on one or more of the mathematicians involved and their contributions to the system. Present your findings to the class.

BASIC LEVEL 1. State an axiom that justifies each of the following. a. 4 × 5 = 5 × 4 b. −(x + 1) = −x − 1 c. If −15 = x, then x = −15. d. If −x = 15, then x = −15. e. If x > a + b and a + b > c, then x > c. f. ******ebook converter DEMO Watermarks*******

INTERMEDIATE LEVEL 2. If x = 2, then = 2. a. Define x and y. b. State the axioms that justify the statement. 3. Antonio was solving the equation . His next step was 3x = 12 − 3x. a. Show how he arrived at the statement 3x = 12 − 3x and state the axiom that justifies the statement. b. Solve for x and state any axiom used. 4. Solve the inequality −4x < 24 and state any axiom or theorem used. 5. Prove that (2a + 3b)(x + 4y) = 2(ax + 6by) + 8ay + 3bx. State any axiom used. ADVANCED LEVEL 6. Amelia is preparing presents for Christmas. She has 6 types of items A, B, C, D, E and F. She is putting items into the presents according to the following axioms. Axiom 1:Each present is a set of three items. Axiom 2:Each type of item is in exactly two presents. Axiom 3:No two types of items may be together in more than one present. Axiom 4:There are exactly 4 presents. Draw a model for Amelia’s system of packing the presents.

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8.2 Axiomatic Structure of Geometry Structure We have learned Geometry in school and solved many problems on Geometry. Now let us study the axiomatic structure of Geometry. Recall that an axiomatic system has four parts: undefined terms, defined terms, postulates (axioms) and theorems.

Undefined and Defined Terms in Geometry In our years of studying Geometry, we have defined many terms such as angle, polygon, prism, etc. For example, we define polygon as a closed plane figure with three or more line segments as its sides. Notice that from the definition itself, we must also define the other terms used. What do we mean by plane figure and line segment? Suppose we define line segment as a part of a line that is bounded by two endpoints. Then we have to define the words line and endpoint. If we define these terms, we will have to use other terms which means that we also need to define them. This will go on forever and we have to stop somewhere. Thus we have the following undefined terms that do not require a definition: points, lines and planes. These terms are said to be the building blocks of geometry.

Most people visualize a point as a tiny dot, a line as a long and seamless linkage of points and a plane as smooth, endless and flat interwoven lines. However, these are not definitions of a point, line and plane.

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Postulates and Theorems involving Points, Lines and Planes The following postulates about points, lines and planes give us the relationship between points and lines, points and planes, lines and planes, and distance and location. P1. P2. P3. P4.

P5. P6. P7. P8. P9. P10.

Given any two distinct points, there is exactly one line that contains them. Every line contains at least two distinct points. (The Distance Postulate) To every pair of distinct points there corresponds a unique positive number. This number is called the distance between the two points. (The Ruler Postulate) The points of a line can be placed in a correspondence with the real numbers such that To every point of the line there corresponds exactly one real (a) number. To every real number there corresponds exactly one point of (b) the line. The distance between two distinct points is the absolute value (c) of the difference of the corresponding real numbers. (The Ruler Placement Postulate) Given two points P and Q of a line, the coordinate system can be chosen in such a way that the coordinate of P is zero and the coordinate of Q is positive. (Segment Addition Postulate) If B is between A and C, then AB + BC = AC. (a) Every plane contains at least three non-collinear points. (b) Space contains at least four non-coplanar points. If two points lie in a plane, then the line containing these points lies in the same plane. Any three points lie in at least one plane, and any three noncollinear points lie in exactly one plane. If two planes intersect, then their intersection is a line.

From the above postulates, we can also draw some theorems, without proof, about points, lines and planes. ******ebook converter DEMO Watermarks*******

T1. T2. T3.

If two different lines intersect, then they intersect at exactly one point. If a point lies outside a line, then exactly one plane contains both the line and the point. If two distinct lines intersect, then exactly one plane contains both lines.

Segment Addition Postulate Case 1 a. Draw a line segment with endpoints A and C. b. Place a point somewhere between A and C and label this point B. c. Measure AB, BC and AC. d. Write a mathematical statement relating AB, BC and AC. Case 2 e. Draw a line segment with endpoints A and C again. f. Extend the line segment AC and label the new endpoint B. g. Measure AB, BC and AC. h. Does the mathematical statement you wrote in Case 1 hold in Case 2? Explain why.

Worked Example

2

(Postulates and Theorems in Geometry)

a. State the postulate or theorem you would use to justify the statement made about each figure. ******ebook converter DEMO Watermarks*******

i. One plane contains points A, B and C. ii.

Lines AB and CD intersect at point E. b. Which postulate tells us that AB is not a line? Explain why.

Solution: a. i. Any three points lie in at least one plane, and any three non-collinear points lie in exactly one plane. (P9) ii. If two different lines intersect, then they intersect at exactly one point. (T1) b. Given any two distinct points, there is exactly one line that contains them. (P1) According to the postulate, there is exactly one line that contains two given points. Hence the line must be straight. AB is a curve and there can be many of the curve AB. Therefore, AB is not a line.

a. State the postulate or theorem you would use to justify the statement made about each figure. i. ******ebook converter DEMO Watermarks*******

Line KL lies in plane P. ii. Only one line contains points P and Q. b. Explain whether the statement “If A, B and C are three points in a plane, then they are collinear” is true or false.

Exercise 8B Questions 1(a)-(d), 2, 3, 4(a)-(e), 5-10

BASIC LEVEL 1. State the postulate or theorem you would use to justify the statement made about each figure. a.

Besides point A, there is another point on line l. b.

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Plane G and plane H intersect along line l. c.

One plane contains lines l1 and l2. d. One plane contains point B and line AC. 2. Points A, B and C are non-collinear.

a. Name all the different lines that can be drawn through these points. b. Write a statement about the intersection of AC and CB. State the postulate or theorem you would use to justify the statement. 3. Planes M and N are two distinct planes and DE is a line.

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a. Write a statement about the figure shown. b. State the postulate or theorem you would use to justify the statement that you have made in (a).

INTERMEDIATE LEVEL 4. Explain whether each of the following statements is true or false using postulates or theorems. a. Two planes can intersect at only one point. b. If A, B and C are three points on a line, then they are collinear. c. It is possible that lines AB and AC intersect at points A and D. d. Exactly one line passes through points A and B. e. A plane contains at least one line and one point not on that line. 5. Points P and Q on a number line have coordinates −5.4 and 2.3. Find the distance PQ. 6. Collinear points P, Q and R have coordinates −7,

and 3.

a. Which point is between the other two? b. Write a mathematical statement relating PQ, QR and PR. 7. Points P, Q, R and S are collinear. If PS = 6PQ, QR = 2PQ and RS = 18 cm, find PS. State the postulate or theorem you used to find the answer.

ADVANCED LEVEL ******ebook converter DEMO Watermarks*******

8. Draw a figure to represent each of the following statement. a. Three coplanar lines that do not intersect. b. Two non-intersecting planes that intersect another plane. c. Three planes that intersect at exactly one point. 9. Lines l1 and l2 are two different lines. Point P lies on l1 and on l2. Point Q lies on l1 and on l2. What can you say about points P and Q? Explain your answer. 10. How many lines can be determined by a. one given point, b. two given points, c. three given points, d. four given points?

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1. The axiomatic structure of a mathematical system is any set of statements that are known to be true and some or all of these statements can be used to logically derive certain results. It consists of the following parts: undefined terms, defined terms, postulates (axioms) and theorems. 2. A postulate or axiom is a mathematical statement that is taken to be selfevidently true without proof. 3. A theorem is any mathematical statement that can be shown to be true using accepted logical mathematical arguments. 4. Points, lines and planes are undefined terms in Geometry. These terms are said to be the building blocks of geometry. 5. Postulates and theorems about points, lines and planes can be formed to give the relationship between points and lines, points and planes, lines and planes, and distance and location.

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1. State an axiom that justifies each of the following. a. 2 − 10 = −10 + 2 b. (x + 7) − 1 = x + (7 − 1) c. If AB = CD and CD = 5 cm, then AB = 5 cm. d. If x > 15, then x + 5 > 20. e. If x = y, then

.

f. −9 + 0 = −9 2. Solve the inequality 8x − 12x > 20 and state any axiom or theorem used. 3. To prove that

, Antonio wrote the following steps:

State any axiom or theorem that he used. 4. Explain whether each of the following statements is true or false using postulates or theorems. a. On line AB, there are only two points A and B. b. The intersection of a line and a plane may be the line itself. c. On a plane, there are at least three collinear points. ******ebook converter DEMO Watermarks*******

d. A plane has at least one line. e. If points A, B and C are non-coplanar, then no plane contains all three of them. f. Two planes can intersect at a line and a point not on the line. 5. PQ and RS are perpendicular lines. a. Are points P, Q, R and S collinear? b. How many different lines can the points form? c. What is the resulting shape formed? Sketch the shape and label it with P, Q, R and S. d. Is it true that the shape lies in at least one plane? Explain using postulates or theorems. 6. The figure shows a rectangular box ABCDEFGH.

a. Are there any collinear points in the figure? Why? b. Name all the sets of coplanar lines. c. CDHG is a plane. What do you observe about points C, D, H and G? State the postulate or theorem in relation to your observation.

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7. The figure shows a pyramid ABCDE.

a. Name all the planes and write statements about the planes and the lines of intersection (if any) shown in the figure. b. State the postulate or theorem you would use to justify the statements that you have made in (a). c. Write a statement about the intersection of the planes and vertex A. What can you say when more than two planes intersect? 8. Point S lies at coordinate −3 and point T lies at coordinate −7 on the number line. Angelo says that ST = −4 units. a. Draw points S and T on the number line. b. Is Angelo right? Explain using postulates or theorems. 9. Points A, B and C lie on the x-axis. A and C have coordinates −2 and 4.8. If B lies between A and C and , a. find the coordinate of B, b. state the postulate or theorem you used to find the answer. 10. Point E has coordinate 1.5. If EF is 2 units, find the possible coordinates of F. 11. Given that AC > AD > AB, AB = 2CD, AC = 8.4 units and CD = 1.2 units, a. draw points A, B, C and D on the number line, b. find BD. c. If the coordinate of A is 0, ******ebook converter DEMO Watermarks*******

i. find the coordinates of B, C and D, ii. state the postulate or theorem in relation to your answer in (c)(i).

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Line l1 intersects planes P1 and P2, but does not lie in P1. Line l2 lies in P1 but does not contain any point on P2. Is it possible for l1 to intersect l2? Explain.

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Chapter Nine Triangle Congruence The photograph shows shophouses which have the same shape and size. This an example of congruence. What are some other examples of congruence that you can find around you?

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LEARNING OBJECTIVES At the end of this chapter, you should be able to: illustrate triangle congruence, illustrate the SAS, AAS and SSS congruence postulates, solve corresponding parts of congruent triangles, prove two triangles are congruent, prove statements on triangle congruence, apply triangle congruence to construct perpendicular lines and angle bisectors.

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9.1 Congruent Figures

Properties of Congruent Figures Fig. 9.1 shows five pairs of scissors.

Fig. 9.1 1. What can we say about the shape, size, orientation and position of the pairs of scissors? 2. If we cut out the pairs of scissors and stack them up, what will we observe? 3. The pair of scissors in (a) can be moved to look like the pair of scissors in (b) by a translation from A1 → A2 and a rotation of 90º about A2. How can we move the pair of scissors in (a) to look like the pairs of scissors in (c), (d) and (e)? From the investigation, we observe that:

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Two figures are congruent if they have exactly the same shape and size. They can be mapped onto each other under translation, rotation and reflection.

You can also investigate the effect of translation, rotation and reflection on a triangle/quadrilateral using the geometry software template ‘Congruence’ at http://www.shinglee.com.sg/StudentResources/

Fig. 9.2 shows two pairs of scissors of different colors.

Fig. 9.2 Are they congruent? Explain your answer. Congruence is a property of geometrical figures. The two pairs of scissors in Fig. 9.2 are congruent because they have exactly the same shape and size. Fig. 9.3 shows some patterns that are formed by congruent figures. These are known as tessellations, which can be found in many real-life objects.

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Fig. 9.3 Another real-life application of congruence is photocopying as the photocopied document is of the same shape and size as the original document. The concept of congruence also plays an important role in the manufacturing sector. The congruence of pen refills allows us to refill our pens when they run dry.

Congruence in the Real World 1. Look around your classroom or school. Find at least 3 different sets of congruent objects. 2. Tessellations, like those shown in Fig. 9.3, can be found on floor tiles. What are some other objects that exhibit tessellations? 3. Discuss with your classmates other real-life applications of congruence.

Search on the Internet for ‘Tessellation Tool’ to make your own tessellations.

Worked Example

1

(Identifying Congruent Shapes) Which shapes are congruent?

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Solution: A, B, C and D are congruent rectangles. F, H and I are congruent triangles.

Which shapes are congruent?

Fig. 9.4 shows two congruent quadrilaterals ABCD and A’B’C’D’. The vertex A corresponds to the vertex A’ because they have the same angle. Similarly, the vertices that correspond to B, C and D are B’, C’ and D’ respectively. ******ebook converter DEMO Watermarks*******

Fig. 9.4 The symbol ‘≡’ means ‘is congruent to’. Thus for the two quadrilaterals in Fig. 9.4, we have ABCD ≡ A’B’C’D’. Notice that the order in which the vertices of A’B’C’D’ are written must correspond to the order in which the vertices of ABCD are written.

We can also write BCDA ≡ B’C’D’A’ because the corresponding vertices match. Can we write CDAB ≡ C’D’A’B’ or DABC ≡ D’A’B’C’? In Fig. 9.4, the side AD corresponds to the side A’D’. Similarly, the sides that correspond to AB, BC and CD are A’B’, B’C’ and C’D’ respectively. Hence, the corresponding angles and the corresponding sides of congruent figures are equal.

Exercise 9A Question 1

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Worked Example

2

(Problem involving Congruent Figures)

Given that ABCD ≡ WXYZ, copy and complete each of the following. i. ∠ABC = ∠WXY = ________˚ ii. ________ = ∠XYZ = ________˚ iii. AD = ________ = ________ cm iv. ________ = WX = ________ cm

Solution: Since ABCD ≡ WXYZ, then the corresponding vertices match: A↔W B↔X C↔Y D↔Z i. ∠ABC = ∠WXY = 80˚ ii. ∠XYZ = ∠BCD = 60˚ (X ↔ B, Y ↔ C, Z ↔ D) iii. AD = WZ = 2 cm (A ↔ W, D ↔ Z) iv. WX = AB = 4 cm (W ↔ A, X ↔ B)

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Given that ABCD ≡ PQRS, copy and complete each of the following. i. PQ = AB = ________ cm ii. SR = ________ = 6 cm iii. PS = ________ = ________ cm iv. QR = ________ = ________ cm v. ∠PQR = ________ = ________˚

Exercise 9A Questions 2-3

Worked Example

3

(Identifying Congruent Triangles and Writing Statement of Congruence) Are the following pairs of triangles congruent? If so, explain your answer and write down the statement of congruence. If not, explain your answer.

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a.

b.

Solution: a. Step 1: Identify the corresponding vertices by comparing the size of the angles. A ↔ P (since ∠A = ∠P = 60˚) B ↔ Q (since ∠B = ∠Q = 70˚) C ↔ R (since ∠C = ∠R = 50˚) Step 2: Write proper statements using the corresponding vertices identified in Step 1. ∠BAC = ∠QPR = 60˚ (notice that the corresponding vertices match) ∠ABC = ∠PQR = 70˚ ∠ACB = ∠PRQ = 50˚ AB = PQ = 3 cm (notice that the corresponding vertices match) BC = QR = 3.39 cm AC = PR = 3.68 cm ******ebook converter DEMO Watermarks*******

∴ The two triangles have the same shape and size and so ∆ABC ≡ ∆PQR. b. In ∆ STU, ∠T = ∠U = 80˚ (base ∠s of isos.ΔSTU) ∠S = 180˚ – 80˚ – 80˚ (∠ sum of ΔSTU) = 20˚ ∴ ∆STU does not have any right angle that corresponds to that in ∆DEF. ∴ ∆STU does not have the same shape as ∆DEF and so it is not congruent to ∆DEF.

Are the following pairs of triangles congruent? If so, explain your answer and write down the statement of congruence. If not, explain your answer.

a.

b.

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c.

Exercise 9A Questions 4(a)-(c)

Worked Example

4

(Problem involving Congruent Triangles) In the figure, ∆ABC ≡ ∆CED. a. Given that ∠BAC = 20˚, ∠CDE = 60˚, AB = 8.8 cm and CD = 10 cm, calculate a. ∠ECD, b. ∠ECB, c. ∠ABC, d. the length of AC, e. the length of AE. b. What can we say about the lines AB and DC?

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Solution: Since ∆ABC ≡ ∆CED, then the corresponding vertices match: A↔C B↔E C↔D a. i. ii.

iii. iv. v. Length of ∴ Length of

b. Since ∠BAC = ∠ECD (= 20˚), then AB // DC (converse of alt. ∠s).

When two lines AB and CD are cut by a transversal PQ, and ******ebook converter DEMO Watermarks*******

if ∠a = ∠b, then AB // CD (converse of corr. ∠s); if ∠a = ∠c, then AB // CD (converse of alt. ∠s); if ∠a + ∠d = 180˚, then AB // CD (converse of int. ∠s).

In the figure, ∆ABC ≡ ∆CDE.

a. Given that ∠ABC = 38˚, ∠DCE = 114˚, AC = 18 cm and DE = 27 cm, find i. ∠CDE, ii. ∠CED, iii. ∠ACB, iv. the length of BC, v. the length of BE. b. What can we say about the lines AC and ED?

Exercise 9A Questions 5-7

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BASIC LEVEL 1. Which

pairs

of

shapes

2.

Given that PQRST ≡ VWXYZ, copy and complete ******ebook converter DEMO Watermarks*******

are

congruent?

each of the following. i. PQ = VW = ________ cm ii. PT = ________ = 2 cm iii. QR = ________ = ________ cm iv. TS = ________ = ________ cm v. SR = ________ = ________ cm vi. ∠PQR = ________ = ________˚ INTERMEDIATE LEVEL 3. Given that EFGH ≡ LMNO, write down all the missing measurements.

4. Are the following pairs of triangles congruent? If so, explain your answer and write down the statement of congruence. If not, explain your answer.

a.

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b.

c.

5. In the figure, ∆ABK ≡ ∆ACK. Given that ∠AKB = 90˚, ∠ACK = 62˚, AB = 17 cm and BK = 8 cm, find i. ∠BAC,

ii. the length of BC.

6. In the figure, ∆ABC ≡ ∆DEC. Given that ∠ACB = 71˚, ∠CDE = 34˚, AC = 6.9 cm and BC = 4 cm, find i. ∠ABC,

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ii. the length of BD.

ADVANCED LEVEL 7. In the figure, ∆ABC is an isosceles triangle where AB = AC, BC = 12 cm and ∠ABK = 58˚. Given that ∆ABK ≡ ∆ACH and ∠AKC = 90˚, find i. the length of CH,

ii. ∠BAH.

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9.2 Congruence Tests Recap In Section 9.1, we have learned that congruent figures have exactly the same shape and size; and they can be mapped onto one another under translation, rotation and reflection. For congruent triangles, this would mean that all the corresponding lengths are equal and all the corresponding angles are equal. For example, ∆ABC is congruent to ∆XYZ (and we write ∆ABC ≡ ∆XYZ) if and only if AB = XY BC = YZ AC = XZ BÂC = YX̂Z AB̂C = XŶZ AĈB = XẐY

Fig. 9.5 Consider ∆EFG, where EF = AB FG = BC ******ebook converter DEMO Watermarks*******

EG = AC FÊG = BÂC EF̂G = AB̂C EĜF = AĈB

Fig. 9.6 ∆EFG is still congruent to ∆ABC even though ∆EFG is laterally inverted, because ∆EFG can be mapped onto ∆ABC by a reflection (and a translation if necessary). In this section, we will investigate whether we need all the 6 conditions (i.e. 3 pairs of corresponding sides are equal and 3 pairs of corresponding angles are equal) to prove that two triangles are congruent.

The vertices of the 2 triangles must match: A↔X B↔Y C↔Z

AB̂C can also be written as ∠ABC. Similarly, XŶZ can be written as ∠XYZ.

SSS Congruence Test

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SSS Congruence Test 1. Cut 3 sticks so that the lengths of the sticks are 5 cm, 8 cm and 10 cm. 2. Try to form a triangle using the 3 sticks in as many ways as possible, such that the lengths of the triangle correspond to the lengths of the 3 sticks. (If you do not have sticks, you can try to construct the triangle using the three given lengths in as many ways as possible.) 3. Do you get the following triangle? If you get a different triangle, flip it over and see if it fits onto this triangle. Is it possible to get other triangles?

