Ch04

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Chapter 4 Algebraic Techniques This chapter deals with operations involving algebraic terms including indices. After completing this chapter you should be able to: ✓ add, subtract, multiply and divide algebraic fractions ✓ apply the index laws to simplify algebraic expressions ✓ establish the meaning of the zero index ✓ define indices for square root and cube root ✓ establish the meaning of negative indices ✓ simplify expressions involving fractional and negative indices ✓ remove grouping symbols and simplify by collecting like terms ✓ factorise by determining common factors.

Syllabus reference PAS5.1.1, 5.2.1 W M: S5. 1. 1–S5. 1. 5, S 5 .2 .1 –5 .2 .5

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

Diagnostic Test

1

2

3

4

5

6

7

Which of the following is not true? 2x 4x 2x 10x A ------ = -----B ------ = --------3 5 3 15 2x 4x 2x 6x C ------ = -----D ------ = -----3 6 3 9 12a When reduced to its simplest form ---------- = 18a 12 2a A -----B -----18 3a 2 4 C --D --3 6 3x 5x ------ + ------ = 11 11 8x 8x A -----B -----22 11 8x 88x C ---------D --------121 11 2x x ------ + --- = 3 4 3x 3x A -----B -----7 12 11x 11x C --------D --------12 7 3m 4 -------- × ------ = 7 5y 34m 12m A ----------B ----------75y 35y 12my 34my C -------------D -------------35 75 12ab 10 When simplified ------------- × ------ = 5 3a 120ab 8ab A ----------------B ---------15a a 40b C ---------D 8b 5 a b --- ÷ --- = 3 2 ab A -----6 2a C -----3b

6 B -----ab 3b D -----2a

8

9

5xy 3x --------- ÷ ------ = 8 2 5y A -----12 2 15x y C --------------16

12 B -----5y 16 D -------------2 15x y Which of the following does not simplify to t20? A t4 × t5

B t 30 ÷ t 10

C (t 4)5

D t16 × t 4

3 4

10

11

12

(a ) ---------------- = 4 2 a ×a A a2 B a4

A 8a7b 9

B 8a12b 20

C 8ab16

D 8(ab)16

(2m 5)3 = 2m 8 B 8m 8

B 3

17

1 --2

1 --2

D 9

B ( 4y ) C 2y

The meaning of y

1 --3

1 --2

D 4y2

is:

1 D ----3y Which of the following is equivalent 1 to ----3- ? 1 a --3 1 A 3a B -----C a D a –3 3a A

16

C 5

4 y may be written in index form as: A 4y

15

C 2m15 D 8m15

3k0 + 2 = A 2

14

D 16

4a 3b 5 × 2a 4b 4 =

A 13

C a6

1 --3

y

B 3y

C

3

y

3m −2 is equivalent to: 1 3 A –6m B ----------2- C ------29m m

1 D -------3m

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

18

8y 5 ÷ 2y 1 --4

A 19

y

6

–1

=

B

21 1 --4

y

4

C 4y

6

D 4y

4

The highest common factor of 6y 2 and 12y is: A 6

B 3y

C 6y

D 12

–2p 2(5p 2 + 3pq) = A –10p 2 – 6pq

B –10p 2 – 6p 3q

C –10p4 + 6p 3q

D –10p4 − 6p 3q

22

When fully factorised 12ab – 3a + 9a2 = A 3a(4b – 1 + 3a) B 3(4ab – a + 3a2) C a(12b – 3 + 9a) D 3ab(4 – 1 + 3a)

20

When expanded and simplified 3(2m – 1) – (m + 5) = A 5m – 8

B 5m + 2

C 5m – 6

D 5m + 4

If you have any difficulty with these questions, refer to the examples and questions in the sections listed in the table. Question

1–8

9, 10

11, 12

13

A

B

C

D

Section

14, 15 16, 17 E

F

18

19, 20 21, 22

G

H

I

A. ALGEBRAIC FRACTIONS Example 1 Complete the following equivalent fractions. 2 ■ a --- = -----3 15

■ 3n b ------ = -----7 14

c

2ab ■ ---------- = -----5 20

2 5 10 a --- × --- = -----3 5 15

3n 2 6n b ------ × --- = -----7 2 14

c

2ab 4 8ab ---------- × --- = ---------5 4 20

Exercise 4A 1

Complete the following equivalent fractions. 7k ■ 2y ■ 4m ■ a ------ = -----b -------- = ---c ------ = -----4 20 5 15 3 6

3t ■ d ------ = ---------10 100

5a ■ e ------ = -----2 12

Example 2 Reduce to its simplest form: 15 a -----20 c

7a -----9a

Dividing the numerator and denominator by the same number is sometimes called cancelling.

8t b -----12 6ab d ---------9a

☞

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

a Dividing the numerator and denominator by 5,

b Dividing numerator and denominator by 4,

3

2

15 15 ------ = --------420 20 3 = --4

8t 8 t ------ = --------312 12 2t = ----3

c Dividing numerator and denominator by a,

d Dividing numerator and denominator by 3 and by a,

1

2 1

7a 7a ------ = --------19a 9a 7 = --9

2

6ab 6 a b ---------- = -------------3 1 9a 9 a 2b = -----3

Reduce to its simplest form: 6x 3m a -----b -------12 9 6a 3p f -----g ---------7a 10p

5t -----20 8x h --------12x

10y d --------15 8ab i ---------4a

c

9b e -----12 12pq j ------------9q

Example 3 Simplify:

3

7 4 a ------ + -----15 15

5x 4x b ------ + -----11 11

7 4 7+4 a ------ + ------ = ------------15 15 15 11 = -----15

5x 4x 5x + 4x b ------ + ------ = ------------------11 11 11 9x = -----11

Simplify: 3x 4x a ------ + -----10 10 9m 6m d -------- – -------10 10

7b 8b b ------ + -----11 11 11k 3k e ---------- – -----12 12

c c

c

7a 2a ------ – -----10 10 7a 2a 7a – 2a ------ – ------ = ------------------10 10 10 5a = -----10 a = --2

2a 3a ------ + -----15 15

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

Example 4 Simplify: 3 2 a --- + --4 3

5n 2n b ------ + -----6 3

3 2 3 3 2 4 a --- + --- = --- × --- + --- × --4 3 4 3 3 4 9 8 = ------ + -----12 12 17 = -----12

4

7m m -------- – ---8 3

5n 2n 5n 2n 2 b ------ + ------ = ------ + ------ × --6 3 6 3 2 5n 4n = ------ + -----6 6 9n = -----6 3n = -----2

5 = 1 ----12

c

c

7m m 7m 3 m 8 -------- – ---- = -------- × --- – ---- × --8 3 8 3 3 8 21m 8m = ----------- – -------24 24 13m = ----------24

Simplify: 2x a ------ + 3 3t f ------ + 10

x --4 2t ----9

5k 3k b ------ + -----6 4 4w w g ------- – -----3 12

7b b ------ – --8 4 3v 4v h ------ + -----2 3 c

3a a d ------ + -----5 10 11e 3e i ---------- – -----10 5

4z 2z e ------ – -----5 3 5x 3x j ------ – -----6 8

Example 5 Simplify:

5

2 5 a --- × --3 9

2a b b ------ × --3 9

2 5 2×5 a --- × --- = -----------3 9 3×9 10 = -----27

2a b 2a × b b ------ × --- = ---------------3 9 3×9 2ab = ---------27

Simplify: m n a ---- × --3 4

k m b --- × ---5 3

c

2p q ------ × --3 5

c c

3h 4 ------ × -------7 5m 3h 4 3h × 4 ------ × -------- = -----------------7 5m 7 × 5m 12h = ----------35m

3a 4 d ------ × -----5 7b

2b 5d e ------ × -----3c 7e

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

Example 6 Simplify: 8 15 a --- × -----9 16

8a 15 b ------ × ---------9 16b 1

5

1

8 15 8 15 a --- × ------ = -----3 × --------29 16 9 16 5 = --6 4 1

c

12ab 10 ------------- × -----5 3a

c

5

8a 15 8 a 15 - × -----------b ------ × ---------- = -------3 2 9 16b 16 b 9 5a = -----6b 2

12ab 10 12 a b 10 ------------- × ------ = -----------------× ----------1 1 1 5 3a 3 a 5 8b = -----1 = 8b

6

Simplify: 3m 10n a -------- × ---------5 7 5y 9 f ------ × -----3 2y

2k 6n b ------ × -----9 5 7 3z g ------ × -----2z 14

4w 9z ------- × -----3 8 2ab 6 h ---------- × -----3 2b c

8a d ------ × 5 8mn i -----------9

15b ---------16 15 × -------3m

3t 10 e ----- × -----5 9u 6pq 25 j ---------- × -----5 3q

Example 7 Simplify: 2 5 a --- ÷ --3 8

a 5 b --- ÷ --4 b

To divide by a fraction, we multiply by its reciprocal. 5 8 a The reciprocal of --- is --- , hence 8 5 2 5 2 8 --- ÷ --- = --- × --3 8 3 5 16 = -----15

5 b b The reciprocal of --- is --- , hence b 5 a 5 a b --- ÷ --- = --- × --4 b 4 5 ab = -----20

1 = 1 ----15

7

Simplify: a b a --- ÷ --5 7

w z b ---- ÷ --6 2

c

p 5 --- ÷ --4 q

3 n d ---- ÷ --m 8

a c e --- ÷ --b d

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

Example 8 Simplify: 2a 6b a ------ ÷ -----3 7

b 1

2a 6b 2 a 7 a ------ ÷ ------ = --------- × -------3 3 7 3 6 b 7a = -----9b 8

Simplify: 2x 8y a ------ ÷ -----3 5 16 8 e ------- ÷ ------9w 3w 4xy 2x i --------- ÷ -----3 5

3a 6b b ------ ÷ -----2 7 6k 7k f ------ ÷ -----5 2 9 6 j --------------- ÷ -------10km 5m

5pq 3p ---------- ÷ -----8 2 1

b

5p 10q ------ ÷ ---------3 9 4m 2m g -------- ÷ -------3 5 c

B. THE INDEX LAWS The index laws for numbers were established in chapter 2: 1 When multiplying numbers with the same base, we add the indices. For example, 36 × 34 = 36 + 4 = 310 2 When dividing numbers with the same base, we subtract the indices. For example, 36 ÷ 34 = 36 − 4 = 32 3 When raising a power of a number to a higher power, we multiply the indices. For example, (36)4 = 36 × 4 = 324

If we use letters to represent numbers then the rules can be generalised: am × an = am + n am ÷ an = am − n (a m)n = a mn

1

5pq 3p 5p q 2 ---------- ÷ ------ = -----------× --------14 8 2 3p 8 5q = -----12

7 3 d ------ ÷ --------5v 10v 7 m h -------- ÷ ---2m 8

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

Example 1 Show by writing in expanded form that: a m4 × m3 = m7

b m5 ÷ m2 = m3

a m4 × m3 = (m × m × m × m) × (m × m × m) =m×m×m×m×m×m×m = m7

c (m4)3 = m12 1

c (m4)3 = (m × m × m × m) × (m × m × m × m) × (m × m × m × m) =m×m×m×m×m×m×m×m×m×m×m×m = m12

Exercise 4B 1

Show by writing in expanded form that: a m 2 × m4 = m6 b m6 ÷ m 2 = m4

c

(m 2)4 = m8

Example 2 a Use a calculator to evaluate the following expressions when a = 3. i a4 × a3 ii a7 b Does the value of a4 × a3 = the value of a7? a i

a4 × a3 = 34 × 33 = 81 × 27

1

m ×m ×m×m×m b m5 ÷ m2 = ----------------------------------------------------1 1 m ×m m×m×m = -------------------------1 3 =m

ii a7 = 37 = 2187

= 2187 b Yes 2

a Use a calculator to evaluate the following expressions when a = 2. ii a9 i a5 × a4 b Does the value of a5 × a4 = the value of a9?

3

a Use a calculator to evaluate the following expressions when m = 5. i m8 ÷ m2 ii m6 8 b Does the value of m ÷ m2 = the value of m6?

4

a Use a calculator to evaluate the following expressions when n = 3. i (n4)2 ii n 8 b Does the value of (n4)2 = the value of n 8?

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

Example 3 Use the index laws to simplify: a y7 × y3

b y 18 ÷ y 17

a y7 × y3 = y7 + 3

b y 18 ÷ y 17 = y 18

= y 10

c – 17

(b5)3

c (b5)3 = b5 x 3

= y1

= b15

=y

5

6

7

8

Use the index laws to simplify: a m3 × m6 b q8 × q7

c t 10 × t 9

d b15 × b × b 4

e v × v5 × v7

Use the index laws to simplify: a a12 ÷ a10 b x15 ÷ x 5

c w8 ÷ w2

d b6 ÷ b 5

e z 20 ÷ z 19

Use the index laws to simplify: a (b4)2 b (h 5)3

c (k 8)2

d (z10)6

e (n 2)4

Use the index laws to simplify: a m4 × m 2 b x9 ÷ x 6 8 7 f n ÷n g b8 ÷ b

c (b4)6 h (y 5)5

d m 3 × m6 × m4 i t10 × t20 × t

e (v 7)10 j a12 ÷ a6

Example 4 Explain why the index laws cannot be used to simplify: a p3 × q4

b m6 ÷ n4

a p3 × q4 = p × p × p × q × q × q × q = p 3q 4 Since the bases are not the same we cannot simplify further. m×m×m×m×m×m b m6 ÷ n4 = -----------------------------------------------------------n×n×n×n 6 m = ------4n Again, since the bases are different we cannot simplify further.