Fig. 9.7 4. Try using 3 other sticks of different lengths, such that the sum of the lengths of the two shorter sticks is greater than the length of the longest stick, and see if you always get a unique triangle (regardless of reflection) no matter how you try to form a triangle. 5. What can you conclude from this investigation?

If the sum of the lengths of the two shorter sticks is less than or equal to the length of the longest stick, then you cannot form a triangle. You will learn ******ebook converter DEMO Watermarks*******

more about this in Chapter 10. From the investigation, we observe the following: SSS Congruence Test: If the 3 sides of a triangle are equal to the 3 corresponding sides of another triangle, then the two triangles are congruent.

Worked Example

5

(Proving that Two Triangles are Congruent using the SSS Congruence Test) Prove that the following two triangles are congruent.

Solution: A↔X B↔Z C↔Y AB = XZ = 4 cm BC = ZY = 9 cm AC = XY = 7 cm ∴ ∆ABC ≡ ∆XZY (SSS)

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How to match the vertices correctly: Step 1: Match the vertex opposite the longest side for both triangles, i.e. A ↔ X. Step 2: Match the vertex opposite the shortest side for both triangles, i.e. C ↔ Y. Step 3: Match the last vertex, i.e. B ↔ Z. In writing the proof, all the vertices must match.

When ∆ABC is congruent to ∆XZY, we can denote the relationship as follows: ∆ABC ≡ ∆XZY.

Exercise 9B Questions 1(a), 2(a), 3(a), 4(a),(b)

1. Copy and complete the proof to show that the following two triangles are

congruent.

Solution: A ↔ ___ B ↔ ___ ******ebook converter DEMO Watermarks*******

C ↔ ___ AB = _____ = 5 m BC = _____ = ___ m AC = _____ (given) ∴ ∆ABC ≡ ∆_____ (_____) 2. The diagram shows a kite WXYZ.

Identify the two congruent triangles in the kite and prove that they are congruent.

Even if two triangles are not congruent, we can still match the corresponding vertices because we must compare the longest side of one triangle with the longest side of the other triangle, etc.

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SAS Congruence Test

SAS Congruence Test Part 1 1. Try to construct ∆XYZ such that XY = 3 cm, YZ = 6 cm and XŶZ = 50° in as many ways as possible. Compare the triangle you have drawn with those drawn by your classmates. 2. Do you get the following triangles (not drawn to scale)? Both triangles are actually congruent to each other: you can map both triangles together by a reflection. Is it possible to get other triangles?

Fig. 9.8 3. Try to construct ∆XYZ for other dimensions, where XY and YZ have a fixed length and XŶZ is a fixed angle, in as many ways as possible and see if you always get a unique triangle (regardless of reflection). 4. Notice that the given XŶZ is between the two given sides XY and YZ: XŶZ is called the included angle. 5. What can you conclude from part 1 of this investigation? Part 2 ******ebook converter DEMO Watermarks*******

6. Try to construct ∆ABC such that AB = 5 cm, AC = 3 cm and AB̂C = 30° in as many ways as possible. 7. Fig. 9.9 shows one possible ∆ABC that satisfies the given dimensions.

Fig. 9.9 Is it possible to construct a different ∆ABC (excluding a laterally inverted triangle)? 8. Notice that the given AB̂C is not in between the two given sides AB and AC, i.e. AB̂C is not the included angle. 9. What can you conclude from part 2 of this investigation? From the investigation, we observe the following: SAS Congruence Test: If 2 sides and the included angle of a triangle are equal to the 2 corresponding sides and the corresponding included angle of another triangle, then the two triangles are congruent.

If the given angle is not the included angle, then SSA may not be a congruence test. For example, in the above investigation, there are two ways to construct ∆ABC as shown in Fig. 9.10.

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Fig. 9.10 Since the two triangles are not congruent, then SSA is not a congruence test in general, although there are exceptions (see RHS Congruence Test later in this section).

Worked Example

6

(Proving that Two Triangles are Congruent using the SAS Congruence Test) Copy and complete the proof to show that the following two triangles are congruent.

Solution: P ↔ ___ Q ↔ ___ R ↔ ___ PQ = _____ = 9 mm QP̂R = _____ = ___° PR = _____ = ___ mm ∴ ΔPQR ≡ Δ_____ (SAS)

It is easier to match the vertices with the given angle first. Then match the vertex opposite either the 9 mm or 12 mm side for both triangles. ******ebook converter DEMO Watermarks*******

1. Prove that the following two triangles, where PQ̂R = SP̂T, are congruent.

2. Determine

whether

the

following

triangles

are

congruent.

Exercise 9B Questions 1(b), 2(b), 3(b), 4(c),(d)

Worked Example

7

(Proving that Two Triangles are Congruent) In the diagram, AOC and BOD are straight lines, OA = OC, OB = OD and AB = 7 cm.

i. Prove that ∆AOB is congruent to ∆COD. ii. Find the length of CD. ******ebook converter DEMO Watermarks*******

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Solution: i. A ↔ C O↔O B↔D OA = OC (given) AÔB = CÔD (vert. opp. ∠s) OB = OD (given) ∴ ∆AOB ≡ ΔCOD (SAS) ii. Since ∆AOB ≡ ΔCOD, then all the corresponding sides are equal. ∴ CD = AB = 7cm

Match O first. Then from OA = OC, match A and C.

1. In the diagram, ABCD is a rectangle, the two diagonals AC and BD intersect at O and CÂB = 25°.

i. Prove that ∆AOB is congruent to ∆COD. ii. Find BD̂C. 2. In the diagram, PQ is equal and parallel to SR, PS = 7 cm and QR̂S = 140°.

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i. Identify two congruent triangles and prove that they are congruent. ii. Find the length of QR and QP̂S.

Exercise 9B Questions 5, 6, 9

The diagonals of a rectangle bisect each other.

AAS Congruence Test

AAS Congruence Test Part 1 1. Try to construct ∆PQR such that PQ = 7 cm, QP̂R = 40° and PQ̂R = 60° in as many ways as possible. Compare the triangles you have drawn with those drawn by your classmates.

Fig. 9.11 ******ebook converter DEMO Watermarks*******

2. Do you always get a unique triangle (regardless of reflection) as shown above (not drawn to scale)? 3. What can you conclude from part 1 of this investigation? Part 2 4. Try to construct ∆ABC such that AB = 6 cm, BÂC = 50° and AĈB = 70° in as many ways as possible.

Fig. 9.12 5. Do you always get a unique triangle (regardless of reflection)? 6. What can you conclude from part 2 of this investigation? 7. Does it matter whether the given side of the triangle is between the 2 given angles (as in part 1) or if it is not between the 2 given angles (as in part 2)? Explain your answer. Hint: See Problem Solving Tip for Question 4 in this investigation.

An easier way to construct the vertex C is to find AB̂C first, and then construct AB̂C. From the investigation, we observe the following: AAS Congruence Test: If 2 angles and 1 side of a triangle are equal to the 2 corresponding angles and the corresponding side of another triangle, then the two triangles are congruent. ******ebook converter DEMO Watermarks*******

Since it does not matter whether or not the given side is between the two given angles, it does not matter whether we write AAS Congruence Test or ASA Congruence Test (unlike SAS Congruence Test where the given angle must be the included angle).

Worked Example

8

(Proving that Two Triangles are Congruent using the AAS Congruence Test) Copy and complete the proof to show that the following two triangles are congruent.

Solution: In ∆DEF, A ↔ ___ B ↔ ___ C ↔ ___ AB̂C = _____ = 80° BÂC = _____ = ___° BC = _____ = ___ mm ∴ ∆ABC ≡ Δ_____ (AAS)

Sometimes there is a need to find the other unknown angle in either triangle. ******ebook converter DEMO Watermarks*******

In each part, prove that the following two triangles are congruent. a.

b.

Exercise 9B Questions 1(c), 2(c), 3(c), 4(e),(f)

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RHS Congruence Test

RHS Congruence Test 1. Try to construct ∆DEF such that DÊF = 90°, DE = 3 cm and DF = 5 cm in as many ways as possible. Compare the triangles you have drawn with those drawn by your classmates.

Fig. 9.13 2. Do you always get a unique triangle (regardless of reflection) as shown above (not drawn to scale)? 3. What can you conclude from the above investigation? 4. Do you notice that the above congruence test is a special case of SSA where the given angle is a right angle? From the investigation, we observe the following: RHS Congruence Test: If the hypotenuse (H) and 1 side (S) of a right-angled (R) triangle are equal to the hypotenuse and 1 side of another right-angled triangle, then the two right-angled triangles are congruent.

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In general, SSA is not a congruence test, but there are exceptions, one of which is the RHS Congruence Test.

The 4 congruence tests covered so far are not the only congruence tests. There are more which are not included in the syllabus. An example is SSA, which is a congruence test if the given angle is obtuse.

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Worked Example

9

(Proving that Two Triangles are Congruent using the RHS Congruence Test) Copy and complete the proof to show that the following two triangles are congruent.

Solution: P ↔ ___ Q ↔ ___ R ↔ ___ PQ̂R = _____ = ___° PR = _____ = ___ cm QR = _____ = ___ cm ∴ ΔPQR ≡ Δ_____ (RHS)

a.

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b.

Exercise 9B Questions 1(d), 2(d), 3(d), 4(g),(h)

Consolidation for Congruence Tests Work in pairs. In each diagram, identify a pair of congruent triangles and prove that they are congruent. a.

b.

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c.

d.

e.

f.

g.

h.

Exercise 9A Questions 7, 8 ******ebook converter DEMO Watermarks*******

Exercise 9B Questions 7, 8

BASIC LEVEL 1. Identify a pair of congruent triangles from the following triangles (not drawn to scale), based on each of the following congruence tests: a. SSS Congruence Test, b. SAS Congruence Test, c. AAS Congruence Test, d. RHS Congruence Test. i.

2.

3.

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4.

5.

6.

7.

8.

9.

2. Copy and complete the proof to show that each of the following pairs of triangles are congruent.

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a.

A ↔ ___ B ↔ ___ C ↔ ___ AB = _____ (given) BC = _____ = 8 cm AC = _____ = ___ cm ∴ ∆ABC ≡ Δ_____ (_____) b.

D ↔ ___ E ↔ ___ F ↔ ___ DE = _____ = 3 m DÊF = _____ = 70° EF = _____ = ___ m ******ebook converter DEMO Watermarks*******

∴ ∆DEF ≡ Δ_____ (_____) c.

L ↔ ___ M ↔ ___ N ↔ ___ LM̂N = _____ = 30° LN̂M = _____ = ___° MN = _____ = ___ cm ∴ Δ_____ ≡ ΔUVW (_____) d.

G ↔ ___ H ↔ ___ I ↔ ___ ******ebook converter DEMO Watermarks*******

GĤI = _____ = ___° GI = _____ = 13 mm HI = _____ = ___ mm ∴ Δ_____ ≡ ΔSTU (_____) 3. Determine whether each of the following pairs of triangles are congruent. a.

b.

c.

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d.

INTERMEDIATE LEVEL 4. In each diagram, identify a pair of congruent triangles and prove that they are congruent. a.

b.

c.

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d.

e.

f.

g.

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h.

5. In the diagram, RTV and STU are straight lines, RT = VT, ST = UT, RS = 4

cm and SR̂T = 80°.

i. Prove that ∆RST is congruent to ∆VUT. ii. Find the length of UV. iii. Find UV̂T. iv. Hence, other than RS = UV, what can you conclude about the lines RS and UV? 6. In the diagram, GH = JI and GI = JH.

i. Identify a triangle that is congruent to ∆GHI and prove that they are congruent. ii. If HĴI = 60° and GÎH = 40°, find GĤI. ADVANCED LEVEL ******ebook converter DEMO Watermarks*******

7. In each diagram, identify a pair of congruent triangles and prove that they are congruent. a.

b.

c.

d.

e.

f.

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8. The diagram shows a parallelogram ABCD.

Use three different ways to prove that ∆ABC and ∆CDA are congruent. 9. In

the

quadrilateral

JADE,

AO

=

JO

Explain why ∆AJD is congruent to ∆JAE.

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and

EO

=

DO.

9.3 Applications of Congruent Triangles In this section, we will apply the concepts of congruent triangles to solve problems in mathematics and in real life.

Worked Example

10

(Application of Congruent Triangles) In Grade 7, we have learned how to construct the bisector of a given angle as shown in the diagram. Prove that OT is the angle bisector of PÔQ.

Draw arcs OR and OS, where OR = OS. Draw arcs RT and ST, where RT = ST. Prove that OT is the angle bisector of PÔQ, i.e. ∠a = ∠b.

Solution: R↔S ******ebook converter DEMO Watermarks*******

O↔O T↔T OR = OS RT = ST OT = OT (common side) ∴ ΔROT and ΔSOT are congruent (SSS Congruence Test). ∴ a = b, i.e. OT is the angle bisector of PÔQ.

In Grade 7, we have learned how to construct the perpendicular bisector of a given line segment as shown in the diagram. Prove that PQ is the perpendicular bisector of AB.

Exercise 9C Questions 1-5

Draw arcs AP, AQ, BP and BQ, where AP = AQ = BP = BQ. Prove that PQ is the perpendicular bisector of AB.

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BASIC LEVEL 1. The diagram illustrates how the length AB (which cannot be measured directly) of a pond is measured. Choose a point C and measure the length of AC and the length of BC. Produce AC and BC to A’ and B’ respectively, so that CA’ = AC and CB’ = BC. By measuring the length of B’A’, we will be able to find the length of AB. Why is this so?

2. To measure the width of the internal trough, AB, of a machine tool which cannot be measured directly, we make use of a device as shown in the diagram. The device is made up of two parts, AA’ and BB’, hinged halfway at O. By measuring the distance between A’ and B’, we will be able to obtain the length of AB. Why is this so?

INTERMEDIATE LEVEL ******ebook converter DEMO Watermarks*******

3. In the diagram, P lies on OA and Q lies on OB such that OP = OQ. Place a set square with one side along PQ and another side passing through O, as shown in the diagram. Explain why OC is the angle bisector of AÔB.

4. The diagram shows Antonio standing at a point A along a river bank. He looks directly across to the opposite bank, adjusting his cap so that his line of vision CB passes through the lowest point at the rim of his cap and falls on the point B. He then turns around without moving his head. His new line of vision CB’ through the lowest point at the rim of his cap now falls on a point B’ on the same side of the river. State which measurement he can make in order to find the width AB of the river. Explain your answer.

ADVANCED LEVEL 5. Using a set square, we can bisect a given angle. In the diagram, P and Q are marked along the arms, OA and OB of AÔB respectively, such that OP = OQ. Move a 90°-45°-45° set square away from O until the 45° edges ******ebook converter DEMO Watermarks*******

coincide with P and Q as shown in the diagram. Explain why OM is the angle bisector of AÔB.

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1. Two figures are congruent if they have exactly the same shape and size. They can be mapped onto each other under translation, rotation and reflection. 2. A figure and its image under a translation, a rotation or a reflection are congruent i.e. all corresponding angles and corresponding sides are equal. 3. The 4 congruence tests covered in this chapter are SSS Congruence Test, SAS Congruence Test, AAS Congruence Test and RHS Congruence Test. SSS Congruence Test

SAS Congruence Test

3 corresponding sides are equal, i.e. AB = XY, BC = YZ, AC = XZ

2 corresponding sides and the included angle are equal, i.e. AB = XY, BC = YZ, AB̂C = XŶZ

AAS Congruence Test

RHS Congruence Test

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1 corresponding side and 2 corresponding angles are equal, i.e. BC = YZ, AB̂C = XŶZ, AĈB = XẐY

1 side and the hypotenuse of the right-angled triangle are equal, i.e. BC = YZ, AC = XZ, AB̂C = XŶZ = 90°

4. In general, SSA is not a congruence test. An exception is the RHS congruence test.

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1. Are the following pairs of triangles congruent? If so, explain your answer and write down the statement of congruence. If not, explain your answer. a.

b.

2. It is given that the quadrilateral ABCD is congruent to the quadrilateral PQRS, ∠A = 100°, ∠B = 70°, ∠C = 95° and PQ = 6 cm. i. Write down the length of AB. ii. Find ∠S. 3. In the figure, ∆ABC ≡ ∆AKH. Given that ∠BAC = 52°, ∠AHK = 36°, AB = 6 cm and AH = 10.2 cm, find i. ∠AKH, ii. the length of KC. ******ebook converter DEMO Watermarks*******

4. In the figure, ∆ABC ≡ ∆AHK. It is given that ∠BAC = 53°, ∠AHK = 90°, AH = 6 cm and the area of ∆ABC is 24 cm2. i. Write down the size of ∠ABC. ii. Write down the length of AB. Hence, find the length of BC. iii. Calculate ∠BXK.

5. Determine whether each of the following pairs of triangles are congruent. If they are congruent, state the congruence test. a.

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b.

c.

d.

6. Identify a pair of congruent triangles and state the congruence test.

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7. Which of the following pairs of triangles are congruent? If they are congruent, state the congruence test and name the other three pairs of equal measurements. a.

b.

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c.

d.

e.

f.

8. In the diagram, AOB and COD are straight lines, AO = BO and CO = DO.

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i. Identify a pair of congruent triangles and state the congruence test. ii. Write down two pairs of equal angles. 9. In the diagram, PQ is equal and parallel to RS, PR = 5 cm and QŜR = 50°.

i. Identify a pair of congruent triangles and state the congruence test. ii. Find the length of QS and QP̂R. 10. In the diagram, P and Q are points along the arms OA and OB of AÔB respectively such that OP = OQ. A set square is used to construct perpendiculars to OA and OB at P and Q respectively. The perpendiculars meet at C. Explain why OC is the angle bisector of AÔB.

11. In the diagram, STU, RTP and RUQ are straight lines, SU is parallel to PQ, RP̂Q = 90°, SR = SP = 9 cm, TU = 5 cm and RU = 7 cm.

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i. Identify two triangles which are congruent. ii. Find the length of UQ and of PQ. 12. In ∆CAT, M is the midpoint of CT, CÂN = PÂN and CP is a straight line

that is perpendicular to NA.

i. Explain why ∆CAN is congruent to ∆PAN. ii. Hence, or otherwise, explain why MTAN is a trapezium.

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The diagram below shows 4 equilateral triangles formed by using 9 toothpicks. By removing 3 toothpicks and rearranging the figure, can you form 4 congruent triangles?

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C1 Revision Exercise 1. Solve the inequality 3(x − 2) ≤ 4x and state any axiom or theorem used. 2. Explain whether each of the following statements is true or false using postulates or theorems. a. Two points can determine a line. b. It is possible that points A and B are in plane P, but line AB is not. 3. Fig. (a) shows planes F and G and points P, Q and R. Fig. (b) shows plane

H

and

points

S,

T

and

U.

i. Write a statement about each of the figures shown. ii. State the postulate or theorem you would use to justify the statements that you have made in (i). 4. It is given that ∆ABC is congruent to ∆PQR, ∠A = 70°, ∠B = 60° and AB ******ebook converter DEMO Watermarks*******

= 8 cm. i. Write down the length of PQ. ii. Find ∠R. 5. The figure shows a parallelogram ABCD. Prove that ∆ABD is congruent to ∆CDB.

C2 Revision Exercise 1. To prove that x + y − 2(5x − y) = 3(y − 3x), Tricia wrote the following steps:

State any axiom or theorem that she used. 2. Points A and B lie on the x-axis at coordinates −2.5 and 5. By the Ruler Placement Postulate, if the coordinate of A is 0, find the coordinate of B. 3. In the figure, PQ = PR and ∠PST = ∠PTS.

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Determine if ∆PQS is congruent to ∆PRT. 4. In the figure, ΔPQR is an equilateral triangle with sides of length 9 cm. A, B and C are points on PQ, QR and PR respectively such that PA = QB = RC = 4 cm.

i. Show that ΔAPC is congruent to ΔBQA. ii. Name the third triangle which is congruent to ΔAPC and ΔBQA and show that ΔABC is an equilateral triangle. 5. In the following figure, ∆ABC and ∆PQR are right-angled triangles. AQBP is a straight line with AQ = 6 cm, QB = 3 cm and BP = a cm. Find the values of a and b.

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Chapter Ten Theorems on Triangle Inequalities Two sisters, Ann and Kate, decided to go to the movies. Ann went straight to the cinema, but Kate had to pick up a parcel from the post office before going to the cinema. Who do you think took the longer route? In this chapter, you will learn the Triangle Inequality Theorem that proves that Kate took the longer route than Ann, regardless of where their home, the post office and the cinema were.