9

Explain why the index laws cannot be used to simplify: a k5 × m 3 b x9 ÷ y6

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

10

Are the following statements true or false? a b4 × b 3 = b7

b m 5 × m 2 = m10

c p 4 × p 5 = p 20

d e 6 × e10 = e16

e a4 × b 5 = ab 9

f

z10 ÷ z 2 = z 8 6 4 p p ----2- = ----q q

g p12 ÷ p 3 = p4

h t8 ÷ t7 = t

k (b7)2 = b14

l

i

w15 ÷ w 3 = w 5

j

(n10)3 = n13

C. APPLYING THE INDEX LAWS Example 1 Simplify: 5

6

5

6

5 4

(a ) b ---------------3 2 a ×a

p ×p a ----------------8 p

5 4 5×4 (a ) a ------------b ---------------= 3 2 3+2 a a ×a

5+6

p ×p p a ----------------= ----------8 8 p p 11

20

p = ------8 p = p11 – 8

a = ------5 a = a20 – 5

= p3

= a15

Exercise 4C 1

Simplify: 5

7

3

x ×x a ---------------6 x 7

6

9

a ×a e ---------------8 2 a ×a k -----------------16 5 k ×k

c

11

f

y ×y -----------------10 8 y ×y

j

(m ) × m

30

i

10

w ×w b --------------------8 w

2 3

16

6

4

k ×k d -----------------8 5 k ×k

10

16

2

x h ---------------3 4 x ×x

14

z ×z g -----------------10 7 z ×z 4 5

5

3 4

k (a ) × (a ) 4

10

a ×a ×a n ------------------------------12 8 a ×a ×a

5 5

y ) m (----------20 y

8

m ×m -------------------10 m

20

30

b ×b ×b o -----------------------------------4 5 (b )

Example 2 Simplify: 4

a 5m × 3m

6

b

7

3

2k × 4k × 3k

5

5 6

l

(t ) ----------10 t

6

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

4

5m × 3m

a

6

4

b

= 5×3×m ×m = 15 × m = 15m

2

6

7

3

2k × 4k × 3k 7

5

3

= 2×4×3×k ×k ×k

4+6

= 24 × k

10

= 24k

5

7+3+5

15

Simplify: 5

a 4m × 3m d 10a

12

7

× 7a

6

b 4

8

g 3z × 4z × 2z

4

6

c

3t × 6t

9

10

f

5b × 6b × b

i

d × 6d × 3d

5p × 2p

e 4w × 6w 3

5

7

h 2q × 5q × 8q

6

8

4

3

4

2

6

4 8

Example 3 Simplify: 8

10

12m a ------------6 3m

b 8

a

10

12m ------------6 3m

b

20a -------------4 16a

12 × m = ------------------6 3×m

8

20 × a = -------------------4 16 × a

10

8

20 a = ------ × ------16 a 4

10

12 m = ------ × ------63 m = 4×m = 4m

3

20a -------------4 16a

6 5 = --- × a 4

2

6

5a = --------4

2

Simplify: 7

6m a ----------23m 10

f

9e ----------66e

12

10a b -------------7 5a 8

2m g ----------36m

10

c

12w --------------8 4w

12

8z d ----------86z

15

6a h -------------10 12a

9

16k e -----------312k

13

i

9t ----------612t

11

j

15b -------------6 20b

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

Example 4 3 5

Simplify ( 2a ) 3 5

( 2a ) 3

3

3

3

= 2a × 2a × 2a × 2a × 2a 3

3

3

3

3

= 2×2×2×2×2×a ×a ×a ×a ×a

3

3 5

5

= 2 × (a ) = 32a

4

15

Simplify: 4 3

b ( 2m )

3 2 5

g (m n )

a ( 3a )

(x y )

f

3 6

c ( 7p )

5 2

4 6 3

h (p q )

2 4

d ( 10k )

7 3 4

i

4 10 2

(a b )

Example 5 Simplify: 2 4

10 8

5 8

a 5m n × 3m n

a

2 4

b

10 8

5 8

5m n × 3m n 2

4

5

7

7 12

= 15m n

12

10

8

5

8

y 12 x = ------ × ------6- × ----28 x y

4

8

=

= 5×3×m ×m ×n ×n = 15 × m × n

12x y ------------------6 2 8x y

b

= 5×m ×n ×3×m ×n 2

12x y ------------------6 2 8x y

3 --2

× x4 × y6 4 6

3x y = -------------2

11 3

e ( 5t ) j

5 2 3

( 2x y )

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

5

Simplify: 3 2

5 3

b 5m n × 2m n

5 8

6 7

d 10x y × 3x y

a 4a b × 2a b

c 3p q × 4p q 10 12

e 2w z

4 5

× 6w z

10 9

4

4 3

6

5 6

f

6a b -------------3 2 4a b 7

15x y g ------------------6 2 5x y

Remember to add indices when multiplying and subtract them when dividing.

12

2k m h ------------------3 6 10k m

11 6

i

6 7

7 8

9a b ----------------8 12a b

j

12m n ------------------6 8 15m n

D. THE ZERO INDEX Example 1 a Use the index laws to simplify a5 ÷ a5. b Hence show that a0 = 1. a Using the index laws, a5 ÷ a5 = a5 − 5 = a0 b But a5 ÷ a5 = 1 (Since any number divided by itself = 1) Hence a0 = 1

Exercise 4D 1

a Use the index laws to simplify a4 ÷ a4.

b Hence show that a0 = 1.

2

a Use the index laws to simplify k 7 ÷ k 7.

b Hence show that k0 = 1.

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

Example 2 Evaluate: a x0

b (3x)0

c 3x 0

a x0 = 1

b (3x)0 = 1

c 3x 0 = 3 × x 0 =3×1 =3

3

Evaluate: a y0 f (6z)0 k 3m0 + 1

b (3y)0 g (10m)0 l 9e0 – 3

c 3y 0 h 10m0 m 6p0 + 7

d 4k 0 i 8b0 n 3a0 + 2b0

e 9t 0 j (7q)0 o 6x 0 – 4y 0

E. INDICES FOR SQUARE ROOTS AND CUBE ROOTS Example 1 2

a Show that ( 6 ) = 6. 1 --2 2

b Use the index laws to show that ( 6 ) = 6. 1 --2

c Hence show that a ( 6)

2

6 = 6 . 1 --2

1 --- × 2 2

=

6 ×

=

6×6

= 61

=

36

=6

6

b

( 6 )2 = 6

=6 1 --2

2

c Since ( 6 ) = 6 and ( 6 )2 = 6 then

1 --2

6 =6 .

Exercise 4E 1

c Hence show that 2

1 --2

a Show that ( 5 )2 = 5.

b Use the index laws to show that ( 5 )2 = 5. 1 --2

5 =5 . 1 --2

a Show that ( a )2 = a. c Hence show that

b Use the index laws to show that ( a )2 = a. 1 --2

a =a .

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

Example 2 Write in index form: a

23

a

23 = 23

1 --2

b

t

b

t =t

1 --2

c

7t

d 7 t

c

7t = ( 7t )

1 --2

d 7 t =7×

t

=7× t = 7t

3

Write in index form: a 3 5k e

12 b f 5 k

x c g 6 y

1 --2

1 --2

m 6y

d h

Example 3 Write down the meaning of: a 5

1 --2

b ( 7z )

1 --2

a 5 =

1 --2

1 --2

b ( 7z ) =

5

7z

c

7z

c

7z

1 --2

1 --2

=7× z =7×

1 --2

z

=7 z

4

Write down the meaning of: a 8

1 --2

e ( 3z )

b 13 1 --2

f

1 --2

( 2m )

c p 1 --2

1 --2

g 5k

d q 1 --2

1 --2

h 4t

1 --2

Example 4 3

a Show that ( 3 5 ) = 5. 1 --3

b Use the index laws to simplify ( 5 )3. c Hence show that

3

1 --3

5 =5 .

☞

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

3

1 --- × 3 3

3

5 ×

=

3

5×5×5

= 51

=

3

125

=5

3

5 ×

1 --3

a (3 5) =

3

b ( 5 )3 = 5

5

=5 1 --3

3

c Since ( 3 5 ) = 5 and ( 5 )3 = 5 then

5

1 --3

5 =5 . 1 --3

3

a Show that ( 3 6 ) = 6. c Hence show that

6

3

3

b Use the index laws to simplify ( 6 )3. 1 --3

6 =6 . 1 --3

3

a Show that ( 3 a ) = a. c Hence show that

3

b Use the index laws to simplify ( a )3. 1 --3

a =a .

Example 5 Write in index form: a

3

7

a

3

7

=7

1 --3

b

3

e

b

3

e

=e

c

3

4z

c

3

4z

1 --3

d 43 z d 43 z = 4 ×

= ( 4z )

1 --3

=4× z = 4z

7

3

z 1 --3

1 --3

Write in index form: a

3

2

b

3

9

c

3

c

d

3

e

3

5y

f

53 y

g

3

9m

h 93 m

w

Example 6 Write down the meaning of: a 8

1 --3

b n

1 --3

c

( 5x )

1 --3

d 5x

1 --3

☞

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

a 8

1 --3

b n

3

=

8

=

1 --3

3

c n

( 5x ) =

3

1 --3

d 5x

1 --3

=5×x =5×

5x

3

1 --3

x

= 53 x

8

Write down the meaning of: a 12

1 --3

b 35

e ( 6m ) 9

c k

1 --3

f

6m

47

b

p

e 5 x

f

63 x

1 --3

g ( 7v )

d d 1 --3

1 --3

h 7v

1 --3

Write in index form: a

10

1 --3

1 --3

c

3

d 83 p

x

h 33 r

g 4 x

Write down the meaning of: a 5 e

1 --3

b 6

( 5p )

1 --2

f

1 --2

( 6r )

c x 1 --3

1 --3

g ( 5xy )

d t 1 --2

h ( 4pq )

F. NEGATIVE INDICES Example 1 4

a a Use the index laws to simplify ----5- . a 1 –1 c Hence show that a = --- . a 4

4

= a–1 4

4

a b Expand and simplify ----5- . a

1

1

1

1

a a ×a ×a ×a b ----5- = -------------------------------------------------1 1 1 1 a ×a ×a ×a ×a a 1 = --a

a a ----5- = a4 – 5 a

4

a a 1 1 c Since ----5- = a–1 and ----5- = --- then a–1 = --- . a a a a

1 --2 1 --3

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Exercise 4F 3

1

2

3

4

5

3

a a Use the index laws to simplify ----4- . a 1 c Hence show that a–1 = --- . a 3 a a Use the index laws to simplify ----5- . a 1 –2 c Hence show that a = ----2- . a 2 a a Use the index laws to simplify ----5- . a 1 –3 ---c Hence show that a = 3 . a 2 a a Use the index laws to simplify ----6- . a 1 c Hence show that a–4 = ----4- . a a a Use the index laws to simplify ----6- . a 1 c Hence show that a–5 = ----5- . a

a b Expand and simplify ----4- . a 3

a b Expand and simplify ----5- . a 2

a b Expand and simplify ----5- . a 2

a b Expand and simplify ----6- . a a b Expand and simplify ----6- . a

From the above exercises, it can be seen that, in general, 1 a–n = ----n a

Example 2 Write down the meaning of:

6

a k–9

b m–15

1 a k–9 = ----9k

1 b m–15 = -------15 m

Write down the meaning of: a y −2 b k –1

c m–3

Example 3 Write with a negative index: 1 a ----5a

1 b ----7y

1 a ----5- = a–5 a

1 b ----7- = y –7 y

d x –6

e t –10

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

7

Write with a negative index: 1 1 a ----8b ----2k a

c

1 -----11 x

1 d ------14 n

1 e -----20 z

Example 4 Write down the meaning of: a 3m–2

b (3m)–2

a 3m–2 = 3 × m–2

1 b (3m)–2 = ---------------2 ( 3m ) 1 = ----------29m

3 1 = --- × ------21 m 3 = ------2m

8

Write down the meaning of: a 3k –1 b (3k)–1 –5 d (2y) e 3t –4

2y –5 (3t)–4

c f

G. FURTHER USE OF THE INDEX LAWS Example 1 Use the index laws to simplify: 1 --2

1 --2

1 --2

1 --2

a 3y × 2y

1 --3

1 --3

1 --3

1 --3

b 10 n ÷ 5 n 1 --2

a 3y × 2y = 3 × y × 2 × y 1 --2

=3×2× y × y =6× y

1 1 --- + --2 2

= 6 × y1 = 6y

1 --2 1 --2

1 --3

10n b 10 n ÷ 5 n = -----------1 5n

--3

1 --3

10 n = ------ × -----1 5 --3 n =2× n

1 1 --- – --3 3

= 2 × n0 =2×1 =2

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Exercise 4G 1

Use the index laws to simplify: 1 --2

a 2z × 6z d 12 a

1 --3

1 --2

1 --2

b 8p ÷ 2p

÷4a

1 --3

1 --2

1 --3

1 --3

2m × 3m × 4m

c

1 --3

3

k e ---------------1 1 --2

k ×k

--2

Example 2 Use the index laws to simplify: a m–6 × m2

b q–2 ÷ q−7

c (x –3)5

a m–6 × m2 = m–6 + 2

b q−2 ÷ q –7 = q–2 – (−7)

c (x –3)5 = x –3 × 5

= m–4 2

= q5

Use the index laws to simplify: –5 –2 –3 7 a a ×a b y ×y e b i

–6

÷b

2

–2 4

(y )