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LEARNING OBJECTIVES At the end of this chapter, you should be able to: illustrate theorems on triangle inequalities (Triangle Inequality Theorem, Exterior Angle Inequality Theorem, Hinge Theorem), apply theorems on triangle inequalities, prove inequalities in a triangle, prove properties of parallel lines cut by a transversal, determine the conditions under which lines and segments are parallel or perpendicular.

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10.1 Triangle Inequality Theorems Recap In Grade 7, we have learned some basic properties of a triangle:

The largest angle, ∠R, of a triangle is opposite the longest side, PQ, and the smallest angle, ∠Q, is opposite the shortest side, PR. The sum of the lengths of any two sides of a triangle must be greater than the length of the third side. This is known as the Triangle Inequality Theorem.

An exterior angle of a triangle, ∠d, is equal to the sum of its interior opposite angles, ∠a and ∠c, i.e. ∠d = ∠a + ∠c.

Triangle Inequality Theorem ******ebook converter DEMO Watermarks*******

To prove that the Triangle Inequality Theorem is always true, consider triangle ABC. We must prove that i. AC + BC > AB, ii. AB + AC > BC, iii. AB + BC > AC.

Extend BC to point D such that AC = CD. Thus ΔCAD is an isosceles triangle with ∠CAD = ∠CDA. ∠BAD = ∠BAC + ∠CAD which implies that ∠BAD > ∠CAD or ∠BAD > ∠CDA. Since ∠BAD > ∠CDA, BD > AB. Also, since BD = BC + CD, we obtain BC + CD > AB. We know that AC = CD. Therefore, we can conclude that BC + AC > AB which is Case (i). Similarly, Cases (ii) and (iii) can be proven.

The larger angle, ∠BAD, is opposite the longer side, BD, and the smaller angle, ∠CDA, is opposite the shorter side, AB. ******ebook converter DEMO Watermarks*******

Using the same approach, prove Cases (ii) and (iii).

Worked Example

1

(Triangle Inequality Theorem) In ΔABC, AB = 5 cm and BC = 3 cm. Find the range of possible values for AC.

Solution: By the Triangle Inequality Theorem, 5 + 3 > AC

5 + AC > 3

AC + 3 > 5

AC < 8

AC > ‒2

AC > 2

∴ The range of possible values for AC is 2 < AC < 8.

Using a number line, we can find the solution that satisfies the three inequalities.

Exercise 10A Questions 1, 2(a)-(f), 7

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In ΔPQR, PQ = 7.5 cm and QR = 12 cm. Find the range of possible values for PR.

Exterior Angle Inequality Theorem

Recall that an exterior angle of a triangle is equal to the sum of its interior opposite angles. Thus an exterior angle of a triangle is greater than either of its interior opposite angles, i.e. ∠d > ∠a and ∠d > ∠c. This is called the Exterior Angle Inequality Theorem.

Worked Example

2

(Exterior Angle Inequality Theorem)

ΔABC is an isosceles triangle. ABD is a straight line. Prove that ∠CBD > ∠ABC. ******ebook converter DEMO Watermarks*******

Solution: By the Exterior Angle Inequality Theorem, ∠CBD > ∠CAB. ∠ABC = ∠CAB (base ∠s of isos. Δ) ∴ ∠CBD > ∠ABC

In the figure, ABC, CDE and BDFG are straight lines. Prove that ∠AFG > ∠ACE.

Exercise 10A Questions 3-5, 12

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Hinge Theorem The Hinge Theorem (or SAS Inequality Theorem) states that: If two sides of one triangle are congruent to two sides of another triangle, but the included angle of the first triangle is larger than the included angle of the second triangle, then the third side of the first triangle is longer than the third side of the second triangle.

Hinge Theorem In this investigation, we shall explore and discover the Hinge Theorem, which describes inequalities for two triangles. A suitable interactive geometry software may be used.

1. Draw a circle with center A (using Circle tool).

2. Place points B and C on the circumference of the circle (using points on Object tool).

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3. Draw lines AB, AC and BC to form ΔABC (using Segment tool). 4. a. Move points B and C along the circumference of the circle without changing ∠BAC. What happens to the length of BC? b. Move point B along the circumference of the circle such that ∠BAC becomes smaller. What happens to the length of BC? c. Move point B along the circumference of the circle such that ∠BAC becomes larger. What happens to the length of BC?

5. Draw ΔADE using steps 1 to 3. 6. What are the corresponding congruent sides of the two triangles? 7. a. Move point D along the circumference of the circle such that ∠DAE is larger than ∠CAB. How does the length of DE compare to the length of BC? b. Move point D along the circumference of the circle such that ∠DAE is smaller than ∠CAB. How does the length of DE compare to the length of BC? The investigation that you have carried out is a special case of the Hinge Theorem where the two triangles are isosceles. Now let us prove the Hinge Theorem. Consider ΔABC and ΔDEF with AB ≡ DE, AC ≡ DF and ∠BAC > ∠EDF.

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Construct ΔARC inside ΔABC such that ΔARC ≡ ΔDEF. Using a compass, construct an arc with AB as the radius and mark point R inside ΔABC. Join R and C to form RC.

We now let P be on BC such that AP bisects ∠BAR. By SAS Congruence Test, we have ΔAPB ≡ ΔAPR. Therefore, PB = PR.

Applying the Triangle Inequality Theorem to ΔCRP, we get CP + PR > CR. Since PB = PR, we have CP + PB > CR. Since CR = EF and CP + PB = BC, we conclude that BC > EF. The converse of the Hinge Theorem is also true: If two sides of one triangle are congruent to two sides of a second triangle, and the third side of the first triangle is longer than the third side of the second triangle, then the included angle of the first triangle is ******ebook converter DEMO Watermarks*******

larger than the included angle of the second triangle.

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Worked Example

3

(Hinge Theorem) The figure shows ΔABD and ΔBCD. ABC is a straight line where AB = BC. ∠ABD = 110°. Prove that AD > CD.

Solution: ∠CBD = 180° (adj. ∠s on a str. line) = 70° AB ≡ CB, BD ≡ BD, ∠ABD > ∠CBD By the Hinge Theorem, AD > CD.

The figure shows a quadrilateral EFGH with diagonal EG. Given that EF = EH = 5 cm, FG = (x + 4) cm, GH = 10 cm, ∠FEG = 38° and ∠HEG = 84°, find the range of values of x.

Exercise 10A Questions 6, 8, 9

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Worked Example

4

((Converse of Hinge Theorem)

PQRS is a quadrilateral with PQ = 4 m, PS = 5 m, QR = RS, ∠PRQ = x° and ∠PRS = 125°. Write an inequality in x.

Solution: QR ≡ SR, PR ≡ PR, PQ < PS By the converse of Hinge Theorem, x < 125 x>0 ∴ 0 < x < 125

The figure shows ΔWYZ with WX = XY = 8 cm, YZ = 16 cm and WZ = 9.5 cm. Is ∠WXZ or ∠YXZ greater? Explain your answer.

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Exercise 10A Questions 10, 11

BASIC LEVEL 1. Determine if the given set of lengths could be those of the sides of a triangle. a. 2 cm, 3 cm, 4 cm b. 6 m, 7 m, 10 m c. 21 m, 21 m, 21 m d. 20 m, 20 m, 4 m e. 8 cm, 12 cm, 3 cm f. (x − 2) cm, (x + 1) cm, 2x cm 2. The lengths of two sides of a triangle are given. Find the range of possible lengths of the third side. a. AB = 4 mm, BC = 7 mm b. AB = 54 cm, AC = 45 cm c. QR = 12.5 m, PR = 21 m d. EF = 5 m, EG = 226 cm 3. In the figure, ABD and BCE are straight lines. Prove that ∠DCE > ∠DAE.

4. The figure shows ΔJKN, ΔJLN and a straight line JKLM. Arrange ∠NKM, ******ebook converter DEMO Watermarks*******

∠NLM and ∠NJM from the smallest to the greatest. Explain your answer.

5. In the figure, ΔACD is isosceles and CDB is a straight line. Show that

∠CAD > ∠DAB.

6. ΔABC and ΔCDE are two triangles with AB = ED = 5 cm and BC = DC = 4 cm. If ∠CDE is ¾ of ∠ABC, show that AC > EC.

INTERMEDIATE LEVEL 7. In the quadrilateral STUV, ST = 12 cm, TU = 10 cm, UV = 3 cm and VS = 6

cm. Find the range of possible lengths of SU.

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8.

The figure shows ΔABD. C is a point on BD such that BC = CD = 6 m. Given that ∠ACB = 45°, m and AD = 11 m, find the range of values of x. 9.

In the quadrilateral WXYZ with diagonal WY, WX = WZ, ∠WXY = 80°, ∠XYW = 35°, ∠WYZ = 25° and ∠WZY = 135°. Write an inequality involving XY and YZ. 10. ABCD is a quadrilateral with sides AB = 20 cm, BC = 16 cm, CD = 20 cm and DA = 23 cm. Show that ∠ACD > ∠BAC.

11. ABCD is a quadrilateral with diagonal AC. AB = 9 cm, BC = CD = 7.5 cm, ******ebook converter DEMO Watermarks*******

AD = 13 cm, AC = 15 cm, ∠ACD = 60° and ∠ACB = (4x + 16)°. Find the

range of values of x.

ADVANCED LEVEL 12. In the figure, CD and BF bisect each other at E. Show that ∠DFA >

∠BCE.

13. The lengths of two sides of a triangle are given. Find the range of possible lengths of the third side. a. PQ = (2x + 2) cm, PR = 3x cm b.

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10.2 Properties of Parallel Lines Recap (Corresponding Angles, Alternate Angles and Interior Angles) In Grade 7, we have learned that when two parallel lines, AB and CD, are cut by a transversal PQ, then

Fig. 10.1 corresponding angles are equal, e.g. ∠a = ∠b (corr. ∠s, AB // CD); alternate angles are equal, e.g. ∠a = ∠c (alt. ∠s, AB // CD); interior angles are supplementary, e.g. ∠a + ∠d = 180˚ (int. ∠s, AB // CD). The converse for each of the above is also true, i.e. when two lines AB and CD are cut by a transversal PQ, and if ∠a = ∠b, then AB // CD; if ∠a = ∠c, then AB // CD; if ∠a + ∠d = 180˚, then AB // CD.

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In this section, we will establish some conditions for lines to be parallel, using Fig. 10.1. When two lines AB and CD are cut by a transversal PQ, and if ∠a = ∠b, then AB // CD. We call this the Converse of the Corresponding Angle Postulate.

Since ∠b and ∠c are vertically opposite angles, ∠b = ∠c. Also, since ∠a and ∠c are alternate angles, ∠a = ∠c. ∴ ∠a = ∠b By the Converse of the Corresponding Angle Postulate, AB // CD. When two lines AB and CD are cut by a transversal PQ, and if ∠a = ∠c, then AB // CD. We call this the Converse of the Alternate Angle Theorem.

Since ∠a and ∠d are interior angles, they are supplementary, i.e. ∠a + ∠d = 180°. Since ∠b and ∠d are adjacent angles on a straight line, ∠b + ∠d = 180°. Comparing the two equations, ∠a = ∠b. By the Converse of the Corresponding Angle Postulate, AB // CD. When two lines AB and CD are cut by a transversal PQ, and if ∠a + ∠d = 180˚, then AB // CD. We call this the Converse of the Interior Angle Theorem.

The Perpendicular Transversal Theorem states that if a transversal is perpendicular to one of two parallel lines, then it is perpendicular to the ******ebook converter DEMO Watermarks*******

second parallel line. Can you prove that the theorem is true?

Worked Example

5

(Properties of Parallel Lines)

In the figure, the lines ST and UV are cut by the transversal ABCD. Find the value of x and of y that will make ST // UV.

Solution: For ST to be parallel to UV, ∠UCB = ∠CBT (alt. ∠s).

For ST to be parallel to UV, ∠UCD = ∠BCV (vert. opp. ∠s) ∠BCV = ∠ABT (corr. ∠s) ∴ ∠UCD = ∠ABT 7y = 80 − y 8y = 80 y = 10 ******ebook converter DEMO Watermarks*******

In the figure, ABC is a straight line. Find the value of x that will make BE // CD.

Exercise 10B Questions 1(a)-(d)

Worked Example

6

(Proving Parallel Lines)

In the figure, the lines ST and UV are cut by the transversal ABCD. Both ∠SBA and ∠VCD are 45°. Prove that ST // UV.

Solution: ******ebook converter DEMO Watermarks*******

∠CBT = ∠SBA = 45° (vert. opp. ∠s) ∠VCD = ∠CBT = 45° (corr. ∠s) ∴ ST // UV

In the figure, the lines EF and GH are cut by the transversal WXYZ. ∠WXF = 130° and ∠WYG = 50°. Prove that EF // GH.

Exercise 10B Questions 2(a)-(d), 3-5

BASIC LEVEL 1. Find the value of x and/or of y that will make WX // YZ. a.

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b.

c.

d.

2. Name a pair of parallel lines. Explain your answer.

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a.

b.

INTERMEDIATE LEVEL 3. In the figure, BDF and CDE are straight lines, ∠ABD = 68°, ∠DBC = 54°, ∠BDC = 68° and ∠DEF = 58°. Prove that a. AB // EC, b. BC // EF.

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4. In the figure, ∠ABC = 123°, ∠BCD = 65° and ∠CDE = 122°. Prove that

AB // DE.

ADVANCED LEVEL 5. In the figure, ∠ABC = 37°, reflex ∠BCD = 285°, ∠CDE = 56° and

∠DEF = 18°. Prove that AB // EF.

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1. Triangle Inequality Theorem The sum of the lengths of any two sides of a triangle must be greater than the length of the third side. 2. Exterior Angle Inequality Theorem An exterior angle of a triangle is greater than either of its interior opposite angles. 3. Hinge Theorem (SAS Inequality Theorem) If two sides of one triangle are congruent to two sides of another triangle, but the included angle of the first triangle is larger than the included angle of the second triangle, then the third side of the first triangle is longer than the third side of the second triangle. The converse of the Hinge Theorem is also true. 4. Conditions for Lines to be Parallel

When two lines AB and CD are cut by a transversal PQ, and if ∠a = ∠b, then AB // CD. We call this the Converse of the Corresponding Angle Postulate. When two lines AB and CD are cut by a transversal PQ, and if ∠a = ∠c, then AB // CD. We call this the Converse of the Alternate Angle Theorem. When two lines AB and CD are cut by a transversal PQ, and if ∠a + ∠d = ******ebook converter DEMO Watermarks*******

180˚, then AB // CD. We call this the Converse of the Interior Angle Theorem.

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1. Determine if the given set of lengths could be those of the sides of a triangle. 1. 2 cm, 5 cm, 7 cm 2. 8 cm, 9 cm, 3 cm 3. 14 m, 30 m, 14 m 4. 2x cm, 4x cm, 5x cm 2. The lengths of two sides of a triangle are given. Find the range of possible lengths of the third side. 1. AB = 16 cm, BC = 26 cm 2. BC = 8.5 cm, AC = 5.5 cm 3. PQ = 4x mm, PR = (2x − 1) mm 4. PQ = x m, QR = 6x m 3. The figure shows ΔABC, ΔBCD and ΔADE. ACD and BDE are straight lines. Prove that ∠ADE > ∠ABC.

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4. The figure shows ΔABC, ΔCDE and ΔCEF. ACD and BCF are straight lines. ∠FCE = ∠ECD = 35°. Prove that AB > DE.

5. The floor area of a gallery is a quadrilateral with sides PQ = 8 m, QR = 7 m, RS = 6 m and PS = 4 m. Find the range of possible lengths of the diagonal QS.

6. ΔHJK has sides HJ = 14 cm, JK = (3x − 2) cm and HK = (4x − 5) cm. I is the midpoint of HJ and ∠HIK = 115°. Find the range of possible values of x.

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7. + 13 7. ΔABC has sides AB = cm, BC = (x + 6) cm and AC = (x + 7) cm. 2 ΔXYZ has sides XY = (x + 9) cm, YZ = (x + 6) cm and XZ = (x + 7) cm. Is ∠ACB larger or smaller than ∠XZY? Explain your answer.

8. Carlo has two sticks of lengths 9 cm and 14 cm. How many different triangles can he form with the sticks if the length of the third stick is an integer? 9. Quadrilateral ABCD has angles ∠BAD = 102°, ∠ABD = (2x + 35)°, ∠ADB = (x − 2y)° and ∠BDC = (5 − 4y)°. Find the value of x and of y that will make AB // CD.

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10. In the figure, ΔJMN and ΔKLN have ∠JMN = 25°, ∠KLN = 44°, ∠NKL = x° and ∠MNL = y°. JNL and MNK are straight lines. 1. Find the value of x that will make JM // KL. 2. Hence find the value of y.

11. In the figure, ∠ABC = 146°, ∠CEF = 34°, ∠HEF = 26° and ∠EHG = 62°. DEF is a straight line. Name a pair of parallel lines and explain why they are parallel.

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12. In the following figures, prove that AB // CD. 1.

2.

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Given the quadrilateral ABCD, prove that AB + BC + CD > AD.

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Chapter Eleven Probability of Simple Events Do you know that casinos make use of probability to set rules to ensure that they will always be on the winning side in the long run? How is this done?

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LEARNING OBJECTIVES At the end of this chapter, you should be able to: illustrate an experiment, outcome, sample space and event, count the number of occurrences of an outcome in an experiment using: (a) a table; (b) a tree diagram; (c) systematic listing; and (d) fundamental counting principle, find the probability of a simple event, illustrate an experimental probability and a theoretical probability, solve problems involving probabilities of simple events.

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11.1 Introduction to Probability We often make statements such as: ‘There is a 50 : 50 chance of our school winning the National Inter-School Basketball Championship.’ ‘I cannot predict whether I will obtain a ‘six’ in my next roll of a die.’ ‘It will probably rain today.’ We make such statements because we are uncertain whether an event will occur. For an uncertain event, we can discuss about its chance of occurrence.

1. The following are some events which we may come across in our everyday life. Event A: The sun will rise from the east every day. Event B: Your friend will win the lottery this year. Event C: You obtain a ‘tail’ when you toss a coin. Event D: It will snow in the Philippines at least once a year. Event E: Babies drink milk every day. Each of these events may or may not happen. Mark these events A to E on the line to show the likelihood they will occur.

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2. We can use values between 0 and 1 inclusive to measure the chance of an event occurring, where an impossible event takes on the value 0 and a certain event takes on the value 1. If there is a 50 : 50 chance that an event will occur, what value does it take? 3. Write down an event that corresponds to each of the five categories above. Mark out each event on the number line based on the estimated chance of occurence.

In our everyday life, we use words such as ‘unlikely’, ‘likely’ or ‘certain’ to describe the chance of an event occurring. The measure of chance, which takes on values between 0 and 1 inclusive, is known as probability. We will learn about probability in this chapter.

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11.2 Sample Space When we perform a scientific experiment, we will obtain a certain result or outcome. However, in probability, the result or the outcome is not certain – it depends on chance. Table 11.1 shows some examples of probability experiments and their possible outcomes.

Probability experiment

Possible outcomes

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Ten identical cards numbered 11, 12, 13, …, 20 are placed in a box. One card is drawn at random from the box. Two black balls and three white balls of the same size are placed in a bag. One ball is drawn at random from the bag.

Table 11.1 Consider the experiment where a coin is tossed. The results are either getting a ‘head’ or a ‘tail’. These results are referred to as the outcomes. The collection of all the possible outcomes of a probability experiment is called the sample space. In the case of tossing a coin, the sample space is a ‘head’ and a ‘tail’. What is the sample space when a die is rolled?

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Worked Example

1

(Sample Space) A fair die is rolled. Write down the sample space and state the total number of possible outcomes.

Solution: A die has the numbers 1, 2, 3, 4, 5 and 6 on its six faces, i.e. the sample space consists of the numbers 1, 2, 3, 4, 5 and 6. Total number of possible outcomes = 6

A spinner is divided into 5 equal sectors of different colors. When the spinner is spun, the color of the sector on which the pointer lands is noted. Write down the sample space and state the total number of possible outcomes.

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Exercise 11A Questions 1, 2(a)-(b)

Worked Example

2

(Sample Space) For each of the following experiments, write down the sample space and state the total number of possible outcomes. a. Drawing a ball at random from a bag containing 2 identical black balls and 3 identical white balls b. Choosing a two-digit number at random

Solution: a. Let B1 and B2 represent the 2 black balls; W1, W2 and W3 represent the 3 white balls. The sample space consists of B1, B2, W1, W2 and W3. Total number of possible outcomes = 5 b. The sample space consists of the integers 10, 11, 12, …, 99.

For (a), as there are two black balls in the bag, there is a difference between drawing the first or the second black ball. Thus we must differentiate between the two black balls by representing them using B1 and B2. Similarly, for the three white balls, we represent them using W1, W2 and W3.