3

f

w ÷w

j

(t )

–2

= x –15

5

c e ×e g z

–2

–7

÷z

4

d n ×n

–4

h k

5 –4

Example 3 Simplify: a 5m–3 × 6m7

b 4y 7 ÷ 5y –2

a 5m–3 × 6m7 = 5 × 6 × m–3 × m7 = 30 × m–3 + 7 = 30 × m4 = 30m4

c (5y –2)3 = 5y –2 × 5y –2 × 5y –2 = 5 × 5 × 5 × y –2 × y –2 × y –2 3

–2 3

= 5 × (y ) = 125y –6

c (5y –2)3 7

4y b 4y 7 ÷ 5y –2 = ---------–2 5y 7 4 y = --- × -----5 y –2 4 = --- × y 7–(–2) 5 4 = --- × y 9 5 9 4 4y = --- y 9 or -------5 5

–6

–3

÷k

–2

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

3

Simplify: a 10a 5 × 9a –3 e 6p–4 ÷ 2p 2 i (3w –6)2

b 6b–5 × 3b–2 f 3k –4 ÷ 8k –2 j 4n–3 × 3n–4 ÷ 6n–5

c 3v –6 × 2v 2 g (5z –4)3

d 8y 5 ÷ 2y –1 h (2m–3)5

Example 4 State whether the following are true or false. a m3 ÷ m5 = m2

b 3y 0 = 1

1 d 2p–3 = --------32p

c 6k4 ÷ 2k4 = 3

a m3 ÷ m5 = m3 – 5

b 3y 0 = 3 × y 0 =3×1

= m–2 Statement is false.

=3 Statement is false.

4

6 k c 6k 4 ÷ 2k 4 = --- × ----42 k = 3 × k4 – 4

d 2p–3= 2 × p–3 1 = 2 × ----3p 2 = ----3p Statement is false.

= 3 × k0 =3×1 =3 Statement is true.

4

State whether the following are true or false. a 6m 0 = 1

b a4 ÷ a7 = a 3

c

8t 9 ÷ 2t 9 = 4

1 d 3c −2 = --------23c 6 g 5x ÷ x 6 = 5x

e 4k 0 = 4

f

b ÷ b6 = b5

4 h 4y −3 = -----3 y

i

8 (2p −1)3 = ----3p

Example 5 By substituting a = 5, show that a–2 ≠ −2a.

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1 a−2 = ----2a 1 = -----2 5 1 = -----25

–2a = −2 × 5 = –10

Hence a–2 ≠ –2a 5

By substituting a = 3, show that: a a 2 ≠ 2a d a–3 ≠ –3a g a2 – a2 ≠ a0

b a3 ≠ 3a e a2 × a ≠ a2 + a h 5a2 × 3a ≠ 5a2 + 3a

c f

a–2 ≠ –2a a2 + a2 ≠ a4

H. REMOVING GROUPING SYMBOLS Example 1 Expand: a 3(x + 5)

b 4(3y – 2z)

Expand means to write the expression without the grouping symbols.

a 3(x + 5) = (x + 5) + (x + 5) + (x + 5) =x+x+x+5+5+5 =3×x+3×5 = 3x + 15 b 4(3y – 2z) = (3y – 2z) + (3y – 2z) + (3y – 2z) + (3y – 2z) = 3y + 3y + 3y + 3y – 2z – 2z – 2z – 2z = 4 × 3y – 4 × 2z = 12y – 8z

From the examples above we can see that, in general: a × (b + c ) = a × b + a × c that is, to remove the grouping symbols we multiply each term inside them by the number at the front. This result is known as the distributive law.

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

Example 2 Use the distributive law to expand: a 5(2y + 3)

b 7(3y – 4w)

a 5(2y + 3) = 5 × 2y + 5 × 3

b 7(3y – 4w) = 7 × 3y – 7 × 4w

= 10y + 15

= 21y – 28w

Exercise 4H 1

Use the distributive law to expand: a 3(2w + 5) b 6(3z – 2) 2 e 10(x + 6) f 7(ab – 2a 2) i 5(4b + 2a + 3) j 3(5x – 3y – 2z)

c 5(4a + 3b) g 4(m 2 + n 2)

d 2(4x – 3y) h 2(m 3 – 3mn)

Example 3 Expand: a 3w(2y + 4z)

b 2a(3a – 4b)

a 3w(2y + 4z) = 3w × 2y + 3w × 4z

c 4m2(m3 + 2m5)

b 2a(3a – 4b)= 2a × 3a – 2a × 4b = 6a2 – 8ab

= 6wy + 12wz c 4m 2(m 3 + 2m5) = 4m 2 × m3 + 4m 2 × 2m5 = 4m5 + 8m7

2

Expand: a 3a(2b + 4c) e y2(y3 – 4) i 2p5(p2 + 3p3)

b 4x(3x – 2y) f 6x(2y – 5x2) j 5x2(2x3 – 3xy)

c 10k(6k – 4m) g 3k2(2k2 + 5)

d m(m2 + 2) h a3(5a2 – 2)

Example 4 Expand: a –3(2w + 5)

b –2(4a – 3b)

c –(4m + 3n)

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a –3(2w + 5) = –3 × 2w + –3 × 5

b –2(4a – 3b) = –2 × 4a – –2 × 3b

= –6w + –15

= –8a – –6b

= –6w –15

= –8a + 6b

c –(4m + 3n) = –1 × 4m + –1 × 3n = –4m + –3n = –4m – 3n

3

Expand: a –2(y + 3) e –(t + 3) i –(7w + 3)

b –5(a + 2) f –(b + 6) j –(4x – 1)

c –3(w + 4) g –3(2k + 5)

d –4(m – 7) h –2(4m – 5)

Example 5 Expand:

4

a –3a(5a2 + 2ab)

b –n3(2n4 – 5n2p)

a –3a(5a2 + 2ab) = –3a × 5a2 + –3a × 2ab

b –n3(2n4 – 5n2p) = –n3 × 2n4 – –n3 × 5n2p

Expand: a –2a(3a 2 + 2ab) d –y3(4y2 – 3xy)