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For each of the following experiments, write down the sample space and state the total number of possible outcomes. a. Drawing a marble at random from a bag containing 5 identical blue marbles and 4 identical red marbles b. Picking a letter at random from a box containing identical cards with letters that spell the word ‘NATIONAL’ c. Selecting a receipt at random from a receipt book with running serial numbers from 357 to 389

Exercise 11A Questions 2(c)-(e)

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11.3 Probability of Single Events In Section 11.1, we have learned that probability is a measure of chance.

Tossing a Coin 1. Are we able to state with certainty whether the outcome is a ‘head’ or a ‘tail’ before a coin is tossed? 2. Toss a coin 20 times. i. Record the outcome of each toss in the following table. See Table.. ii. Write down the fraction of obtaining a ‘head’ or a ‘tail’ in the table above. iii. Compare your results with those of your classmates. Are they the same? What can you deduce about the results of tossing a coin? 3. a. In groups of 4 or 5, add and record the total number of ‘heads’ obtained by your group members. Repeat for the total number of ‘tails’. Compute the fraction of obtaining a ‘head’ or a ‘tail’.

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1. a. Outcome for _____ tosses

Total number of ‘heads’ or ‘tails’

Fraction of obtaining a ‘head’ or a ‘tail’

Head Tail b. As a class, add and record the total number of ‘heads’ obtained by all students. Repeat for the total number of ‘tails’. Compute the fraction of obtaining a ‘head’ or a ‘tail’. Outcome for _____ tosses

Total number of ‘heads’ or ‘tails’

Fraction of obtaining a ‘head’ or a ‘tail’

Head Tail 2. Look at the last column in the three tables. Do you notice that the probabilities of obtaining a ‘head’ or a ‘tail’ approach ½ when there are more tosses? 3.

If we toss a coin 1000 times, would we expect to obtain exactly 500 ‘heads’ and exactly 500 ‘tails’? Explain your answer.

When a coin is tossed, if the chance of obtaining a ‘head’ is the same as the chance of obtaining a ‘tail’, we say that the coin is fair or unbiased. This means that for a fair coin, there are two equally likely outcomes, i.e. ******ebook converter DEMO Watermarks*******

obtaining a ‘head’ and obtaining a ‘tail’. Thus the chance of obtaining a ‘head’ is 1 out of 2. We say that the probability of obtaining a ‘head’ is ½. What is the probability of obtaining a ‘tail’?

Search on the Internet for interactive applets on probability that involve tossing a coin.

Rolling a Die Go to http://www.shinglee.com.sg/StudentResources/ spreadsheet ‘Rolling a Die’.

and

open

the

Fig. 11.1 1. Click on the button ‘Roll the Die’ and it will roll the die once. Repeat for a total of 20 rolls. Record the number of ‘1’, ‘2’, ‘3’, ‘4’, ‘5’ and ‘6’ obtained in the following table. Compute the fraction of obtaining each outcome.

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Outcome

Number of corresponding outcomes for 20 rolls

Fraction of obtaining each corresponding outcome for 20 rolls

‘1’ ‘2’ ‘3’ ‘4’ ‘5’ ‘6’

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2. As a class, add and record the total number of ‘1’, ‘2’, ‘3’, ‘4’, ‘5’ and ‘6’ obtained by all students. Compute the fraction of obtaining each outcome. Outcome for _____ rolls

Total number of corresponding outcomes

Fraction of obtaining each corresponding outcome

‘1’ ‘2’ ‘3’ ‘4’ ‘5’ ‘6’ 3. Look at the last column in the two tables. Do you notice that the probabilities of obtaining any one of the six outcomes approach when there are more rolls? 4. If we roll a die 600 times, would we expect to obtain exactly 100 ‘6’? Explain your answer. When a die is rolled, there are six possible outcomes, i.e. 1, 2, 3, 4, 5 and 6. If the die is fair, then each of the six outcomes is equally likely to occur. Thus the chance of obtaining a ‘six’ is 1 out of 6. We say that the probability of obtaining a ‘six’ is . ******ebook converter DEMO Watermarks*******

In general, in a probability experiment with m equally likely outcomes, if k of these outcomes favor the occurrence of an event E, then the probability, P(E), of the event happening is given by:

This is known as theoretical probability, the probability that is obtained based on mathematical theory. In the investigation, you conducted a probability experiment to determine the chance of obtaining a ‘head’ or a ‘tail’ when a coin is tossed. Based on theoretical probability, the probability of obtaining a ‘head’ is ½, i.e. 10 ‘heads’ in 20 tosses. However, it is unlikely that all your classmates obtained 10 ‘heads’ in 20 tosses. The probability that you obtained in the experiment is known as experimental probability. Thus if you obtained 11 ‘heads’ in 20 tosses, your experimental probability of getting a ‘head’ is . From the investigation, we can conclude that as the number of trials increases, the experimental probability of an outcome occurring tends towards the theoretical probability of the outcome happening.

Worked Example

3

(Probability involving Number Cards) A card is drawn at random from a box containing 12 cards numbered 1, 2, 3, …, 12. Find the probability of drawing i. a ‘7’, ii. an even number, iii. a prime number, iv. a perfect square, v. a negative number,

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vi. a number less than 13.

Solution: Total number of possible outcomes = 12 i. ii. There are 6 even numbers from 1 to 12, i.e. 2, 4, 6, 8, 10 and 12.

iii. There are 5 prime numbers from 1 to 12, i.e. 2, 3, 5, 7 and 11. iv. There are 3 perfect squares from 1 to 12, i.e. 1, 4 and 9.

v. There are no negative numbers from 1 to 12.

vi. All the 12 numbers from 1 to 12 are less than 13.

A prime number is a positive integer that has exactly 2 different factors, 1 and itself. Thus 1 is not a prime number.

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When the probability of an event occurring is 0, we say that it is an impossible event. When the probability of an event occurring is 1, we say that it is a certain event.

A ball is drawn at random from a bag containing some balls numbered 10, 11, 12, …, 24. Find the probability of drawing i. a ‘21’, ii. an odd number, iii. a composite number, iv. a perfect cube.

Exercise 11A Questions 3-4, 11, 18-20

A composite number is a positive integer that has more than 2 different factors. In Worked Example 3, we observe that the probability of drawing a negative number from the 12 cards numbered 1, 2, 3, …, 12 is 0. This means that we will never be able to draw a negative number from the 12 cards. In the same worked example, we also notice that the probability of drawing a number less than 13 from the 12 cards is 1. This means that we will definitely be able to draw a number less than 13 from the 12 cards.

1. In the Thinking Time in Section 11.1, the event D ‘It will snow in the ******ebook converter DEMO Watermarks*******

Philippines at least once a year.’ is an impossible event, i.e. it will never occur. What can we say about the probability of D occurring? 2. In the Thinking Time in Section 11.1, the event A ‘The sun will rise from the east every day.’ is a certain event, i.e. it will definitely occur. What can we say about the probability of A occurring? 3. Is it possible that the probability of an event occurring is less than 0 or greater than 1? From the above explanation and Thinking Time, we can conclude that: For any event E, 0 ≤ P(E) ≤ 1. P(E) = 0 if and only if E is an impossible event, i.e. it will never occur. P(E) = 1 if and only if E is a certain event, i.e. it will definitely occur.

Probability theory was first used primarily in gambling problems. Girolamo Cardano (1501 – 1576), an Italian, wrote a gambler’s manual which made use of probability theory. In 1654, Chevalier de Mere (1607 – 1684), a Frenchman, posed a gambling problem to his fellow countryman, Blaise Pascal (1623 – 1662). In response to this problem, Pascal and another French mathematician, Pierre Fermat (1601 – 1665), laid the foundations for the theory of probability. This theory has widespread applications in business and in the sciences. Its applications range from the determination of life insurance premiums to the description of the behavior of molecules in a gas. In fact, it can also be used to predict the outcome of an election. Search on the Internet for other real-life applications of probability theory. Present your findings to the class. ******ebook converter DEMO Watermarks*******

Worked Example

4

(Probability involving Playing Cards) A card is drawn at random from a standard pack of 52 playing cards. Find the probability of drawing i. a black card, ii. a red ace, iii. a diamond, iv. a card which is not a diamond.

There are 4 suits in a standard pack of 52 playing cards, i.e. club , heart and spade .

, diamond

Each suit has 13 cards, i.e. Ace, 2, 3, …, 10, Jack, Queen and King. All the clubs and spades are black in color. All the diamonds and hearts are red in color. All the Jack, Queen and King cards are picture cards.

Solution: Total number of possible outcomes = 52 i. There are 26 black cards in the pack.

ii. There are 2 red aces in the pack, i.e. the ace of hearts and the ace of ******ebook converter DEMO Watermarks*******

diamonds.

iii. There are 13 diamonds in the pack.

iv. Since there are 13 diamonds in the pack, there are 52 – 13 = 39 cards which are not diamonds.

Notice that, P(drawing a card which is not a diamond) = 1 – P(drawing a diamond). In general, for any event E, we have:

Casinos make use of probability theory to set rules to ensure that they will always be on the winning side in the long run so that they will not go out of business. Search on the Internet for examples of how casinos use probability to their advantage.

A card is drawn at random from a standard pack of 52 playing cards. Find the probability of drawing i. a red card, ii. an ace, ******ebook converter DEMO Watermarks*******

iii. the three of clubs, iv. a card which is not the three of clubs.

Exercise 11A Questions 5, 12-14

Worked Example

5

(Probability involving Letters of the English Alphabet) A letter is chosen at random from the word ‘MATHEMATICS’. Find the probability that the letter is i. an ‘A’, ii. a vowel, iii. not a vowel.

Solution: Total number of letters = 11 i. There are 2 ‘A’s. ii. There are 4 vowels, i.e. 2 ‘A’s, 1 ‘E’ and 1 ‘I’. iii. Method 1: There are 7 consonants, i.e. 1 ‘C’, 1 ‘H’, 2 ‘M’s, 1 ‘S’ and 2 ‘T’s.

Method 2:

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There are 26 letters in the alphabet. 5 of them are vowels, namely ‘a’, ‘e’, ‘i’, ‘o’, and ‘u’.

1. A letter is chosen at random from the word ‘CHILDREN’. Find the probability that the letter is i. a ‘D’, ii. a consonant, iii. not a consonant. 2. A marble is drawn at random from a bag containing 9 red marbles, 6 yellow marbles, 4 purple marbles and 5 blue marbles. Find the probability of drawing i. a purple marble, ii. a red or a blue marble, iii. a white marble, iv. a marble that is not white. 3. A box contains 24 balls, some of which are red, some of which are green and the rest are blue. The probabilities of drawing a red ball and a green ball at random from the box are and respectively. Find the number of blue balls in the box.

Exercise 11A Questions 6-10, 15-17 ******ebook converter DEMO Watermarks*******

The first three children of a couple are boys. What is the probability that their next child will be a girl?

BASIC LEVEL 1. A dart board is divided into 6 equal sectors. When a dart lands on it, the number of the sector on which it lands is noted. Write down the sample space and state the total number of possible outcomes.

2. For each of the following experiments, write down the sample space and state the total number of possible outcomes. a. Tossing a fair tetrahedral die with faces labeled 2, 3, 4 and 5 respectively b. Drawing a card at random from a box containing ten identical cards labeled A, B, C, D, E, F, G, H, I, J c. Drawing a disc at random from a bag containing 5 identical red discs, 3 identical blue discs and 2 identical green discs d. Picking a letter at random from a box containing identical cards with letters that spell the word ‘TEACHER’ e. Choosing a three-digit number at random 3. An 8-sided fair die with faces labeled 2, 3, 3, 4, 7, 7, 7 and 9 is rolled once. ******ebook converter DEMO Watermarks*******

Find the probability of getting i. a ‘7’, ii. a ‘3’ or a ‘4’, iii. a number less than 10, iv. a number which is not ‘2’. 4. A card is drawn at random from a box containing some cards numbered 10, 11, 12, …, 22. Find the probability of drawing i. an even number, ii. a number between 13 and 19 inclusive, iii. a prime number that is less than 18, iv. a number greater than 22, v. a number that is divisible by 4. 5. A card is drawn at random from a standard pack of 52 playing cards. Find the probability of drawing i. the ace of spades, ii. a heart or a club, iii. a picture card, iv. a non-picture card. 6. Each of the letters of the word ‘PROBABILITY’ is written on a card. All the cards are well-shuffled and placed face down on a table. A card is turned over. Find the probability that the card shows i. the letter ‘A’, ii. the letter ‘B’, iii. a vowel, iv. a consonant. 7. A spinner is divided into 5 equal sectors. When the spinner is spun, what is the probability that the pointer will stop at a sector whose label is ******ebook converter DEMO Watermarks*******

i.

?

ii. a letter of the English alphabet? iii. a vowel? iv. a consonant? 8. A bag contains 4 pieces of candy – caramel, chocolate, gummy and licorice. A piece of candy is removed at random from the bag. Find the probability that the candy is i. a caramel, ii. either a chocolate or a gummy, iii. not a licorice. 9. An envelope contains 40 shopping vouchers, of which 25 vouchers each have a value of ₱1500 and 15 vouchers each have a value of ₱3000. Gemma picks a voucher at random from the envelope. Find the probability that the voucher has a value of ₱3000. 10. A group of 30 people consisting of 9 men, 6 women, 12 boys and 3 girls are waiting to get their passport photographs taken. A person is selected at random from the group. Find the probability that the person is i. a male, ii. either a woman, a boy or a girl. INTERMEDIATE LEVEL 11. A two-digit number is chosen at random. Find the probability that the number is ******ebook converter DEMO Watermarks*******

i. less than 20, ii. a perfect square. 12. Two Joker cards are added to a standard pack of 52 playing cards. A card is then drawn at random from the 54 cards. Find the probability of drawing i. a red card, ii. a two, iii. a joker, iv. a queen or a king. Note: A Joker card is neither a black nor a red card. 13. All the clubs are removed from a standard pack of 52 playing cards. A card is drawn at random from the remaining cards. Find the probability of drawing i. a black card, ii. a diamond, iii. a picture card, iv. a card which is not an ace. 14. Sam wakes up in the morning and notices that his digital clock reads 07 25.

After noon, he looks at the clock again. What is the probability that i. the number in column A is a 4? ii. the number in column B is an 8? iii. the number in column A is less than 6? iv. the number in column B is greater than 5? 15. A box contains 2 dozen pairs of contact lenses, of which 8 pairs are tinted. ******ebook converter DEMO Watermarks*******

A pair of contact lenses is drawn at random from the box. Find the probability that it is not tinted. 16. The table shows the number of each type of school personnel at a school. See Table.. a. If a school personnel is selected at random, find the probability that the school personnel is i. a teacher, ii. a management staff, iii. an administrative or a maintenance staff. b. Two teachers and an administrative staff resign from the school. A school personnel is selected at random from the remaining staff. Find the probability that the school personnel is i. an administrative staff, ii. not a laboratory staff. 17. There are a total of 117 pairs of socks in a clothes bin. Each pair of socks is placed in a bag. The probabilities of selecting a yellow pair of socks and a grey pair of socks at random from the bin are and respectively. Find the number of pairs of socks in the bin which are i. yellow, ii. neither yellow nor grey. ADVANCED LEVEL 18. An IQ test consists of 80 multiple-choice questions. A question is selected at random. Find the probability that the question number i. contains only a single digit, ii. is greater than 67, iii. contains exactly one ‘7’, ******ebook converter DEMO Watermarks*******

iv. is divisible by both 2 and 5. 19. Each of the numbers 2, 3, 5 and 7 is written on a card. Two of the cards are drawn at random to form a two-digit number. Find the probability that the two-digit number is i. divisible by 4, ii. a prime number. 20. A biased tetrahedral die with faces labeled 1, 2, 3 and 4 is rolled once. The chance of getting a ‘3’ is twice that of getting a ‘1’. The chance of getting a ‘2’ is thrice that of getting a ‘3’. There is an equal chance of getting a ‘2’ and a ‘4’. Find the probability of getting a prime number.

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11.4 Further Examples on Probability of Single Events In this section, we will take a look at more examples that involve probability.

Worked Example

6

(Probability involving Groups of People) In a class of 30 students, there are 12 girls and 2 of them are short-sighted. 6 of the boys are not short-sighted. If a student is chosen at random, find the probability that the student is i. a boy, ii. short-sighted.

Solution: i. Number of boys P(student chosen is a boy)

ii. Number of girls who are short-sighted = 2 Number of boys who are short-sighted Number of students who are short-sighted P(student chosen is short-sighted)

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In a class of 40 students, there are 24 boys and 16 of them are not shortsighted. 4 of the girls are short-sighted. If a student is chosen at random, find the probability that the student is i. a girl, ii. not short-sighted.

Exercise 11B Questions 1-2, 6-7

Worked Example

7

(Probability involving Angles of Sectors) A circle is divided into sectors of different colors. A point is selected at random in the circle. Find the probability that the point lies in the i. yellow sector,

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ii. green sector, iii. black sector.

Solution: i.

ii.

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iii.

Since a point is selected at random, any point in the circle will have the same chance of being selected. We assume that the point will not fall on any of the lines separating the four sectors.

The number of points in a sector is proportional to the area of the sector, which is proportional to the angle of the sector.

A circle is divided into sectors of different colors. A point is selected at random in the circle. Find the probability that the point lies in the

i. red sector, ii. blue sector, iii. purple sector, iv. green or white sector. ******ebook converter DEMO Watermarks*******

Exercise 11B Questions 3-4

Worked Example

8

(Probability involving Algebra) A box contains x red marbles, (x + 3) yellow marbles and (4x – 15) blue marbles.

i. Find an expression, in terms of x, for the total number of marbles in the box. ii. A marble is drawn at random from the box. Write down an expression, in terms of x, for the probability that the marble is blue. iii. Given that the probability in (ii) is , find the value of x.

Solution: i. Total number of marbles ii. P(drawing a blue marble) iii. Given that

1. There are 12 green balls and (x + 2) yellow balls in a box. i. Find an expression, in terms of x, for the total number of balls in the box. ******ebook converter DEMO Watermarks*******

ii. A ball is drawn at random from the box. Write down an expression, in terms of x, for the probability that the ball is yellow. iii. Given that the probability in (ii) is , find the value of x. 2. There are 28 boys and 25 girls in a school hall. After y girls leave the hall, the probability of selecting a girl at random becomes . Find the value of y.

Exercise 11B Questions 5, 8-13

BASIC LEVEL 1. A box of 30 marbles and balls consists of 8 blue marbles, 3 green marbles, 1 yellow marble, 11 blue balls, 4 green balls and 3 yellow balls. If an item is chosen at random, find the probability that the item is i. a marble, ii. not blue, iii. not a yellow ball, iv. red. 2. Ann has 5 novels and 5 comic books in her bag. Three of her books are in Tagalog, 2 of which are comic books. The rest of her books are in English. If a book is chosen at random from her bag, find the probability of choosing i. a book in Tagalog, ii. a novel which is in English. 3. A survey is conducted to find out which of the four fruits, apple, papaya, guava and mango, the students in a class prefer. The pie chart shows the ******ebook converter DEMO Watermarks*******

results of the survey. A student is selected at random. Find the probability

that the student prefers

i. apple, ii. mango, iii. papaya or guava. 4. A regular octagon is divided into 4 regions, where O is its center. A point is selected at random in the octagon. Find the probability that the point lies

in

i. region R, ii. region S, iii. region P or Q. 5. There are 15 girls and x boys at a school parade square. i. Write down an expression, in terms of x, for the total number of students at the school parade square. ii. A student is selected at random. Write down an expression, in terms of x, for the probability that the student is a girl. iii. Given that the probability in (ii) is , find the value of x.

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INTERMEDIATE LEVEL 6. A class of 38 students went on a short trip to Bangkok. Of the 18 boys, 6 of them checked in their luggage at the airport. 8 of the girls did not check in their luggage. If a student is chosen at random, find the probability that the student i. is a girl who did not check in her luggage, ii. checked in his/her luggage. 7. a. A class has 16 boys and 24 girls. Of the 16 boys, 3 are left-handed. Of the 24 girls, 2 are left-handed. If a student is chosen at random to clean the whiteboard, find the probability that the student is i. a boy, ii. left-handed. b. The student chosen to clean the whiteboard in (a) is a girl who is not left-handed. Another student is selected at random from the remaining students to borrow the visualizer from the class next door. Find the probability that the student is i. a boy who is left-handed, ii. a girl who is not left-handed. 8. Santa Claus has (3h + 11) red presents and (h + 5) white presents in his stocking. Carlo selects a present at random from the stocking. Given that the probability that he obtains a red present is , find the value of h. 9. Some patients participated in a clinical trial for a new drug to treat osteoporosis. A patient is selected at random. The probability that the patient had no change in his bone mass density is , the probability that he had a slight reduction in his bone mass density is

and the probability that

he had a significant reduction in his bone mass density is of k. ******ebook converter DEMO Watermarks*******

. Find the value

10. A carton contains 15 toothbrushes, of which p have soft bristles. After 5 more toothbrushes with soft bristles are added to the carton, the probability of drawing a toothbrush with soft bristles becomes . Find the value of p. 11. There are 23 boys and 35 girls on the school’s track and field team. After q boys and (q + 4) girls graduate at the end of this year, the probability of selecting a boy at random to represent the school for an event becomes . Find the value of q. ADVANCED LEVEL 12. A bag contains 40 balls, some of which are red, some of which are yellow and the rest are black. The probabilities of drawing a red ball and a yellow ball at random from the bag are and respectively. i. Find the probability of drawing a black ball at random from the bag. (2x + 1) red balls and (x + 2) yellow balls are added to the bag while (x – 3) black balls are removed from the bag. The probability of drawing a yellow ball at random from the bag is now . Find ii. an expression, in terms of x, for the total number of balls in the bag now, iii. the number of yellow balls in the bag now. 13. There are 50 students in an auditorium, of which 2x are boys and y are girls. After (y – 6) boys leave the auditorium and (2x – 5) girls enter the auditorium, the probability of selecting a girl at random becomes . Find the value of x and of y.

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11.5 Simple Combined Events, Possibility Diagrams and Tree Diagrams Possibility Diagrams In this section, we will learn how to list the sample space of an experiment involving two or more objects (e.g. rolling two dice), and calculate probabilities for simple combined events. The possible outcomes for rolling a fair die are 1, 2, 3, 4, 5 and 6, and we write the sample space as {1, 2, 3, 4, 5, 6}. How do we write the possible outcomes for rolling two fair dice? We can represent a possible outcome by using an ordered pair, e.g. (2, 3) means that the first die shows a ‘2’ and the second die shows a ‘3’; which is different from (3, 2). So what does (3, 2) mean? How can we write the sample space for rolling two fair dice? Is {(1, 1), (1, 2), (1, 3), …, (6, 6)} clear enough? Listing out all the outcomes would be very tedious and we may miss out some outcomes. Therefore, there is a need to use a different method to represent the sample space. Fig. 11.2 shows one way of drawing a possibility diagram to represent the sample space for rolling two fair dice.

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Fig. 11.2 A possibility diagram is used when each outcome of the sample space has two components. For example, in the above case, an outcome (represented by a red dot ) is determined by the values displayed by the first and second dice. From the above possibility diagram, we observe that the total number of possible outcomes is 6 3 6 = 36. We can also calculate the probability of certain events using a possibility diagram, as shown in Worked Example 9.

Worked Example

9

(Use of Possibility Diagram) Two fair dice are rolled. What is the probability that i. both dice show the same number, ii. the number shown on the first die is greater than the number shown on the second die?

Solution:

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i. P(both dice show the same number)

ii. P(number shown on first die is greater than the number shown on second die)

Mark out the favorable outcomes on the possibility diagram. Count the number of

for (i) and the number of

for (ii).

1. A fair tetrahedral die (4-sided die) and a fair 6-sided die are rolled simultaneously. The numbers on the tetrahedral die are 1, 2, 5 and 6 while the numbers on the 6-sided die are 1, 2, 3, 4, 5 and 6. a. Display all the outcomes of the experiment using a possibility diagram. b. Using the possibility diagram or otherwise, find the probability that i. both dice show the same number, ii. the number shown on the tetrahedral die is greater than the number shown on the 6-sided die, iii. the numbers shown on both dice are prime numbers. ******ebook converter DEMO Watermarks*******

2. A bag contains five cards and the cards are numbered 1, 2, 3, 4 and 5. A card is drawn at random from the bag and its number is noted. The card is then replaced and a second card is drawn at random from the bag. Using a possibility diagram, find the probability that i. the number shown on the second card is greater than the number shown on the first card, ii. the sum of the two numbers shown is greater than 7, iii. the product of the two numbers shown is greater than 10. There is another way to draw a possibility diagram to represent the sample space for rolling two fair dice, as shown in Worked Example 10.

Exercise 11C Questions 4, 11, 20

Worked Example

10

(Use of Possibility Diagram) Two fair dice are rolled. Find the probability that the sum of the numbers shown on the dice is i. equal to 5, ii. even.

Solution:

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1. P(sum is equal to 5)

2. P(sum is even)

The sum of the numbers is shown in each cell. In your own possibility diagram, you can draw double lines to avoid accidentally counting the numbers in the first row and in the first column, when counting the number of favorable outcomes.

Mark out the favorable outcomes on the possibility diagram. Count the number of

for (i) and the number of

for (ii).

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1. The numbers on a fair tetrahedral die are 1, 2, 5 and 6 while the numbers on a fair 6-sided die are 1, 2, 3, 4, 5 and 6. The two dice are rolled at the same time and the scores on both dice are recorded. The possibility diagrams below display separately some of the values of the sum and product of the two scores.

a. Copy and complete the possibility diagrams. b. Using the possibility diagrams, find the probability that the sum of the scores is i. even, ii. divisible by 3, iii. a perfect square, iv. less than 2. c. Using the possibility diagrams, find the probability that the product of the scores is i. odd, ii. larger than 12, iii. a prime number, iv. less than 37. 2. A circular card is divided into 3 equal sectors with scores of 1, 2 and 3. ******ebook converter DEMO Watermarks*******

The card has a pointer pivoted at its center. The pointer is spun twice. Each time the pointer is spun, it is equally likely to stop at any of the sectors.

a. With the help of a possibility diagram, find the probability that i. each score is a ‘1’, ii. at least one of the scores is a ‘3’. b. In a game, a player spins the pointer twice. His final score is the larger of the two individual scores if they are different and their common value if they are the same. The possibility diagram below shows the player’s final score.

i. Copy and complete the possibility diagram. ii. Using the diagram, find the probability that his final score is even. iii. Using the same diagram, find the probability that his final score is a prime number.

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Exercise 11C Questions 5-6, 12-15, 21

Tree Diagrams The sample space for tossing a fair coin is {H, T}. The sample space for tossing two fair coins can be represented by a possibility diagram, as shown in Fig. 11.3. How can we represent the sample space for tossing three fair coins?

Fig. 11.3 We use a different type of diagram called a tree diagram to represent the sample space, as shown in Fig. 11.4. The following steps show how the tree diagram is constructed. 1. When the first coin is tossed, there are two possible outcomes, head (H) or tail (T), so we start with a point and draw two branches H and T.

2. The second coin is then tossed. Regardless of the outcome of the first toss, the second coin would also yield either a H or a T, thus we draw two branches after the H and the T from the first toss as shown below. There are a total of 2 × 2 = 4 branches, i.e. there are 4 possible outcomes at this stage.

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3. The third coin could also yield two outcomes when the first two outcomes are HH, HT, TH or TT. Thus we obtain the tree diagram as shown in Fig. 11.4.

Fig. 11.4 From Fig. 11.4, we observe that there are a total of 2 × 2 × 2 = 8 branches, i.e. the total number of possible outcomes is 8. In summary, Example of

Components of

Representation of

Experiment

Each Outcome

Sample Space

Tossing 1 coin

1

List of outcomes in a set

Tossing 2 coins

2

Possibility diagram or tree diagram

Tossing 3 coins

3

Tree diagram

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In a family, there are two children and one of them is a boy. What is the probability that the other child is a girl?

Worked Example

11

(Use of Tree Diagram) Three fair coins are tossed. Find the probability that i. there are two heads and one tail, ii. there is at least one tail.

Solution:

i. P(two

heads

and

one tail) (see shaded regions)

ii. P(at least one tail)

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=

1. Carlo is a darts player. There is an equal probability that he will hit or miss the bull’s-eye. He aims for the bull’s-eye and attempts 3 throws. Using a tree diagram, find the probability that i. he misses the bull’s-eye once, ii. he hits the bull’s-eye at least once. 2. Box A contains 4 pieces of paper numbered 1, 2, 3 and 4. Box B contains 2 pieces of paper numbered 1 and 2. One piece of paper is removed at random from each box. a. Copy and complete the following tree diagram.

b. Find the probability that i. at least one ‘1’ is obtained, ii. the sum of the two numbers is 3, iii. the product of the two numbers is at least 4, iv. the sum is equal to the product.

Exercise 11C Questions 7, 16-18

Students may also use a possibility diagram to solve, if no diagram is provided.

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BASIC LEVEL 1. A fair coin and a fair 6-sided die are tossed and rolled respectively. Using set notations, list the sample space of the experiment. 2. A box contains 7 pens, 3 of which are faulty. A pen is drawn from the box at random. Using set notations, list the sample space of this probability experiment and find the probability that the pen drawn is not faulty. Suppose that the first pen drawn is not faulty and it is not replaced in the box. A second pen is now drawn from the box. Using set notations, find the probability that the second pen drawn is faulty. 3. Each letter of the word ‘POSSIBILITY’ is written on identical cards. One card is chosen at random. Using set notations, find the probability that the letter on the chosen card is i. a ‘S’, ii. a ‘P’ or an ‘I’, iii. a vowel, iv. a consonant. 4. A box contains three cards bearing the numbers 1, 2 and 3. A second box contains four cards bearing the numbers 2, 3, 4 and 5. A card is chosen at random from each box. a. Display all the possible outcomes of the experiment using a possibility diagram. b. With the help of the possibility diagram, calculate the probability that i. the cards bear the same number, ii. the numbers on the cards are different, iii. the larger of the two numbers on the cards is 3. ******ebook converter DEMO Watermarks*******

5. Six cards numbered 0, 1, 2, 3, 4 and 5 are placed in a box and well-mixed. A card is drawn at random from the box and the number on the card is noted before it is replaced in the box. The cards in the box are thoroughly mixed again and a second card is drawn at random from the box. The sum of the two numbers is then obtained. a. Copy and complete the possibility diagram below, giving all the possible sums of the two numbers. Some of the possible sums are shown.

b. How many possible outcomes are there in the sample space of this experiment? c. What is the probability that the sum of the two numbers i. will be 7, ii. will be a prime number, iii. will not be a prime number, iv. will be even, v. will not be even? d. Which sum is more likely to occur, the sum of 7 or the sum of 8? 6. It is given that X = {4, 5, 6} and Y = {7, 8, 9}. An element x is selected at random from X and an element y is selected at random from Y. The possibility diagrams below display separately some of the values of x + y and xy. ******ebook converter DEMO Watermarks*******

a. Copy and complete the possibility diagrams. b. Find the probability that the sum x + y is i. prime, ii. greater than 12, iii. at most 14. c. Find the probability that the product xy is i. odd, ii. even, iii. at most 40. 7. A fair coin is tossed three times. Display all the possible outcomes of the experiment using a tree diagram. From your tree diagram, find the probability of obtaining i. three heads, ii. exactly two heads, iii. at least two heads. INTERMEDIATE LEVEL 8. Bag P contains a red, a blue and a white marble while bag Q contains a blue and a red marble. The marbles are identical except for their color. A marble is picked at random from both bag P and bag Q. List all the possible outcomes of the sample space. Find the probability that the two marbles selected are ******ebook converter DEMO Watermarks*******

i. of the same color, ii. blue and red, iii. of different colors. 9. A two-digit number is formed using the digits 1, 2 and 3. Repetition of digits is allowed. a. List the sample space. b. Find the probability that the two-digit number formed is i. divisible by 3, ii. a perfect square, iii. a prime number, iv. a composite number. 10. The three daughters-in-law of Mrs Ramos are happily awaiting the arrival of their bundles of joy within the year. List the sample space of the sexes of the three babies, given that the babies are equally likely to be either a boy or a girl. Hence, find the probability that Mrs Ramos will have i. three grandsons, ii. two grandsons and one granddaughter, iii. one grandson and two granddaughters. 11. In an experiment, two spinners are constructed with spinning pointers as shown in the diagrams below. Both pointers are spun. Each time the pointer is spun, it is equally likely to stop at any sector.

a. Find the probability that the pointers will point at i. numbers on the spinners whose sum is 6, ii. the same numbers on both spinners, ******ebook converter DEMO Watermarks*******

iii. different numbers on the spinners, iv. two different prime numbers. b. What is the probability that the number on the first spinner will be less than the number on the second spinner? 12. In a game, the player throws a fair coin and a fair 6-sided die simultaneously. If the coin shows a head, the player’s score is the score on the die. If the coin shows a tail, then the player’s score is twice the score on the die. Some of the player’s possible scores are shown in the possibility diagram below.

a. Copy and complete the possibility diagram. b. Using the diagram, find the probability that the player’s score is i. odd, ii. even, iii. a prime number, iv. less than or equal to 8, v. a multiple of 3. 13. Two fair 6-sided dice were thrown together and the difference of the resulting numbers on their faces was calculated. Some of the differences are shown in the possibility diagram below.

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a. Copy and complete the possibility diagram. b. Using the diagram, find the probability that the difference of the two numbers is i. 1, ii. non-zero, iii. odd, iv. a prime number, v. more than 2. 14. A bag contains 5 identical balls which are numbered 1, 2, 4, 5 and 7. Two balls are drawn at random, one after another and without replacement. Find the probability that the i. numbers obtained on both balls are prime, ii. sum of the numbers obtained is odd, iii. product of the numbers obtained is greater than 20, iv. difference in the numbers obtained is less than 7, v. product of the numbers obtained is divisible by 9. 15. The diagrams below show two circular cards, each with a pointer pivoted at its center. The first card is divided into 4 equal sectors with scores 1, 2, 4 and 5. The second card is divided into 4 equal sectors with scores 0, 1, 3 and 5. In a game, both pointers are spun. Each time the pointer is spun, it is ******ebook converter DEMO Watermarks*******

equally likely to stop at any sector.

Find the

probability that the i. scores on both cards are the same, ii. scores on both cards are prime, iii. sum of the scores is odd, iv. sum of the scores is divisible by 5, v. sum of the scores is 6 or less, vi. product of the scores is not 0, vii. product of the scores is greater than 11. 16. A spinner with three equal sectors (shown below) and a fair coin are used in a game. The spinner is spun once and the coin is tossed once. Each time the pointer is spun, it is equally likely to stop at any sector.

Calculate the probability of getting i. red on the spinner and tail on the coin, ii. blue or yellow on the spinner and head on the coin. 17. A bag contains 3 cards numbered 1, 3 and 5. A second bag contains 3 cards numbered 1, 2 and 7. One card is drawn at random from each bag. Calculate the probability that the two numbers obtained i. are both odd, ii. are both prime, ******ebook converter DEMO Watermarks*******

iii. have a sum greater than 4, iv. have a sum that is even, v. have a product that is prime, vi. have a product that is greater than 20, vii. have a product that is divisible by 7. 18. A fair die is made from a tetrahedron such that each of its four faces is printed with one number. The numbers are 1, 2, 3 and 4. a. When the die is rolled, what is the probability that i. it will land with the face printed ‘4’ down, ii. it will land such that the sum of the three upper faces is an odd number? b. If the same die is rolled and a fair coin is tossed at the same time, list all the possible outcomes using a tree diagram. ADVANCED LEVEL 19. A box contains 7 electrical components. The box was dropped in transit and 1 of the components became defective, but not visibly.The components are taken out from the box at random and tested until the defective component is obtained. What is the probability that the defective component is the first component tested? 20. Hotel Y is a two-storey hotel, with rooms (and their respective room numbers) arranged as shown in the diagram below. Rooms are allocated at random when guests arrive and each guest is allocated one room. Kate and Kristel arrive at Hotel Y on a particular day. Upon their arrival, none of the rooms in Hotel Y are occupied.

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a. With the help of the possibility diagram, find the probability that Kristel and Kate i. stay next to each other, ii. stay on different storeys, iii. do not stay next to each other. b. Suppose the hotel accepts Kate’s request that she only wants to be allocated rooms on the second floor, what will be the probability that she will be staying next to Kristel? 21. Two fair tetrahedral dice and a fair 6-sided die are rolled simultaneously. The numbers on the tetrahedral dice are 1, 2, 3 and 4 while the numbers on the 6-sided die are 1, 2, 3, 4, 5 and 6. What is the probability that the score on the 6-sided die is greater than the sum of scores of the two tetrahedral dice?

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1. Probability is a measure of chance. 2. A sample space is the collection of all the possible outcomes of a probability experiment. 3. In a probability experiment with m equally likely outcomes, if k of these outcomes favor the occurrence of an event E, then the probability, P(E), of the event happening is given by:

4. For any event E, 0 ≤ P(E) ≤ 1. P(E) = 0 if and only if E is an impossible event, i.e. it will never occur. P(E) = 1 if and only if E is a certain event, i.e. it will definitely occur. 5. For any event E, P(not E) = 1 – P(E). 6. Example of Experiment

Components of Each Outcome

Representation of Sample Space

Tossing 1 coin

1

List of outcomes in a set

Possibility diagram or tree diagram ******ebook converter DEMO Watermarks******* Tossing 2 coins

2

Tossing 3 coins

3

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Tree diagram

1. Each of the numbers 5, 6 and 8 is written on a card. One or more of these cards are drawn at random to form a one-, two- or three-digit number. a. For this experiment, i. write down the sample space, ii. state the total number of possible outcomes. b. Find the probability that the number formed i. consists of two digits, ii. is a multiple of 5. 2. A 6-sided fair die is rolled once. Find the probability of getting i. an even number, ii. a composite number, iii. a number that is divisible by 4. 3. All the 26 red cards from a standard pack of playing cards are mixed thoroughly. A card is then drawn at random. Find the probability of drawing i. the queen of hearts, ii. the jack of clubs, iii. either the six of hearts or the seven of diamonds, iv. a card which is not a nine. 4. In a shopping mall, if a customer spends a minimum of ₱1600 in a single receipt, he has the chance to spin a wheel to win a prize. The wheel is ******ebook converter DEMO Watermarks*******

divided into 6 equal sectors. The prizes correspond to the letters on the wheel. A: ₱1000 shopping voucher B: ₱1500 supermarket voucher C: watch worth ₱2000 D: umbrella E: ₱1200 dining voucher F: 1 kg cheesecake Find the probability that a customer who spins the wheel wins i. an umbrella, ii. a voucher, iii. ₱3000 cash. 5. A reel of a fair slot machine has 22 symbols, of which 4 are cherries, 7 are oranges, 9 are peaches and the rest are grapes. Find the probability that the reel will stop to show i. the symbol ‘orange’, ii. the symbol ‘grape’, iii. the symbol ‘pineapple’, iv. either the symbol ‘cherry’ or the symbol ‘peach’. 6. A bag contains 20 sweets, of which 7 are toffee wrapped in green paper, 6 are mints wrapped in green paper, 3 are toffee wrapped in red paper and 4 are mints wrapped in red paper. If a sweet is drawn at random from the bag, find the probability that the sweet is i. a mint wrapped in red paper, ii. a toffee, iii. wrapped in green paper. ******ebook converter DEMO Watermarks*******

7. A bag contains 7 white staplers and 11 orange staplers. a. If a stapler is drawn at random from the bag, find the probability that the stapler is i. green, ii. either white or orange. b. 12 red staplers are added to the bag. A stapler is then drawn at random from the bag. Find the probability that the stapler is i. red, ii. not orange. 8. A survey is conducted to find out how the students in a class travel to school. The students either take the bus, the train, the car or walk to school. The pie chart shows the results of the survey. A student is selected at random. Find the probability that the student travels to school

i. by car, ii. by train or on foot, iii. by bicycle. 9. A bed of flowers consists of 100 stalks of flowers, of which 20 are lilies, h are roses and the rest are tulips. i. Given that the probability of picking a stalk of tulip at random is , find the value of h. ii. 10 stalks of lilies are removed from the bed. A stalk of flower is picked at random from the remaining stalks of flowers. Find the probability that ******ebook converter DEMO Watermarks*******

a stalk of rose is picked. 10. In a car park, there are 125 cars, 3p motorcycles, 2q lorries and 20 buses. One of the vehicles leaves the car park at random. i. Given that the probability that the vehicle is a motorcycle is equation in p and q. ii. Given that the probability that the vehicle is a bus is equation in p and q.