= –15a3 + –6a2b

= –2n7 – –5n5p

= –15a3 – 6a2b

= –2n7 + 5n5p

b –4x(2x 2 – 3xy) e –3m4(2m 2 + 5mn)

Example 6 Expand and simplify by collecting like terms. a 3(a + 2) + 7

b 3 + 2(3n – 5)

a 3(a + 2) + 7 = 3a + 6 + 7 = 3a + 13

Remember to multiply before adding!

b 3 + 2(3n – 5) = 3 + 6n – 10 = –7 + 6n or 6n – 7

c

–3p 2(3p 2 + 4pq)

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

5

Expand and simplify: a 4(a + 3) + 6 e 6(3z – 1) + 4 i 4 + 3(2w – 4)

b 2(3b – 12) + 12 f 10 + 2(4x + 3) j 16 + 5(4e – 6)

c 3(4w + 2) – 7 g 12 + 2(3b – 5)

d 5(2y – 3) – 2 h 13 + 4(y + 5)

Example 7 Expand and simplify 5 – 2(4y – 3). 5 – 2(4y – 3) = 5 – 8y + 6 = 11 – 8y or – 8y + 11 6

Expand and simplify: a 12 – 2(a + 5) e 20 – 3(2w + 5) i 5 – 3(3 + 4z)

b 8 – 3(y – 2) f 2 – 5(3t – 4) j 3 – 10(1 – 2w)

c 9 – 4(b + 3) g 4 – 3(5x + 2)

d 7 – 2(v – 6) h 10 – 2(3k – 1)

Example 8 Expand and simplify 4(2m – 3) + 3(m – 2). 4(2m – 3) + 3(m – 2) = 8m – 12 + 3m – 6 = 11m – 18 7

Expand and simplify: a 5(2k + 3) + 3(k – 2) c 4(2p – 1) + 2(3p + 5) e 2(5x – 3) + 5(3x – 1) g (6v – 1) + 3(2v – 5) i 7(2a – 3b) + 3(3a + 4b)

b d f h j

2(6m + 7) + 3(m – 1) 3(3a + 2) + 4(a – 3) 3(4y – 2) + (2y + 7) 4(3x + 2y) + 2(5x – 3y) 3a(2a + 6) + 4a(3a – 5)

Example 9 Expand and simplify: a 2(3p + 4q) – 4(2p – 3q)

b 3(4m – 1) – (m + 4)

a 2(3p + 4q) – 4(2p – 3q) = 6p + 8q – 8p + 12q = –2p + 20q or 20q – 2p b 3(4m – 1) – (m + 4) = 12m – 3 – m – 4 = 11m – 7

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8

Expand and simplify: a 3(2k + 5) – 2(k + 3) c 2(6t + 1) – 3(t + 4) e 2(a + 5) – 4(a – 1) g 4(3x + y) – (2x – 7y) i 2q(q – 5) – 4(q – 5)

b d f h j

5(w + 4) – 3(w – 2) 3(5z – 1) – (2z + 5) 5(d – 3) – 3(2d + 1) 3(2a – 3b) – (2a + 3b) 4z(3z + 2) – (z – 1)

I. FACTORISING Use the distributive property to expand 2(3a + 5) = 2 × 3a + 2 × 5 = 6a + 10 Reversing the process, 6a + 10 = 2 × 3a + 2 × 5 = 2 × (3a + 5) = 2(3a + 5) The reverse process to expanding is called factorising.

Note that the 2 is the HCF of 6a and 10.

Example 1 Factorise: a 3y + 12

b 20k – 8

a The HCF of 3y and 12 is 3, hence

b The HCF of 20k and 8 is 4, hence

3y + 12 = 3 × y + 3 × 4

20k – 8 = 4 × 5k – 4 × 2

= 3 × (y + 4)

= 4 × (5k – 2)

= 3(y + 4)

= 4(5k – 2) Note that as 2 is a common factor of 20k and 8, 20k – 8 = 2 × 10k – 2 × 4 = 2(10k – 8) While this is a correct equivalent expression for 20k – 8, it has not been fully factorised. 20k – 8 = 4(5k – 2).

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

Exercise 4I 1

Factorise: a 8a + 10 e 4w – 12 i 24k – 18n

b 6x – 4 f 16m – 8 j 16x 2 + 24y2

c 3a + 6b g 12ab + 8

d 5x + 10y h 10m – 20n

Example 2 Factorise: a 3u + 12v + 9w

b 12a – 8b + 20c

a The HCF of 3u, 12v and 9w is 3, hence 3u + 12v + 9w = 3 × u + 3 × 4v + 3 × 3w = 3(u + 4v + 3w) b The HCF of 12a, 8b and 20c is 4, hence 12a – 8b + 20c = 4 × 3a – 4 × 2b + 4 × 5c = 4(3a – 2b + 5c)

2

Factorise: a 5x + 15y + 10z d 12m – 6n – 18r

b 3p – 6q + 9r e 20xy + 50z + 30

c

4a + 12b – 8c

Example 3 Complete the following factorisations and check by expanding. a y 2 + 2y = y (__ + __)

b 12a2 – 8ab = 4a (__ – __)

a Since y 2 + 2y = y × y + y × 2, then

b

y 2 + 2y = y(y + 2) Check: y(y + 2) = y × y + y × 2 = y 2 + 2y

3

Since 12a 2 – 8ab = 4a × 3a – 4a × 2b, then 12a 2 – 8ab = 4a(3a – 2b) Check: 4a(3a – 2b) = 4a × 3a – 4a × 2b = 12a 2 – 8ab

Complete the following factorisations and check by expanding. a y 2 + 7y = y (__ + __) b m 2 – 3m = m (__ – __) c 3mn + 4m = m (__ + __) d 9p – 5pq = p (__ – __) 2 f 2bc – b 2 = b (__ – __) e x + 5xy = x (__ + __)

☞

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g 6m 2 – 3m = 3m (__ – __) i 16pq – 12p 2 = 4p (__ – __)

h 12a 2 – 10ab = 2a (__ – __) j 12k 2 +18k = 6k (__ + __)

Example 4 Complete the following factorisations and check by expanding. a y2 + 7y = __ (y + 7)

b 2m2 – 8m = __ (m – 4)

a The HCF of y2 and 7y is y, hence

b The HCF of 2m2 and 8m is 2m, hence

y2 + 7y = y(y + 7)

2m2 – 8m = 2m(m – 4)