, form an

, form another

iii. Hence, find the value of p and of q. 11. A man throws a die and a coin. Find the probability that he will get i. the number ‘3‘ followed by a head, ii. an even number followed by a tail. 12. Two balanced dice are thrown together. Find the probability that they will show i. the same number, ii. two even numbers, iii. two odd numbers, iv. one odd and one even number. 13. 50 discs, numbered from 1 to 50, are placed in a bowl. One disc is picked at random. Find the probability that the number on the disc i. is greater than 28, ii. includes the digit `3`, iii. is prime, iv. is divisible by 4. 14. Six discs, with the numbers 1 to 6 written on each of them, are placed in a bag. Two discs are drawn at random from the bag and placed side by side to form a two-digit number. By drawing a possibility diagram, find the probability that the number formed is ******ebook converter DEMO Watermarks*******

i. divisible by 2, ii. divisible by 5, iii. a prime number, iv. a perfect square. 15. A three-digit number is formed using the digits 1, 2 and 3. a. Display all the possible outcomes using a tree diagram. b. Find the probability that the three-digit number formed i. contains at least one digit 3, ii. is odd. 16. Bag A contains 2 red cards, 4 yellow cards and 2 green cards. Bag B contains 3 red cards and 5 yellow cards. A card is picked at random from both bag A and bag B. By drawing a tree diagram, find the probability of getting a. a red card, b. two yellow cards, c. two cards of different colors.

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1. A standard pack of 52 playing cards is randomly divided into two unequal piles. Given that the probability of drawing a picture card from the smaller pile is while the probability of drawing a non-picture card from the bigger pile is

, find the number of cards in each pile.

2. Chris writes 3 letters to 3 of his friends, Daniel, Angelo and Miguel. He types each of their addresses on each of the 3 envelopes and puts the letters into the envelopes randomly before he sends them out. Find the probability that i. exactly one of his friends receives the correct letter, ii. exactly two of his friends receive the correct letters, iii. all three of his friends receive the correct letters. 3. Antonio rolls two identical fair six-sided dice but does not know the result. He was told that one of the scores is a ‘3’. Given this information, what is the probability that both of the scores are ‘3’s? Hint: Analyze the possible outcomes using a possibility diagram.

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D1 Revision Exercise 1. The length of two sides of an isosceles triangle are 24 cm and 12 cm. What is the length of the third side? Explain why. 2. The figure shows ΔBCD with straight lines ABC, CDE and FBD. Prove that ∠ABF < ∠BDE.

3. Angelo has a long stick and breaks it to form a pentagon of sides JK = 4.5 cm, KL = 9 cm, LM = 7.5 cm, MN = 12 cm and JN = 9 cm. He uses the remaining stick to join up point K to point M and point K to point N. Given that KM = KN, find the range of length of the remaining stick that he needs.

4. A two-digit number is chosen at random. Find the probability that the number is i. an even number, ii. a prime number less than 18, iii. a multiple of 10, ******ebook converter DEMO Watermarks*******

iv. divisible by both 3 and 4. 5. All the hearts are removed from a standard pack of 52 playing cards. A card is drawn at random from the remaining cards. Find the probability of drawing i. the ten of hearts, ii. a red card, iii. either a king or an ace, iv. a card which is not a two.

D2 Revision Exercise 1. In the diagram, ABC and PBQ are straight lines, AB̂R = 47°, BR̂S = 147°, PB̂C = 80° and BĈD = 82°. Name a pair of parallel lines and explain why they are parallel.

2. 28 students from the Ecology Club are planning to visit the Marine Life Park during the December holidays. Of the 16 girls, 2 of them have visited the park. 6 boys have not visited the park. If a student is chosen at random, find the probability that the student i. is a boy who has visited the park, ******ebook converter DEMO Watermarks*******

ii. has not visited the park. 3. There are 48 boys and 2 girls on the country’s badminton team. After a sports fair, x boys and 2x girls join the team. Given that the probability of selecting a boy at random to represent the country for a competition is now , find the number of girls who join the team. 4. Two coins are tossed at the same time. If S is the sample space, list all the possible outcomes. Find the probability of obtaining a. two tails, b. a head and a tail. 5. Two four-sided dice are thrown together and the product of the resulting numbers is calculated. Some of the products are shown in the possibility diagram. a. Copy and complete the possibility diagram. b. Using the diagram, find the probability that the product of the two numbers is i. odd, ii. even, iii. more than or equal to 6, iv. a multiple of 3, v. not a prime number. See Table. 1. A bag P contains 3 red sweets, 1 yellow sweet and 2 green sweets. Another bag Q contains 2 red sweets and 4 green sweets. A fair coin is tossed. If the result of the toss is head, a sweet is drawn from bag P, otherwise a sweet is drawn from bag Q. By constructing a tree diagram, find the probability of getting 1. a red sweet, 2. a yellow sweet, ******ebook converter DEMO Watermarks*******

3. a green sweet.

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Problems in Real-World Contexts PROBLEM 1: Purchasing an Apartment (Stamp Duty) i. To purchase an apartment, Mr Ramos has to pay stamp duty to the property developer. The apartment costs 16 000 000 and his property agent tells him that the formula used to calculate the stamp duty is Stamp Duty = 3% × Selling Price of Apartment − 180 000. How much stamp duty does Mr Ramos have to pay? ii. Mr Ramos checks the property developer’s website on the Internet and finds that the formula used to calculate the stamp duty is different. First 6000 000 : 1% × Selling Price of Apartment Next 6000 000 : 2% × Selling Price of Apartment Thereafter : 3% × Selling Price of Apartment Using this formula, calculate the stamp duty that Mr Ramos has to pay. Is it the same as the answer you obtained in (i)? iii. Mr Ramos is curious to know why the formula in (i) works. How do you prove to him that both formulae can be used to calculate the stamp duty? iv. What is the minimum selling price of the apartment, below which the formula in (i) will not work? Explain your answer. v. Mr Ramos notices that, according to the formula in (ii), the stamp duty for the first 12 000 000 is 180 000, which also appears in the formula in (i). He wonders whether this is a coincidence or whether it will always work. Think of a counterexample to show that this is just a coincidence. ******ebook converter DEMO Watermarks*******

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PROBLEM 2: Climate Change Climate change is defined as significant variations in the statistical observations of the weather conditions, which occur over an extended period of time. a. An indication of climate change is the overall increase in the annual mean surface temperature. Studies show that the annual mean surface temperature in Country X has increased from 26.8 ˚C in 1948 to 27.6 ˚C in 2011. Find the percentage increase in the annual mean surface temperature from 1948 to 2011, giving your answer correct to 2 decimal places. b. One of the contributors of climate change is greenhouse gas emissions. Country X’s emissions were 41 million tonnes (MT) in 2005. Its BusinessAs-Usual (BAU) emissions are expected to reach 77.2 MT in 2020. The pie chart shows a projection of its BAU emissions for the year 2020 for various sectors.

Note: Others include waste, water and other electricity use. Which of the above accounts for the majority of Country X’s projected emissions in 2020? Find the amount of emissions contributed by that sector. c. Find out some measures that have been put in place to reduce emissions and to mitigate the effects of climate change. ******ebook converter DEMO Watermarks*******

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PROBLEM 3: Smartphone Price Plans Smartphone price plans usually include a fixed monthly component and a variable component depending on the monthly usage. The user is entitled to a certain amount of free talk time and data usage. There are three telecommunication companies providing mobile services. In groups of three or four, work out how you can advise your classmates to choose one of the following plans. See Table. Guiding Questions 1. Are there any assumptions that you may need to make? 2. Other than the basic monthly subscription fee, what are some other factors you should consider? 3. How do you solve this problem? Explain your method clearly.

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PROBLEM 4: Conical Wine Glass The photograph shows a conical glass.

A problem commonly faced by bartenders is when their customers request for only half a glass of cocktail. Bartender Sam: “I always fill the cocktail to half the depth of the glass.” Bartender Carlo: “I disagree. I always fill the cocktail to two-thirds of the depth of the glass.” Bartender Miguel: “I believe that the cocktail should be poured to threequarters of the depth of the glass.” Bartender Chris: “No, I think that the cocktail should be poured to four-fifths of the depth of the glass.” i. Which of the bartenders’ method is obviously wrong? Explain your answer. ii. Make a guess whose method is the closest to getting half a glass of cocktail. iii. Formulate the problem mathematically to determine which of the bartenders’ method is the most accurate.

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Problem 5: Constructing a 3-Dimensional Star To raise funds for charity, your class has been tasked to make and sell starshaped paper lanterns. You will have to design 3-dimensional stars by creating a 2-dimensional template of each star and determining the amount of materials that are required.

You may have to consider the following: Creation of a 2-dimensional template Materials for lantern, e.g. colored paper, glue, string, lightbulbs Size and color of paper to be used and amount of paper needed Amount of money to be raised Guiding Questions: 1. How can we obtain the dimensions of the 2-dimensional template? Consider a star which is about 20 cm wide and 6 cm deep, with cross section and a 3-dimensional section APBH as shown in Fig. (a). TA = BQ = 8 cm, AB = 4 cm and height of star above centre O, OH = 3 cm. Note that TABQ is not a straight line. Fig. (b) shows the template used to make one part of the 3-dimensional star with cross section OAPB. You can fold the template along the lines and glue the flap on side PA’ to PA to make this part of the 3-dimensional star. You need to make 5 such parts and glue each to the other along the remaining 2 flaps to make the 3-dimensional star.

The depth of the star is 6 cm because OH = 3 cm and OH’ = 3 cm, where H’ is the point directly below centre O on the other surface of the 3******ebook converter DEMO Watermarks*******

dimensional star.

2. How do you calculate the amount of paper needed for each star? What size of paper is appropriate and available? How many sheets of paper would be needed? 3. What is the total cost of materials? 4. Propose a price for a lantern. Is the price proposed reasonable? How many lanterns have to be sold to raise the amount of money?

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Practice Now Answers Chapter 1 Practice Now 1 (a)

–5(2x – 5)

(b)

8(3x + 2)

(c)

x(2 + k)

(d)

2ab(1 + 2c)

(e)

9a(2 – 6y + 4z)

Practice Now (Page 7) (a)

(x + 1)(x + 5)

(b)

(x – 1)(x – 5)

(c)

(x + 2)(x + 6)

(d)

(x – 2)(x – 6)

(e)

(x – 5)(x + 3)

Practice Now 2 (a)

(x + 1)(x + 7)

(b)

(x – 4)(x – 7)

(c)

(x − 1)(x + 2)

(d)

(x + 1)(x − 8)

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Practice Now 3 (a)

(2x + 3)(x + 4)

(b)

(5x − 3)(x − 2)

(c)

(−2x + 3)(x − 3)

(d)

3(3x – 8)(x – 1)

(e)

(–3x – 1)(x – 2)

Practice Now 4 1.

2.

(a)

(x + 3y)(x – 5y)

(b)

(6x + 5y)(x + y)

(c)

(3a – b)2

(a)

(xy + 4)2

(b)

(3xy – 8)(xy – 2)

Practice Now 5 1.

2.

(a)

(x + 6)2

(b)

(2x + 5)2

(c)

(3x + 1)2

(a) (b)

Practice Now 6 1.

(a)

(2 − 9x)2

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2.

(b)

(5x − y)2

(c)

(3x − 2y)2

(a) (b)

Practice Now 7 1.

(a)

(6x + 11y)(6x – 11y)

(b)

(x + 8)(x – 8)

(c)

(9 + 2x)(9 – 2x)

2. 3.

(2x + 9)(2x – 5)

Practice Now 8 (a)

81 600

(b)

41 200

(c)

26

(d)

11 000

(e)

6800

Practice Now 9 (a)

4x(2xy + 1)

(b)

pr(r + l)

(c)

–a3by + a2y = a2y(–ab + 1)

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(d)

3c2d + 6c2d2 + 3c3 = 3c2(d + 2d2 + c)

(e)

p2q – 2pq2 + 4p2q2 = pq(p – 2q + 4pq)

Practice Now 10 (a)

(x + 1)(2 + a)

(b)

(x + 2)(9 – b)

(c)

3(2x – 3)(c – 2d )

(d)

(4 – x)(7h + 1)

(e)

2(3x – 2)(c + d)

Practice Now 11 (a)

(y + 4)(x + 3)

(b)

(x + 3y)(4a + b)

(c)

(x – 1)(x2 + 1)

(d)

(3y – 2)(2x + z)

(e)

(z – y)(x2 + 4)

Practice Now 12 (a)

(6p + 7q2)(36p2 – 42pq2 + 49q4)

(b)

(x + 3)(x2 – 3x + 9)

(c)

3(x – 4)(x2 + 4x + 16)

(d)

(1 – 2x)(37 + 20x+ 4x2)

(e)

(p + 8q)(7p2 – 23pq + 43q2)

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Chapter 2 Practice Now 1 (a) (b) (c) (d) (e) Practice Now 2 (a) (b) (c) (d) Practice Now 3 1.

(a) (b) (c)

2.

(a) (b)

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Practice Now 4 1.

(a) (b) (c) (d)

2.

(a)

-

Practice Now 5 1.

(a) (b) (c)

2.

(a) (b) (c)

Practice Now 6 (a) (b) (c) (d) (e) ******ebook converter DEMO Watermarks*******

Practice Now 7 1.

(i) (ii)

2.

a = 10

(i) (ii)

4

Practice Now 8 1.

(i) (ii)

2.

(i) (ii)

k = −27

Practice Now 9 1.

(i) (ii)

2.

(i) (ii)

3.

x=1

x = ±3

(i) (ii)

Practice Now 10 1.

(i)

y=2

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(ii) 2.

y=1

(i) (ii)

3.

b = 13

4. Practice Now 11 (a) (b) (c)

x= −3

(d) (e)

b=– 27

Chapter 3 Practice Now 1 1.

(ii)

q = 2.5

Practice Now 2 (a)

−

(b)

3

(c)

−1

(d) ******ebook converter DEMO Watermarks*******

(e)

−6

Practice Now 3 (a) (b)

−1

(c)

0

(d) (e)

−2

Practice Now 4 (a)

h = 12

(b)

k=2

(c)

p=4

(d)

w=2

(e)

r = −4

Chapter 4 Practice Now (Page 78) (a)

y = 1, y = –3.5

Practice Now (Page 80) (a)

x = 4, x = –1.2

Practice Now 1 (a) ******ebook converter DEMO Watermarks*******

(b)

y=4

(c)

x = –3

(d)

6y = x + 6

(e)

y = –2x – 2

Practice Now 2 (a)

y = −2x + 7

(b)

y = 3x + 2

(c)

y=3

(d)

2y = −x − 8

(e)

y = ax − a2

Practice Now 3 (a)

2y = −x + 1

(b)

y = 2x + 4

(c)

y = −3x − 3

(d)

2y = x − 8

(e)

y = 6x + 2

Practice Now 4 (a)

7

(c)

4

(d)

(ii) 0.5

(e)

y = –3x + 1. They are the same line.

Practice Now 5 ******ebook converter DEMO Watermarks*******

1.

x = 1, y = 2

2.

x = –1, y = 2

Practice Now 6 1.

(a) x = 4, y = 1 (b) x = 3, y = –1 (c) x = 1, y = –1 (d) x = 3, y = –1

2.

x = –3, y = 5

Practice Now 7 (a)

x = 3, y = 4

(b)

x = 3, y = –1

Practice Now 8 (a)

x = 1, y = –2

(b)

x = 1, y = –3

Practice Now 9 x = 10, y = 3 Practice Now 10 x = –1, y = 2 Practice Now 11 x = 2, y = –1 ******ebook converter DEMO Watermarks*******

Practice Now 12 (a)

x = 9, y = 15

(b)

x = –3, y = 6

Practice Now 13 1.

13.5, 22.5

2.

34°, 146°

3.

30 cm

Practice Now 14

Practice Now 15 1. 10, 40 2. 12 years old, 16 years old 3. ₱70 Practice Now 16 65 Chapter 5 Practice Now 2 y < x, x + 2y ≤ 8 Practice Now 3 (a) x + y ≤ 1000, y ≥ 2x, x ≥ 100, ******ebook converter DEMO Watermarks*******

y ≤ 800 (c)

330 cans of Coola and 670 cans of Shiok

Practice Now (Page 145) 1.

2.

(i)

y=5

(ii)

x = –1

(i) (ii

Chapter 6 Practice Now 1 (a)

Yes

(b)

No

Practice Now 2 (a)

11

(b)

5a + 6

(c)

x=1

Practice Now 3 (b)

a = 2, b = –3

Practice Now (Page 158) 1.

6, 4, 0

2.

(a) −5, 5

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3.

(a)

Practice Now 4 –5 < g(x) ≤ –1 Practice Now 5 5 ≤ x ≤ 10 Chapter 7 Practice Now 1 (a)

True statement

(b)

False statement

(c)

True statement

Practice Now 2 (a)

True

(b)

False

(c)

True

(d)

False

Practice Now 3 (a)

False

(b)

True

(c)

False

(d)

True

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Practice Now 6 False Practice Now 7 False Practice Now 8 True Practice Now 9 (a)

Inductive reasoning

(b)

Deductive reasoning

Chapter 8 Practice Now 1 x = 3; Addition Axiom Practice Now 2 (b)

False

Chapter 9 Practice Now 1 A, H; B, E; C, F; D, G, I Practice Now 2 (i)

5

(ii) DC ******ebook converter DEMO Watermarks*******

(iii)

AD, 2

(iv)

BC, 5.3

(v)

∠ABC, 90

Practice Now 3 (a)

Not congruent

(b)

ΔDEF ≡ ΔTSU

(c)

Not congruent

Practice Now 4 (a)

(b)

(i)

38°

(ii)

28°

(iii)

28°

(iv)

27 cm

(v)

9 cm

AC and ED are parallel lines.

Practice Now 5 1.

E, F, D; EF, FD, 11, ED; EFD, SSS

2.

ΔWXY ≡ ΔWZY

Practice Now 6 2.

No

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Practice Now 7 1.

(ii)

25°

2.

(i)

ΔPQS ≡ ΔRSQ

(ii)

7 cm, 140°

Chapter 10 Practice Now 1 4.5 < PR < 19.5 Practice Now 3 x<6 Practice Now 4 ∠YXZ Practice Now 5 x = 35 Chapter 11 Practice Now 1 Red, Orange, Purple, Yellow, Green; 5

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Practice Now 2

(a)

B1, B2, B3, B4, B5, R1, R2, R3, R4; 9

(b)

N1, A1, T, I, O, N2, A2, L; 8

(c)

357, 358, 359, …, 389; 33

Practice Now 3 (i) (ii) (iii) ******ebook converter DEMO Watermarks*******

(iv)

0

Practice Now 4 (i) (ii) (iii) (iv) Practice Now 5 1.

(i) (ii) (iii

2.

(i) (ii)

3.

(iii

0

(iv)

1

12

Practice Now 6 (i) (ii) Practice Now 7 ******ebook converter DEMO Watermarks*******

(i) (ii) (iii)

0

(iv) Practice Now 8 1.

(i)

14 + x

(ii) (iii 2.

6

4

Practice Now 9 1.

(b)

(i)

(ii)

(iii) 2.

(ii)

(i) (iii)

Practice Now 10 1.

(a)

2, 6, 7; 3, 4, 7, 8; 4, 5, 9; 5, 6, 9, 10; 6, 7, 10, 11; 8, 11, 12 1, 2, 5, 6; 2, 10, 12; 3, 6, 15; 4, 8, 20, 24; 5, 10, 25, 30; 6, 12, 36

(b)

(i)

(ii) (iii)

(iv) 0

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(c)

2.

(i)

(ii)

(iii)

(iv) 1

(a)

(i)

(ii)

(b)

(ii)

(iii)

Practice Now 11 1.

(i)

2.

(b)

(ii) (i) (iii)

(ii) (iv)

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Answers Exercise 1A 1. (a) (b) (c) (d) (e) 2. (a) (b) (c) (d)

3(4x – 3) –5(5y + 7) 9b(3 – 4y) 4a(2x + 3 – z) 2m(2 – 3y – 9z) a(3 − x) d(8ce − 2c − e)

(e) 3. (a) (b) (c) (d) (e) (f)

4a(2b − bc + 3c − 1) –13b(3b + a) 5x(1 + 2b + 2c) 3x(–y + 2z) 2x(–6y – 7) 3a(5b – 8) 4y(2x – 7)

Exercise 1B 1. (a) (b) (c) (d) (e) (f) (g)

(a + 1)(a + 8) (b + 3)(b + 5) (c – 4)(c – 5) (d – 2)(d – 14) (f – 2)(f + 8) (h – 10)(h + 12) (k + 2)(k – 6)

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(h) 2. (a) (b) (c) (d) (e) (f) (g) (h) 3. (a) (b) (c) (d) (e) (f) (g) (h) 5. (a) (b)

(m + 1)(m – 21) (3n + 7)(n + 1) (2p + 1)(2p + 3) (3q – 4)(2q – 3) (4r – 3)(r – 1) (4s – 5)(2s + 3) (6t – 5)(t + 4) (2u + 3)(2u – 7) (9w + 13)(2w – 3) (–a + 7)(a + 5) (–3b + 1)(b – 25) 2(2c + 1)(c + 2) 5(d – 5)(d – 24) 4(2f – 5)(f + 3) 3(8h + 3)(h – 1) 2(–k + 5)(2k + 3) 5n(7m – 6)(m + 1) (4p – 3)(p + 3) –0.2r(16q – 3)(4q + 1)

Exercise 1C 1. (a) (a + b)(a + 2b) (b) (x + 2y)(2x + y) (c) (2c + d)(c + 3d) (d) (p + 3q)(3p + q) 2. (a) (a – b)(a + 4b) (b) (c + 3d)(c – 7d) (c) (2h – 3k)(h + 5k) (d) (3m + 2n)(m – 6n) (e) 3(p + 2q)(p + 3q) (f) t(2r – 5s)(r – 2s) 3. (a) (xy – 3)(xy + 5) (b) (4xy + 5)(3xy – 8) (c) 2z(2xy – 3)(xy – 4) ******ebook converter DEMO Watermarks*******

(d)

(3x – 2y)(2x + 3y)

Exercise 1D 1. (a) (a + 7)2 (b) (2b + 1)2 (c) (c + d)2 (d) (2h + 5k)2 2. (a) (m – 5)2 (b) (13n – 2)2 (c) (9 – 10p)2 (d) (7q – 3r)2 3. (a) (b) (c) (d) 4. (a) (b) 5. (a)

(s + 12)(s – 12) (6t + 5)(6t – 5) (15 + 7u)(15 – 7u) (7w + 9x)(7w – 9x) 1800 54 3(a + 2)2

(b) (c) (d) (h2 + k)2 6. (a) 4(3m – 2n)2 (b) (c) (d) (5 – tu)2 7. (a) 2(4a + 7b)(4a – 7b) (b) ******ebook converter DEMO Watermarks*******

(c) (d) (m + 8n2)(m – 8n2) 8. (a) a(a + 6) (b) –(5b + 19)(5b + 11) (c) (c + d + 2)(c – d – 2) (d) (2h – 1 + 2k)(2h – 1 – 2k) (e) (5m + n – 1)(5m – n + 1) (f) 4p 9. (i) (x + 2) cm (ii) (x3 + 6x2 + 12x + 8) cm3 10. (a) (b) (c) (d)

–(11x + 7)(7x + 11) (4x + 1 + 3y)(4x + 1 – 3y) (2x + y – 2)(2x – y + 2) 13(x + y + 1)(x + y – 1)

Exercise 1E 1. (a) 9x (5x – 9y) (b) 3x(13y – 5xz) (c) xy2(z2 – xy) (d) –5 x3(3y + 2) 2. (a) (b) (c) (d) 3. (a) (b) (c) (d) 4. (a) (b) (c) (d)

(x – 2y)(6a + 5) (x + 3y)(2b – c) (5x – y)(3d – 4f ) 5(x + 3y)(h + 2k) (x – 5)(a + 4) (a + b)(x + y) (1 + y)(x + 2y) (x – 3)(x + 2y) (a + b)(x – z) (c + 2d)(–2c + 9d) (2h – k)(x – 3y) 2(4m – n)(3x + y)

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5. (a) (b) (c) (d) (e) (f) (g) (h) 6. (a)

(3x + 4y)(a + 7b) (3y + 5)(4c – 3) (d + f)(y – z) (x + 2y)(3x – 4z) (y – 4)(2x – 3) 5(y – 5x)(x – 2) xy(y – 5)(x + y) (k – h)(x – y) 24(y – 5x2)(6p + q)

(b) 2(x + 2y)(x – 2y)(5x – 10y – 6) 7. (i) (ii) 48 8. (i) (x2 + 3)(x – 1) (ii) (x4 – 6x2 + 10)(x2 – 4) Exercise 1F 1. (a) (x − 2)(x2 + 2x + 4) (b) (2x + 1)(4x2 − 2x + 1) (c) (5 − 3x)(25 + 15x + 9x2) (d) 2(x + 4)(x2 − 4x + 16) 2. (a) (5x − 4y)(25x2 + 20xy + 16y2) (b) (3x2 + 2y3)(9x4 − 6x2y3 + 4y6) (c) (7a + 6x2)(49a2 − 42ax2 + 36x4) (d) (6a − 9x3)(36a2 + 54ax3 + 81x6) 3. (a) (10 − 3x)(9x2 − 6x + 28) (b) (x2 + 4x + 9)(x4 + 8x3 + 19x2 + 12x + 21) (c) (2a2 − 2x + 3)(4a4 + 4a2x − 6a2 + 4x2 − 12x + 9) ******ebook converter DEMO Watermarks*******

(d) (a − 2x − 9)(a2 + 2ax − 3a + 4x2 + 12x + 21) 4. 10 cm, 12 cm Review Exercise 1 1. (a) (b) 2. (a) (b) (c) (d) (e) (f) 3. (a) (b) (c) (d) (e) (f) 4. 5. (a) (b) (c) (d) 6. (a) (b)

7q(3p + 2 − 4r) 4(x − 2y + 4z) (a + 4)(a + 9) (b – 7)(b – 8) (c – 3)(c + 17) (d + 3)(d – 15) (q + 2)(q – 9) (y – 3)(y + 8) (3f – 2)(3f + 8) (3h + 2)(h – 7) 7(2k + 1)(k + 3) 3(3m – 2)(2m – 3) c(e – 1)(e + 6) p(n – 4)(n + 7) (x – 7y)(x + 9y) (2x + 3y)(x + y) (3xy – 4)(2xy + 1) z(3 – 2xy)(1 – 2xy) (1 + 11x)(1 – 11x) (x + 3y)2

(c) 25(x – 2y)2 (d) (6y + 7x + 7)(6y – 7x – 7) (e) (6x + y4)(6x – y4) (f) 7. (a) –7y(2x + 3y) (b) 9xy(y – 4x) ******ebook converter DEMO Watermarks*******

(c) (d) (e) (f)

(a + b)(3x – 4y) (x – 2y)(5 – x + 2y) (x + 3y)(x + 2) (3x – 2)(x2 + 1)

(g) 2(2x – 3y)(c – 2d) (h) (y – 2)(5x + 6y) 8. (x + 1)(x + 2)(x – 2) 9. (a) 4600 (b) 318 000 10. (a) (2x2 – 7)(4x4 + 14x2 + 49) (b) (3a + 4y2)(9a2 – 12ay2 + 16y4) (c) 2(4 – x)(28x2 – 26x + 19) (d) 2(2x + 3)(7x2 + 24x + 21) 11. (i) (4p + 2q + 9) marks (ii) (6p + 3q − 9) marks (iii) 3(2p + q − 3) Challenge Yourself – Chapter 1 2 or 16

2. Exercise 2A 1. (a) (b) (c) (d) (e) (f) 2. (a) (b)

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(c) (d) (e) (f) 3. (a) (b) (c) (d) (e) (f) 4. (a) (b) (c) (d) 2c 5. (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l) (m) 6. (a) ******ebook converter DEMO Watermarks*******

(b) (c) (d) (e) (f) (g) (h) d (i) (j) (k) b(a – 2b) (l) 7. Exercise 2B 1. (a) (b) (c) (d) (e) (f) 2. (a) (b) (c) (d) (e) (f) (g) ******ebook converter DEMO Watermarks*******

(h) 3. (a) (b) (c) (d) (e) (f) (g) (h) 4. (a) (b) (c) (d) (e) (f) 5. (a) (b) (c) (d) 6. Exercise 2C 1. (a) (b) (c) (d) ******ebook converter DEMO Watermarks*******

2. (a) a = m(b + c) (b) (c) (d) 3. (a) h = m3 + k (b) D = b2 – 4ac (c) (d) 4. ±4 5. (a) (b) 6. (a) (b) (c) (d) (e) (f) (g) 7. (a) (b)

3 3 1 6

(c) (d) 8. (a) (b) (c) (d) 9. (a) (b) (c) ******ebook converter DEMO Watermarks*******

(d) 10. (i) (ii) –3.08 11. (i) (ii) 25 12. (a) (b) (c) 13. (a) (b) 14. (a) (b) (c) (d) (e)

126 ± 11.5 2210 19.0 –1 3

15. (i) (ii) 60 cm 16. (i) (ii) 0.0912 m 17. (i) (ii) 19.0 m s–1 18. 19. (a) 17.6 (b) 688 (c) ±27.5 20. (a) (b) (c) (i) 0.0005 (ii) 0.577 cm ******ebook converter DEMO Watermarks*******

Review Exercise 2 1. (a) (b) (c) (d) (e) (f) (g) (h) 2. (a) (b) (c) 6 (d) (e) – n (f) (g) (h) 3. (a) (b) (c) (d) (e) (f) (g) (h) 4. (a) (b) ******ebook converter DEMO Watermarks*******

(c) (d) (e) (f) 5. (a) (b) 6. (a) (b) 7. (a) (b) (c) (d) 8. (a) (b) 9. (i)

4 ±10 7860 16.6 ±6.44 9.81

(ii) 1.2 ohms 10. (i) (ii) 7 or –1 Challenge Yourself – Chapter 2 1. 2.

n = 6, k = –5

Exercise 3A 1. A(–4, –3), B(–2, 4), C(3, 4), D(4, 2), E(1, 1), F(3, –3) 2. K(−3, −3), L(3, 0), M(1, 2), N(−1, 1) 4. (a)Rectangle (b) Rhombus (c) Isosceles triangle (d) Quadrilateral ******ebook converter DEMO Watermarks*******

(e) Trapezium 5. 24 units2 6. The points lie on a straight line. 7. (c) D(−4, 2) 8. –5 or 7 Exercise 3B 1. 2. 3. 5. 6.

(b) (b) (b) (ii) (ii)

They are parallel lines. They are parallel lines. They are parallel lines. r=3 a = 2, b = 12, c = 1.5

Exercise 3C 1. 0, undefined 2. (a) − ½ (c) (e) 3. 4. 5. 6. (a) (b) (c) (d) (e)

(b) –10 (d) –3 (f) 0

1, 0 1, –1 –2, 8 –3, –3

(f) (g) 7. (h) 8. 4 ******ebook converter DEMO Watermarks*******

9 3 3 (i) (ii) They are equal. 13. 0, –3, undefined, 1 2 9. 10. 11. 12.

Review Exercise 3 Rectangle Square Trapezium Kite A(–5, 0), B(–4, 3), C(–3, 4), D(0, 5), E(3, 4), F(4, 3), G(4, –3), H(3, –4), I(–3, –4), J(–4, –3) (b) (i) H (ii) G 3. (i) 1. (a) (b) (c) (d) 2. (a)

(ii) (iii) 4. (i) 5 (ii) –10 (iii) 70 5. (ii) a = −2, b = 0 6. (a) 5, 7 (b) –7, –4 7. (i) y = 2x – 3 (ii) 5 8. (i) –3 (ii) 14 9. y = –8x Exercise 4A 1. (a) y = 6, y = –2 ******ebook converter DEMO Watermarks*******

2. 3. 4. 5.

(a) 3 15 (a) (c) (e) (g) 6. (a)

x = 0.5, x = –2

y = –x (b) y = 2x + 1 (d) y = –x – 6 (f) y = 0 (h) x = 0 (b) y = 3x – 2

(c) y = –3x + 1 (d) (e) y = 4 (f) y = ax + a 7. y = 2x 8. (a) y = 1 − x (b) 3y = 3 − x (c) y = 4x + 4 (d) 2y = −3x − 6 (e) y = 6x − 4 (f) 8y = x + 10 9. (a) 0, 1; y = 1 (b) undefined; no y-intercept, x = 1.5 (c) 1, –1; y = x – 1 (d) 10. (i) 4 units2 (ii) (iii) y= x 11. –4 or 3 12. (i) (iii) (3, 0) 13. (i)

(ii) (ii) 0

14. 15. (i) y = 3x − 8 (ii) (4, 4) ******ebook converter DEMO Watermarks*******

16. (i) (−6, 0) (ii) (iii) (iv) y = −1 17. (a) (i) 3 (ii) y = 3x + 3 (b) (6, 3) 18. (i) y = 2x − 6 (ii) (2, −2) 19. (a) (c) −3 (d) (ii) (e)

; They are the same line.

20. m = 0; n = 0 21. (a) −5, −3, 1 (c) y = 2x − 3; They are the same line. (d) (ii) Exercise 4B 1. (a) x = –1, y = –3 (b) x = –5, y = –2 (c) x = 3, y = 1 (d) x = 0, y = 2 (e) x = 5, y = 3 (f) x = 3, y = –4 2. (a) x = 4, y = 2 (b) x = 1, y = –1 (c) x = 2.6, y = –2.6 (d) x = –1.5, y = –0.5 3. (a) (i) –7, 9, 17 (b) (i) 0, 2, 3 (c) x = –4, y = 1 ******ebook converter DEMO Watermarks*******

4. (a) (b) (c) (d) 5. (a) (b)

Infinite number of solutions No solution Infinite number of solutions No solution No solution Infinite number of solutions

Exercise 4C 1. (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l) 2. (a) (b) (c) (d) (e) (f) 3. (a) (b) (c) (d) (e) (f)

x = 8, y = 8 x = 12, y = 7 1 x = 1, y = 4 x = 3, y = 2 x = 1, y = 2 x = 1, y = 1

a = –7, b = –13

x = 3, y = 2 x = 4, y = –5 x = –1, y = 2 x = –8, y = 19 x = 4, y = 5 x = –2, y = –4 x = 3, y = 1 x = 7, y = –11 x = 1, y = 2 x = 3, y = –1 x = 2, y = 3

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4. (a) (b) (c) (d) (e) (f) (g) (h)

5.

6.

7.

8.

x = 6, y = 1 x = 1, y = 3 x = –1, y = –1 x = 3, y = 10 x = 1, y = 2 x = 5, y = –1 x = 0, y = 1

(i) x = 7, y = 6 (j) x = 0, y = 4 (k) (l) x = 5, y = 1 (m) x = 5, y = −1 (n) (a) x = 0.75, y = –0.25 (b) x = 3, y = 5 (c) x = 2, y = –1.5 (d) x = 0.5, y = 1.5 (a) x = 2, y = 1 (b) x = 1 , y = 10 (c) x = 1, y = 1 (d) x = 2, y = –1 (e) x = 2, y = –2 (f) x = –6, y = 2 (a) x = 5, y = 6 (b) x = 1, y = –1 (c) x = 6, y = –4 (d) x = 13, y = 11 (a) x = –4, y = 4 (b) x = 4, y = –3 (c) (d) x = 1, y = –2 (e) x = –3, y = 5

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(f) 9. (a) (b) (c) (d) 10. (a)

11.

12. 13. 14.

x = 15, y = –5 x = –2, y = 11 x = 9, y = 4 x = 3, y = 6

(b) x = 3, y = 12 (c) x (d) (e) x = 1, y = 3 (f) x = 15, y = −6 (a) x = –1, y = 3 (b) x = 2, y = 1 (c) x = 1, y = 4 (d) p = 1, q = –2 p = 2, q = 3 19 m, 6 s

Exercise 4D 1. 2. 3. 4. 5.

25, 113 5, 15 ₱495, ₱885 ₱66, ₱79

6. 7. 8. 9.

8, 40 46°, 74° 21 cm 875 cm2

10. 11. 12.

32 cm 35 7, 6

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₱44 7, 20 10 ₱1040, ₱1560 ₱427 960, ₱395 040 (a) 16 years (b) 50 years 6, 14 72 ₱15, ₱50 (i) 4000 (ii) ₱166

13. 14. 15. 16. 17. 18. 19. 20. 21. 22.

Review Exercise 4 1. (a) 10, 2, –6 (c) 2 (d) (ii) (–0.5, 3) 2. (i) (ii) (iii) 3. (i) 4. (a) (b) (c) 5. (a) (b)

(iv) y = 0.1x + 10 (i) p = –9, q = 11 (i) 5.5, –0.5, –3.5 x = 1, y = 1 x = 2, y = –2 x = 2, y = 2 1 2 x = 1, y = –4

(c) (d) (e) x = 3, y = –4 (f) x = 2, y = 1.5 6. 17, 14 7. 44 cm 8.

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9. 48 10. (i) 11 years (ii) 30 years 11. ₱90, ₱120 12. 21, 15 13. ₱5 14. 6 kg, 14 kg 15. ₱676 16. 38, 37 Challenge Yourself – Chapter 4 1. (i) p = 4, q = 12 (ii) p = 4, q ≠ 12 (iii) p ≠ 4, q is any real number 2. 3. 2, 3 4. (i) 8 (ii) 7 (iii) 5 5. x = 4, y = 18, z = 78 or x = 8, y = 11, z = 81 or x = 12, y = 4, z = 84 Revision Exercise A1 1. (a) (b) (c) (d) 2. (a)

5(3x − 4) 11(2 + y) −16a(4 + x) 3y(3b − 1 + 4c) 5(2 − 3x + 4x2)

(b) a2x(5 − 3ax + 6x) (c) 2a2(3 + 4a − 5a3) (d) 3xy(4x2 − 3xy + 2y2) 3. (a) 2(f − 1)(2f − 3) ******ebook converter DEMO Watermarks*******

(b) (6hk − 1)2 (c) 5mn(m − 3n − 5) (d) (x − y)(2p − 3q) (e) (5x − 2)2 (f) (x − 4)(x − 7) 4. (a) (2cd − 3)(cd + 4) (b) (5hk + 1)2 (c) −4m(m + 4) (d) (3r − s)(p + 2q) 5. (a) 9140 (b) 806 000 6. (a) (b) 7. (i) (ii) x = 3 8. (a) (b) 3d − 2cd2 − 4c2 9. (i) (ii) 10 ms−1 Revision Exercise A2 1. (ii) D(−2, 0) (iii) 2. 3. 4. 5. 6. 7.

(0, 2) x = 1, y = 2 p = 3, q = −4 42 chickens and 28 rabbits

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8. Exercise 5A 2. (a) x ≥ 0, y ≥ 0, y ≤ 3, x + y ≤ 4 (b) x ≤ 3, 2y > x, 3x + 2y ≥ 6, 3y ≤ x + 9 (c) x < 4, 4y ≥ x, 4x + y ≥ 4, 2y ≤ x + 4 (d) 2y > x, 2x + y ≤ 4, y ≤ 2x + 2 (e) y ≥ 1, 2y ≥ x + 2, 4x + 3y ≤ 12, y ≤ 2x + 4 (f) x ≥ 1, x ≤ 4, 2y + x ≥ 4, 2y < x + 4 3. (a) x ≥ 0, y ≤ 4, x + y ≤ 5, 2y ≥ x (b) y ≤ 2, x + y ≤ 4, y ≤ 4x + 4, 4y ≥ x (c) 4x + 3y ≤ 12, 2y ≥ x − 2, y ≤ 3x + 3 (d) 2x + 3y ≤ 6, y ≤ 2x + 2, 2y ≥ x − 2 (e) y ≥ 0, 3y < 2x + 6, 2x + y < 4 (f) 3y > 2x − 3, 2x + y < 7, 2y ≤ x + 4, 2x + y ≥ 0 Exercise 5B 1. (a) x + y ≤ 30, x ≥ 8, y ≥ 2x (c) 600 cm 2. (a) x + y ≤ 40, x ≥ 12, x ≥ 2y (c) 266 g 3. (a) x + y ≤ 25, x ≥ 5, 2y ≥ x ******ebook converter DEMO Watermarks*******

(c) 20.5 liters 4. (a) x + y ≤ 200, x ≥ 2y, y ≥ 50, x ≤ 140 (c) 140 bottles of Power Clean and 60 bottles of Disappear 5. (a) 60x + 100y ≤ 10 000, 3x + 6y ≥ 300, 40 ≤ x ≤ 80 (c) 40 “Economy” packets and 30 “Giant” packets 6. (a) 40x + 50y ≥ 300, 3000x + 2000y ≥ 20 000 (c) 8 7. (a)

(b) y > x, x ≤ 2300, y ≤ 5000 (d) 1500 kg of Fragrant and 5000 kg of Instant Review Exercise 5 2. (a) y ≥ 0, x ≤ 2, y ≤ 2, 2x + y ≥ 2 (b) x ≤ 3, 2y ≤ x + 4, 3x + 2y ≥ 6, 2y ≥ x (c) x + 2y ≤ 10, 2y − 4x + 3 ≤ 0, 2y − x + 3 ≥ 0 (d) 3x + y ≥ 3, 3y ≤ 4x + 9, 5x + 3y ≤ 36, 5y + 2 ≥ 2x (e) x + 2y ≤ 11, 11y + 7 ≥ 8x, 8x + y + 5 ≥ 0, y ≤ x + 4 (f) y ≤ 7x, x + y ≤ 8, 2x + y ≤ 12, 3y + 19 ≥ 5x, 3x + 2y ≥ 0 (g) x ≤ 3, y ≤ 2x + 1, x + y > 4 ******ebook converter DEMO Watermarks*******

3.

4.

5.