Check: y(y + 7) = y × y + y × 7

Check: 2m(m – 4) = 2m × m – 2m × 4

= y2 + 7y

4

= 2m2 – 8m

Complete the following factorisations and check by expanding. a p2 + 3p = __ (p + 3) b k2 – 2k = __ (k – 2) 2 c 3w + 2w = __ (3w + 2) d 2z2 – z = __ (2z – 1) e 4mn – 3m2 = __ (4n – 3m) f 2x2 + 8x = __ (x + 4) 2 g 4pq – 12p = __ (q – 3p) h 8pq – 12pr = __ (2q – 3r) 2 i 10z – 5z = __ (2z – 1) j 6km – 8m2 = __ (3k – 4m)

Example 5 Explain why the following factorisation is incorrect. 18x2y + 12xy2 = 6xy(3x + 2) 6xy (3x + 2) = 6xy × 3x + 6xy × 2 = 18x2y + 12xy ≠ 18x2y + 12xy2

5

Explain why the following factorisations are incorrect. a y2 + 7y = 7(y2 + y) b 24ab + 8a = 8a(3b – 1) d 15x2y + 12xy 2 = 3xy(5x + 4) c x2 − 6xy = 6x(x − y) e 10ab2 – 5a2b = 5ab(2b2 – a)

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

Example 6 Factorise: a w 2 + 5w

b 3y 2 – 6y

a The HCF of w 2 and 5w is w, hence

c 4a 2 + 8ab b The HCF of 3y 2 and 6y is 3y, hence

w 2 + 5w = w × w + w × 5

3y 2 – 6y = 3y × y – 3y × 2

= w(w + 5)

= 3y(y – 2)

c The HCF of 4a 2 and 8ab is 4a, hence 4a 2 + 8ab = 4a × a + 4a × 2b = 4a(a + 2b)

6

Factorise: a x 2 + 4x e 2k 2 + 4k i 12pq – 18p 2

b y 2 – 7y f 3y 2 – 12y j 16km + 24k 2

c a 2 + ab g 10b 2 + 5ab

d m 2 – 5mn h 9w 2 – 6w

Example 7 Factorise 18ab – 3a + 9a2. The HCF of 18ab, 3a and 9a 2 is 3a, hence 18ab – 3a + 9a 2 = 3a × 6b – 3a × 1 + 3a × 3a = 3a(6b – 1 + 3a)

7

Factorise: a 2ab + 4a + 4a 2 d 5xy – 10x – 5x 2

b 6x 2 + 3x + 9xy e 8k 2 – 6k – 10km

c

2m – 6m 2 + 4mn

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Language in Mathematics

1

Three of the words in the following list have been spelt incorrectly. Find these words and write the correct spelling. reduse, simplify, subistute, aply, numerator, equivalent

2

Complete the following words used in this chapter by replacing the vowels: a f __ ct __ r b z __ r __ c __ nd __ x d __ lg __ br __ __ c e f __ ct __ r __ s __

3

Using an example, explain the meaning of: a reciprocal b highest common factor

4

Write in words:

c

x3

5

Write down the names of the following grouping symbols: a () b []

c

6

7

a

x2

b

x

d

3

x

{}

Match each word with its meaning. a equivalent

A go backwards

b expand

B the same as

c reverse

C a letter used to represent numbers

d evaluate

D remove the grouping symbols

e pronumeral

E find the value of

How many words of three or more letters can you make from the word DISTRIBUTIVE. (No proper names or plurals allowed.)

Glossary algebraic

apply

base

cancel

check

common

cube root

denominator

distributive law

equivalent

evaluate

expand

expression

factor

factorise

grouping symbols

index

indices

negative

numerator

power

reciprocal

reduce

reverse

simplify

square root

substitute

value

zero

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

121

CHECK YOUR SKILLS 1

Which of the following is not true? 3x 9x A ------ = -----5 15

2

3

7

5a C -----81

45a D ---------9

11x B --------15

11x C --------8

3x D -----15

45a B ---------21b

28ab C ------------15

20ab D ------------21

xy C -----6y

x D --6

5x B -----3y

5y C -----3x

xy D -----15

14a B ---------15

35ab C ---------------54

x y --- ÷ --- = 5 3

7ab 5b ---------- ÷ ------ = 9 6

2

54 D ---------------235ab

Which of the following does not simplify to k15? A (k5)3 7

10

5a B -----18

5 3xy When simplified ------ × --------- = 9y 10 15xy 15x A -----------B --------90 90

15 A ---------14a 9

3xy D --------2x

4a 5 ------ × ------ = 3 7b

3x A -----5y 8

✓

3y C -----2

2x x ------ + --- = 5 3

20a A ---------21b 6

3x 6x D ------ = -----5 8

4a a ------ + --- = 9 9

3x A -----8 5

3x 15x C ------ = --------5 25

9xy When reduced to its simplest form --------- = 6x 9y A -----B 3y 6

5a A -----9 4

3x 6x B ------ = -----5 10

B k

10

×k

5

C k

30

÷k

2

3 5

D (k )

9

b ×b ----------------= 2 4 (b ) A b55

B b2

C b8

D 18

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9 8

11

12x y ----------------= 6 2 8x y 3 6

A 4x3y6 12

17

18

1 --2

20

21

22

D 9m7

B 20

C 3

D 8

B ( 9a )

1 --2

C 3a

1 --2

The meaning of 2a is: 3

2a C -----3

B 23 a

2a

1 Which of the following is equivalent to ------2- ? m 1 A -------B m–2 2m

2 C ---m

2c–2 is equivalent to: 2 A ----2c

B −4c

1 C --------22c

3xy -----------2

B 3m–2

1 C --- m2 3

✓ D 9a

2

D 2a

3

D m

6m –3 ÷ 2m–5 = A 3m2

19

C 3m7

1 --3

A 16

B 9m10

9 a may be written in index form as: A 9a

15

D

5y 0 + 3 = A 4

14

9

3x y C -------------2

(3m5)2 = A 3m10

13

3 4

3x y B -------------2

−3k(2k 2 – 4km) =

1 --2

1 D --------24c

1 D --- m–2 3

A –6k 2 + 12km

B

–6k 2 – 12km

C –6k 3 + 12k 2 m

D –6k 3 + 12k 2 m

When expanded and simplified 2w(w – 6) + 3(w – 5) = B 2w2 – 9w – 15 C –7w – 15 A 2w2 – 3w – 15 The highest common factor of 10pq and 12p2 is: A 2p B 2

C p

When fully factorised 4x2 – 6x = A 2x2 (2x2 – 3) B 2(2x2 – 3x)

C x(4x – 6)

D –7w2 – 15

D 2pq

D 2x(2x – 3)

If you have any difficulty with these questions, refer to the examples and questions in the sections listed in the table. Question