6. (c)

(h) y ≥ 0, y ≤ 2x + 2, x + y < 5 (i) 3y < 4x (a) x > 20, y ≥ 40, 70 ≤ x + y ≤ 80 (c) 40 hardback copies and 40 paperback copies (a) 2000x + 5000y ≥ 100 000, 4000x + 3000y ≥ 100 000 (c) 29 (a) y > 10, 20 ≤ x ≤ 50, x + y < 70, x ≥ y (c) 15 (a) x + y ≥ 10, 240x + 160y ≤ 2400, 200x + 80y ≥ 1600 12.5 kg

Challenge Yourself – Chapter 5 ₱675 000 in government bond fund, ₱225 000 in bank’s fund, ₱100 000 in high-risk account Exercise 6A 1. (i) (ii) 2. (i) (ii) 3. (a) (b) (c) (d) 4. 6. (a)

y = 17 y = −3 x = –3 x = 10 Yes Yes No, 2 has two images. Yes 8, −28, −2, −7 (i) (ii)

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(b) (i) x = 1 (ii) 7. (a) 3 (b) 9 (c) 5 (d) 10 8. (a) 18 (b) −17 (c) 8 (d) 1 (e) 2 9. (a) x = 5 (b) x = −2 (c) (d) x = 3 (e) x = 1 (f) 10. 11. (a) a2 + 5 (b) a2 + 2a + 6 (c) 4a (d) a4 + 5 (e) a6 + 5 (f) a6 + 2a3 + 6 12. (a) −5a (b) (c) −5a2 − 5a + 8 13. (a) a = 4, b = −26 (b) −10 14. (a) a = −3, b = 13 (b) 7 ******ebook converter DEMO Watermarks*******

15. (a) (b) (c) (d)

13, 17, 21 No No No No

16. (a) No (b) No (c) (d) (e) (f) 17. (a) (b) (c)

39 a=4 a=3 a=3

Exercise 6B 1. 1, 5, 7 2. (a) x-intercept = −5, y-intercept = −5 3. (a) 4. (a) (b) (c) (d) 5. (a) (b) (c)

−1 ≤ f(x) ≤ 1 −4 < f(x) < −2 −5 ≤ f(x) ≤ 3 f(x) ≥ 1

(d) 6. (a) (b) 7. (a)

−3 ≤ x ≤ 4 a = −4, b = 1 9, 1 a = 3, b = −4

–8 ≤ x ≤ –1 −2 ≤ x ≤ 1 1

2

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8. (a) f(x) = −x + 2 9. (a) 1 ≤ g(x) ≤ 10 (b) −1 ≤ g(x) ≤ 8 (c) −5 ≤ g(x) ≤ 4 (d) −1 ≤ g(x) ≤ 8 10. (a) −1 ≤ x ≤ 1 (b) 0 ≤ x ≤ 2 (c) 2 ≤ x ≤ 6 Review Exercise 6 1. (i) y = 50 (ii) y = 32 (iii) y = 14 (iv) y = −4 2. (i) x = 10 (ii) x = −3 (iii) x = 5 (iv) 3. (a) No, 4 has no image. (b) Yes 4. 6. (a) 2 (b) 11 (c) 3 (d) 8 7. (a) x = 8 (b) x = 3 (c) x = 0 (d) x = 9 (e) x = −9 (f) x = 1 8. (a) 21a + 12 (b) 7a ******ebook converter DEMO Watermarks*******

(c) 9. (a) a = 2, b = −2 (b) 14 10. 11. (a) x-intercept = 2, y-intercept = −16 12. (a) 13. (a) −1 ≤ f(x) ≤ 3 (b) 4 ≤ f(x) ≤ 14 (c) (d) f(x) ≤ −3 14. (a) (b) (c) (d) −7 ≤ x ≤ −2 15. (a) a = 1, b = 4 (b) 5, 4 16. (a) −3 ≤ g(x) ≤ 1 (b) −1 ≤ g(x) ≤ 0 (c) g(x) ≤ 2 Challenge Yourself – Chapter 6 (b) Exercise 7A 1. (a) (b) (c) (d) (e) (f)

False statement Not statement Not statement True statement Not statement False statement

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2. (a) (b) (c) (d) 3. (a) (b) (c) 4. (a) (b) (c) 5. (a) (b) (c) (d) (e)

False False True False False True True True False True False False False True True

Exercise 7B 3.

(a) (b) (c) (d)

True True True False

Exercise 7C 1. (a) (b) (c) (d) (e) (f) 2. (a) (b) (c) (d)

True, true, true True, true, true False, false, true False, false, false False, false, false False, false, false True, true, true False, false, true False, false, true True, true, true

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(e) (f) 3. (a) (b) (c) (d)

False, false, false False, false, true False, false, true False, false, false False, false, true True, true, true

Exercise 7D 1. (a) Inductive reasoning (b) Deductive reasoning Review Exercise 7 1. (a) (b) (c) (d) (e) (f) 4. (a) (b) (c) (d) (e) (f)

True statement False statement Not statement Not statement False statement True statement True, true, true False, false, true True, true, false True, false, false True, false, false True, true, false

Revision Exercise B1 2. (a) y≥5−x (b) y ≤ 2x + 4, y ≥ −x − 2 3. (ii) 8 5. (b) 25 units2 ******ebook converter DEMO Watermarks*******

Revision Exercise B2 1. k = −20 2. (b) 3 ≤ f(x) ≤ 13 3. (b) −1 ≤ x < 1 4. 5. (a) The converse is false. The inverse is false. The contrapositive is true. (b) The converse is true. The inverse is false. The contrapositive is true. Exercise 8A 1. (a) Commutative Axiom of Multiplication (b) Distributive Axiom (c) Symmetric Axiom (d) Multiplication Axiom (e) Transitive Axiom (f) Associative Axiom of Multiplication 2. (b) Existence of Multiplicative Inverse, Multiplication Axiom 3. (a) Multiplication Axiom (b) Addition Axiom, Multiplication Axiom 4. Addition Axiom, Multiplication Axiom 5. Distributive Axiom Exercise 8B 2. (a) AB, BC, AC ******ebook converter DEMO Watermarks*******

4. (a) False (b) True (c) False (d) True (e) True 5. 7.7 units 6. (a) Q 7. 36 cm 10. (a) 0 (b) 1 (c) 3 (c) 6 Review Exercise 8 1. (a) Commutative Axiom of Addition (b) Associative Axiom of Addition (c) Transitive Axiom (d) Addition Axiom (e) Multiplication Axiom (f) Existence of Additive Identity 2. Addition Axiom, Multiplication Axiom 3. Commutative Axiom of Addition, Existence of Multiplicative Inverse, Reflexive Axiom 4. (a) False (b) True (c) False (d) False (e) True (f) False 5. (a) No (b) 4 or 6 ******ebook converter DEMO Watermarks*******

6.

7. 8. 9. 10. 11.

(c) Triangle or square or rhombus or kite (d) True (a) No (b) {AD, DH, HE, EA}, {DH, HG, GC, CD}, {GC, CB, BF, FG}, {BF, FE, EA, AB}, {AB, BC, CD, DA}, {EF, FG, GH, HE} (c) Non-collinear and coplanar (a) ABC, ACD, ADE, AEB, BCDE (b) No (a) −0.3 −0.5 or 3.5 (b) 4.8 units (c) (i) 2.4, 8.4, 7.2

Challenge Yourself – Chapter 8 Yes Exercise 9A 1. A, F; B, J; C, E; D, G; I, K 2. (i) 3.5 cm (ii) VZ (iii) WX, 3.5 cm (iv) ZY, 2.1 cm (v) YX, 2 cm (vi) ∠VWX, 90 3. EF = 3.4 cm, GH = 2.4 cm, ∠FEH = 100°, ∠FGH = 75°, MN = 5 cm, OL = 3 cm, ∠LMN = 65°, ∠NOL = 120° 4. (a) ΔABC ≡ ΔPQR (b) ΔDEF ≡ ΔTSU ******ebook converter DEMO Watermarks*******

(c) 5. (i) (ii) 6. (i) (ii) 7. (i) (ii)

Not congruent 56° 16 cm 75° 10.9 cm 6 cm 96°

Exercise 9B (ii) and (vii) (b) (iii) and (v) (i) and (ix) (d) (vi) and (viii) P, Q, R; PQ, QR, PR, 6; PQR, SSS Z, Y, X; ZY, ZŶX, YX, 5; ZYX, SAS W, V, U; WV̂U, WUV, 70, VU, 7; NML, AAS (d) U, T, S; UTS, 90, US, TS, 5; ΔIHG, RHS 3. (a) No (b) No (c) No (d) No 4. (a) ΔABD ≡ ΔCBD (b) ΔABD ≡ ΔCDB (c) ΔABC ≡ ΔEDC (d) ΔABC ≡ ΔCDA (e) ΔADE ≡ ΔCDB (f) ΔBCD ≡ ΔEFD (g) ΔABD ≡ ΔCBD (h) ΔABC ≡ ΔCDA 5. (ii) 4 cm (iii) 80° (iv) RS is parallel to UV. 6. (i) ΔJIH (ii) 80° 7. (a) ΔABC ≡ ΔCDA (b) ΔEFG ≡ ΔGHE (c) ΔIJK ≡ ΔKLI ******ebook converter DEMO Watermarks******* 1. (a) (c) 2. (a) (b) (c)

(d) ΔMNO ≡ ΔOPM (e) ΔQRS ≡ ΔSTQ (f) ΔUVW ≡ ΔWXU Exercise 9C 4.

Length of AB´

Review Exercise 9 1. (a) Not congruent (b) ΔDEF ≡ ΔTUS 2. (i) 6 cm (ii) 95° 3. (i) 92° (ii) 4.2 cm 4. (i) 90° (ii) 6 cm; 8 cm (iii) 53° 5. (a) Yes (b) Yes (c) No (d) Yes 6. ΔDEF ≡ ΔJLK, AAS 7. (a) ΔABC ≡ ΔDEC AĈB = DĈE, BC = EC, AC = DC (b) ΔFGH ≡ ΔFIJ FĜH = FÎJ, FG = FI, FH = FJ (c) ΔKLN ≡ ΔMNL KLN = MN̂L, KN̂L = MLN, KL = MN (d) ΔSQP ≡ ΔRPQ SQ̂P = RP̂Q, SQ = RP, SP = RQ (e) ΔEBF ≡ ΔECD EB̂F = EĈD, BF = CD, EF = ED (f) ΔFHG ≡ ΔFIJ FĤG = FÎJ, FH = FI, GH = JI ******ebook converter DEMO Watermarks*******

8. (i) (ii) 9. (i) (ii) 11. (i) (ii)

ΔOAD ≡ ΔOBC AÔD = BÔC, OÂD = OB̂C ΔPQR ≡ ΔSRQ, SAS 5 cm, 50° ΔSTR ≡ ΔSTP UQ = 7 cm, PQ = 10 cm

Challenge Yourself – Chapter 9 Yes Revision Exercise C1 1. x ≥ −6; Distributive Axiom, Addition Axiom 2. (a) True (b) False 4. (i) 8 cm (ii) 50° Revision Exercise C2 Distributive Axiom, Addition Axiom, Commutative Axiom of Addition, Reflexive Axiom 7.5 ΔPQS ≡ ΔPRT (ii) ΔCRB a = 6, b = 29

1.

2. 3. 4. 5. Exercise 10A 1. (a) (b) (c) (d)

Yes Yes Yes Yes

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(e) No (f) No 2. (a) 3 < AC < 11 (b) 9 < BC < 99 (c) 8.5 < PQ < 33.5 (d) 274 < FG < 726 4. ∠NJM < ∠NKM < ∠NLM 7. 3 < SU < 9 8. 9. XY > YZ 11. x < 11 Exercise 10B 1. (a) (b) (c) (d) 2. (a) (b)

x = 22, y = −1 x = 36 x = −8, y = 20 CD // GH EF // GH

Review Exercise 10 1. (a) No (b) Yes (c) No (d) Yes 2. (a) 10 < AC < 42 (b) 3 < AB < 14 5. 4 < QS < 12 6. x > 3 7. ∠ACB < ∠XZY 8. 17 9. x = 7, y = −11 ******ebook converter DEMO Watermarks*******

(i) x = 25 (ii) y = 69 AB // DF

10. 11. Exercise 11A

1. 1, 2, 3, 4, 5, 6; 6 2. (a) 2, 3, 4, 5; 4 (b) A, B, C, D, E, F, G, H, I, J; 10 (c) R1, R2, R3, R4, R5, B1, B2, B3, G1, G2; 10 (d) T, E1, A, C, H, E2, R; 7 (e) 100, 101, 102, …, 999; 900 3. (i) (ii) (iii) 1 (iv) 4. (i) (ii) (iii) (iv) 0 (v) 5. (i) (ii) 12 (iii) (iv) 6. (i) (ii) (iii) (iv) 7. (i) ******ebook converter DEMO Watermarks*******

(ii) (iii) (iv) 8. (i) (ii) 12 (iii) 9. 38 10. (i) (ii) 11. (i) (ii) 12. (i) (ii) (iii) (iv) 13. (i) (ii) (iii) (iv) 14. (i) (ii) (iii) 1 (iv) 15. 16. (a) (i) (ii) (iii) (b) (i) ******ebook converter DEMO Watermarks*******

(ii) 17. (i) 26 (ii) 64 18. (i) (ii) (iii) (iv) 19. (i) 1 4 (ii) 1 3 20. Exercise 11B 1. (i) (ii) (iii) (iv) 0 2. (i) (ii) 3. (i) (ii) (iii) 4. (i) (ii) (iii) 12 5. (i) 15 + x (ii) (iii) 60 6. (i) (ii) 7. (a) (i) ******ebook converter DEMO Watermarks*******

(ii) (b) (i) (ii) 8. 9 9. 10. 10 11. 7 12. (i) (ii) 2x + 46 (iii) 30 13. x = 16, y = 18 Exercise 11C 1. {(1, H), (2, H), (3, H), (4, H), (5, H), (6, H)} {(1, T), (2, T), (3, T), (4, T), (5, T), (6, T)} 2. F: Faulty pen; N: Nonfaulty pen 3. (i) (iii) 4. (b)

(ii) (iv) (i) (ii) (iii)

0, 1, 2, 3, 4, 5, 6; 2, 3, 5, 6; 2, 3, 4, 5, 6, 7; 3, 4, 5, 6, 7, 8; 4, 6, 7, 8, 9; 5, 6, 7, 8, 9, 10; (b) 36 ******ebook converter DEMO Watermarks******* 5. (a)

(c)

(i) (ii) (iii)

(iv)

(v) (d) 6. (a) (b)

Sum of 7 12, 13; 12, 13, 14; 13, 15 28, 35; 32, 48; 36, 45, 54 (i) (ii) (iii)

(c) 7. (i) (iii) 8. (i) (iii) 9. (a) (b) 10.

(i) (iii) 11. (a) (b) 12. (a) (b)

(i) (ii) (iii) (ii) {R, BB, WB, RR, BR, WR} (ii) {11, 12, 13, 21, 22, 23, 31, 32, 33} (i) (ii) 0 (iii) (iv) B: Boy; G: Girl S = {BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG} (ii) (i) (ii) (iii) (iv) 2, 3, 4, 5, 6; 2, 4, 8, 10, 12 (i)

(ii)

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(iii)

(iv)

(v) 13. (a)

(b)

1, 2, 3, 4, 5; 1, 0, 2, 3; 2, 1, 0, 1, 2, 3; 3, 2, 1, 0, 1, 2; 4, 3, 2, 1, 0, 1; 5, 3, 2, 1 (i) (ii) (iii) (iv) (v)

14. (i) (iii) (v) 15. (i) (iii)

(ii) 11 20 1 5 (iv) 1 0 (ii) (iv)

(v)

(vi)

(vi) 16. (i)

(ii)

17. (i)

(ii)

(iii)

(iv)

(v)

(vi)

(vi) 18. (a)

(i) (ii)

19. 20. (a)

(i)

(ii)

(iii) ******ebook converter DEMO Watermarks*******

(b) 21. Review Exercise 11 1. (a) (i) 5, 6, 8, 56, 58, 65, 68, 85, 86, 568, 586, 658, 685, 856, 865 (ii) 15 (b) (i) (ii) 2. (i) (ii) (iii) 3. (i) (ii) 0 (iii) 4.

5.

6.

7.

(iv) (i) (ii) (iii) 0 (i) (ii) (iii) 0 (iv) (i) (ii) (iii) (a) (i) 0 (ii) 1 (b) (i) (ii)

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8. (i)

9. 10.

11. 12. 13.

(ii) (iii) 0 (i) 55 (ii) (i) 37p – 2q – 145 = 0 (ii) 3p + 2q – 55 = 0 (iii) p = 5, q = 20 (i) (ii) 1 4 1 (i) (ii) 4 (iii) 1 1 (iv) 2 4 (i) (ii) (iii) (iv)

14. (i) 1 (ii) 2 (iii) (iv) 15. (b) (i) (ii) 16. (a) (b) (c) Challenge Yourself – Chapter 11 1. 2.

3.

22, 30 1 (i) 2 (ii) 0 (iii) 1 11

Revision Exercise D1 ******ebook converter DEMO Watermarks*******

1. 24 cm 3. 9 < l < 27 4. (i) (ii) (iii) (iv) 5. (i) 0 (ii) (iii) (iv) Revision Exercise D2 1. RS // PBQ 2. (i) (ii) 3. 280 4. S = {HH, HT, TH, TT} (a) (b) 5. (b) (i) (ii) (iii) (iv) (v) 6. (a) (b) ******ebook converter DEMO Watermarks*******

(c) Problems in Real-World Contexts 1. (i) ₱300 000 (ii) ₱300 000; Yes (iv) ₱12 000 000 2. (a) 2.99% (b) Industry; 46.5516 MT 4. (i) Sam’s method (iii) Chris’s method

******ebook converter DEMO Watermarks*******

A

B

C 7(ay – 7y)

F G −3x − 4y K L 7ay – −25x – 9y 49y

H

D E 3(x – 2y) – 2(3x – y) I J 29x − 3y

M N 2x – 3[5x – y – 2(7x – y)] Table 1.1

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7y(a – 7) O −3x – 8y

A

B (x – y)2

(x + y)(x + y)

C

D x2 + y2

F G 2wz – 6wy + 3xy (x + y)2 – xz K L (x + y)(x – y)

E

(2w – x)(z – –5x2 + 28x – 3y) 24 I J

H (2w + x)(z – (x – y)(x – y) x2 – y2 3y) M N O 2x – (x – 4)(5x 2 –5x2 – 24x + 2 2 x – 2xy + y x + 2xy + y2 – 6) 24 Table 1.2

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Coordinates of Endpoints

(a)

A(–2, 1) and B(0, 5)

(b)

C(7, 5) and D(4, 8)

(c)

E(–2, 6) and F(–4, 3)

(d)

G(1, 1) and H(3, 1)

(e)

I(–4, 3) and J(–4, 6)

Slope of Line Segment

Sign of Slope

positive

Table 3.1

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y2 – y1

x2 – x1

2x – 5y = 32

2x + 3y = 0

x

–4

1

6

x

–3

0

3

y

–8

–6

–4

y

2

0

–2

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x

–5

3

y

******ebook converter DEMO Watermarks*******

7

Case 1

Coordinates of Point

Case 2 x + 2y

Coordinates of Point

Observation Table 5.1

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Case 3 x + 2y

Coordinates of Point

Total Weight (kg)

Fragrant

x

Instant

y

Weight of A (kg)

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Weight of B (kg)

x y

–7

2 0

5

Table 6.1

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4 9

x y

–1

2 0

–1 Table 6.2

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3 –7

x

1

2

3

4

5

y = 2x

2

4

6

8

10

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x

–2

–1

0

1

2

y = 2x + 1

–3

–1

1

3

5

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x

–1

0

y

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1 2

2

x f(x)

–2

–1 3

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0

1

x g(x)

–2

–1 5

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0

1 -3

x

–4

–2

f(x)

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0 3

2

x

–1

0

g(x)

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1

2

5

8

x

y

LHS

RHS

LHS = RHS?

−1

0

1

1

Yes

−1

−1

0

0

Yes

0

−1

−1

−1

Yes

0

1

−1

−1

Yes

1

0

1

1

Yes

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Outcome

Tally

Number of ‘heads’ or ‘tails’ for 20 tosses

Head Tail

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Fraction of obtaining a ‘head’ or a ‘tail’

Personnel

Management

Teaching

Laboratory

Administrative

Number of personnel

26

62

8

9

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1 1

2

1

2 3

3

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3

× 1 2

1

2

3

4

2 6

3 4

12

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Price Plan

A

B

C

Monthly subscription

1200

1300

1250

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Unlimited

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100 minutes If the outgoing calls exceed 100 minutes, the excess usage is charged at 0.09/second.

Free local data bundle

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12 GB

12 GB

12 GB

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