1–8

9, 10

11, 12

13

Section

A

B

C

D

14, 15 16, 17 E

F

18 G

19, 20 21, 22 H

I

LEY_bk9_04_finalpp Page 123 Wednesday, January 12, 2005 10:32 AM

Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

REVIEW SET 4A

2

Reduce to simplest form: 12a 9xy a ---------b --------15 6x

3

Simplify: 4a 7a a ------ + -----13 13

2m 5m b -------- + -------3 8

Simplify: a y 10 × y 7 e (5m4 )3

b k 11 ÷ k 5 f 3a 5 b 3 × 2ab 6

c (p7 )2

t ×t d -------------3 4 t ×t

Evaluate: a v0

b 5v 0

c (5v)0

d 2v 0 + 1

5

6

8

9

10

11

12

c

4w 2w ------- × ------5 3

5x y d ------ ÷ --6 3 7

8

Write the meaning of: a x

7

b

3mn ■ ------------ = -----4 20

Complete:

4

a

2x ■ ------ = -----9 18

1

1 --2

b 3x

1 --2

c ( 3x )

Write the meaning of: a z –3

b 2z –3

Simplify: a y –3 × y5 d 6b–2 × 3b7

b e6 ÷ e–2 e 4k–5 ÷ 2k–3

1 --2

State whether the following are true or false. a 4q 0 = 4 b a 5 ÷ a7 = a2 4 –2 d 4b = ----2e n2 × n = n2 + n b Expand: a 5(2v – 4w) b a3 (2a2 + 4a) Expand and simplify: a 4(m – 2) + 3(2m + 5)

b 3(3a – b) – (2a – b)

Factorise: a 8w + 20

b x2 + 9x

d x

1 --3

c (2z)–3

c

(n–4 )5

c

6m5 ÷ 3m5 = 2m

c

–3(4x + 5)

c

4pq – 12q2

e 2x

1 --3

123

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

REVIEW SET 4B 3x ■ a ------ = -----7 21

Complete:

2

Reduce to simplest form: 8m 4ab a -------b ---------12 2a

3

Simplify: 2a 7a a ------ + -----9 9

3y 2y b ------ + -----4 3

Simplify: a m14 × m6 e (2m7 )4

b t25 ÷ t5 f 4p3 q7 × 5p4q

c (z6 )4

(b ) d ----------------7 4 b ×b

Evaluate: a s0

b 4s 0

c (4s)0

d 4s 0 − 1

4

5

6

8

9

10

11

12

c

3k 2m ------ × -------5 3

3w 2w d ------- ÷ ------5 3 5 6

Write the meaning of: a c

7

b

2ab ■ ---------- = -----3 18

1

1 --2

b 2c

1 --2

c ( 2c )

Write the meaning of: a e–4

b 3e–4

Simplify: a k –6 × k2 d 5n−3 × 4n8

b m–4 ÷ m–1 e 2a−5 ÷ 4a−8

1 --2

State whether the following are true or false. b b6 ÷ b9 = b–3 a 3w 0 = 1 1 d 2t –2 = -------2e p –2 = –2p 2t Expand: a –10(4p + 3) b m2 (3m5 − m3) Expand and simplify: a 3(2q + 6) − 2(q − 7)

b 2a(2a – 5) – (3a + 1)

Factorise: a 12x − 18

b 2y2 − 7y

d c

1 --3

c (3e)–4

c

(n3)–5

c

a7 ÷ a7 = a

c

–a(2a − 5)

c

4a2 – 3ab + 2a

e ( 2c )

1 --3

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

REVIEW SET 4C

2

Reduce to simplest form: 8w 5mn a ------b -----------24 7n

3

Simplify: 2d 6d a ------ + -----7 7

7b 3b b ------ – -----8 4

Simplify: a p6 × p8 e (3v7 )3

b y16 ÷ y 8 f 3x8 y9 × 6x2y5

c (t5 )6

c ×c d -----------------4 5 (c )

Evaluate: a a0

b 7a0

c (7a)0

d 7a0 + 5

5

6

8

9

10

11

12

c

6x 10y ------ × --------5 9

3w 2w d ------- ÷ ------4 5 12

8

Write the meaning of: a q

7

b

4xy ■ --------- = -----3 12

Complete:

4

a

3h ■ ------ = -----5 20

1

1 --2

b 4q

1 --2

c ( 4q )

1 --2

d q

Write the meaning of: a b −5

b 3b −5

c (3b)−5

Simplify: a d –5 × d –3 e 6m3 ÷ 9m–4

b n –2 ÷ n–3

c (k –2)–3

State whether the following are true or false. a 7h0 = 7 b a ÷ a5 = a−5 3 −4 d 3s = ----4e n2 + n2 = n4 s Expand: a 4(5w + 2x) b –k4 (2k3 − 4k) Expand and simplify: a 2(5t − 1) + 3(2 − 3t)

b 10(x – 2y) – 5(2x − y)

Factorise: a 15n − 20

b 4b2 + 6b

1 --3

c

1 2p5 ÷ 6p5 = --- p 3

c

–(4s − 7)

c

12m2 + 14mn

e 4q

1 --3

d 5a–6 × 3a3

125

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

REVIEW SET 4D

2

Reduce to simplest form: 20h 4pq a ---------b ---------16 6q

3

Simplify: 10k 5k a ---------- – -----11 11

2w w b ------- – ---9 6

Simplify: a y8 × y5 e (9h5 )2

b k16 ÷ k10 f 6a7 b8 × a6b 3

c (p5 )10

c ×c d -----------------4 2 (c )

Evaluate: a w0

b 8w 0

c (8w) 0

d 8w 0 − 9

5

6

8

9

10

c

3a 4b ------ × -----2 5

9x 4x d ------ ÷ -----10 3 3

13

Write the meaning of: a m

7

b

7pq ■ ---------- = -----2 10

Complete:

4

a

5b ■ ------ = -----8 24

1

1 --2

b 6m

1 --2

c ( 6m )

Write the meaning of: a x –2

b 2x –2

Simplify: a y –2 × y –5 d 5z−4 × 3z7

b n–6 ÷ n–8 e 6m−4 ÷ 9m−7

1 --2

State whether the following are true or false. b a9 ÷ a7 = a–2 a 2b0 = 1 7 d 7n–3 = ----3e (2m)2 = 4m2 n Expand: a 2x(6y − 3z) b –m3(3a3 − 4m2)

11

Expand and simplify: a 5(3 + 5n) + 2(4 − 3n) b a(6a + 1) – 2a(a + 1)

12

Factorise: a 24x + 18

b h2 − 8h

d m

1 --3

c (2x)–2

c

(p–3)5

c

2p6 ÷ 10p6 = 2

c

–2(5 − 2a)

c

3y2 + 6y − 9

e ( 6m )

1 --3