Chapter 2 Further algebraic skills

Further algebraic skills

The main mathematical ideas investigated are:

▶ adding and subtracting like algebraic terms

▶ multiplying and dividing algebraic terms

▶ adding and subtracting algebraic fractions

▶ establishing and applying the index laws in

algebra

▶ solving linear equations.

ALGEBRA AND MODELLING

Syllabus references: AM3CEC

Outcomes: MG1H-3, MG1H-9, MG1H-10

UNCORRECTED PAGE PROOFS

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Insight Mathematics General 12 HSC course 124

2A Simplifying algebraic expressions Simplifying algebraic expressionsThis section begins with a review of some of the simplifi cations used in the Preliminary General Mathematics course.

WORKED EXAMPLE 1Simplify, where possible, by collecting like terms.

a 9a − a b 3xy − 5xy c 4x − x2 + 7x

Solve Think Apply

a 9a − a = 8a 9a and a are like terms. Collect the like terms by carrying the

sign in front of each term. Like terms

have exactly the same pronumerals,

hence x and x2 are not like terms.

b 3xy − 5xy = −2xy 3xy and 5xy are like terms.

c 4x − x2 + 7x = 11x − x2 4x and 7x are like terms.

EXERCISE 2A1 Simplify, where possible, by collecting like terms.

a 3x + 7x b 6x − 3x c x + x d 3p − 2p

e x2 − 7x2 f x2 + 3x g 7ab + 5ba h x + 7 + 2x

i 8a2 + 14a2 − 6a2 j p + 3p − 8 k p2 + 2p + 4p2 l 7k + k − 8

WORKED EXAMPLE 2Simplify the following by collecting like terms.

a 4x − 9x + 3 − x b 4 − 2a + 3a + 7 c a + 8 − 3a + 7

Solve/Think Apply

a 4x − 9x + 3 − x = 4x − 9x − x + 3

= −6x + 3

Collect the like terms by carrying the

sign in front of each term.

b 4 − 2a + 3a + 7 = −2a + 3a + 4 + 7

= a + 11

c a + 8 − 3a + 7 = a − 3a + 8 + 7

= −2a + 15

2 Complete the following.

a 5x + 7 + 2x − 3 = 5x + 2x + 7 − 3 = ___

b 4a + 3b − 2a − 5b = 4a − 2a + 3b − 5b = ___

c −3m − 4 − 2m + 3 = −3m − 2m − 4 + 3 = ___

d x2 + 3x − 5x + 4x2 = x2 + 4x2 + 3x − 5x = ___

3 Simplify the following by collecting like terms.

a x + 2 + 3x + 6 b 4 + 5a + 7 − 2a c 2a + 4b + 4a − 2b

d n2 − 2n + 3n + 2n2 e 8 − 7x − 6 + 2 f x2 + 2x + 3 − 5x

g a2 + a + a + a2 h 5t − 4t + 7 + t i −5a − 3a + 3 − 5



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25Chapter 2 Further algebraic skills

4 Simplify, where possible, by collecting like terms.

a −8l − 4 + 2l − 7 b x2 + 2x − 7x − x2 c 4x − 2y − (−x) + y

d ab + b − 2ab − 5b e −x − 6 − 2x − 3 f 5t − (−t) + 6 − 3t

g 3a − 2 − a + 2 − a h a2b + a2b − 3a2b + 8a i 5d − 2c + d − 2c + 4

j 4p5 + 5p4 − 2p5 − 6p4 k 8m2 − 5n2 + 7n2 − 4m2 l 6s2t + 5s2 − 8s2t − 6s2

WORKED EXAMPLE 3Simplify:

a −2 × 5x b 5a × 3b

Solve Think Apply

a −10x −2 × 5x = −2 × 5 × x

= (−2 × 5) × x

= −10x

Multiply the coeffi cients and

multiply the pronumerals. The

coeffi cient of an algebraic terms is

the number by which the pronumeral

is being multiplied.b 15ab 5a × 3b = 5 × a × 3 × b

= (5 × 3) × (a × b)

= 15ab

5 Complete the following to simplify the expressions.

a 4x × 5y = 4 × x × 5 × y b −3a × 2b = −3 × a × 2 × b

= (4 × 5) × (x × y) = ___ = (−3 × ___) × (a × ___) = ___

c 5 × 3x2 = 5 × 3 × x2 d −4ab × −3c = −4 × a × b × −3 × c

= ___ = (−4 × ___) × (a × ___ × ___)

= ___

6 Simplify:

a 5 × 3x b 4y × 7 c −3 × 8x d −5 × 7a

e 4p × 5q f −5a × 3b g −2k × −3m h −6w × −z

WORKED EXAMPLE 4Simplify:

a 10x ÷ 2 b 20pq ÷ −5p c 8ab ÷ 10b

Solve Think Apply

a 5x 10x ÷ 2 = 10x

____ 2

= 510 × x

_______ 12 = 5x

Expand each term. Divide the

coeffi cients and the pronumerals.

b −4q 20pq ÷ −5p = 20pq

_____ −5p

= 420 × 1p × q

___________ 1−5 × 1p

= 4q

___ −1 = −4q

c 4a

___ 5 8ab ÷ 10b =

8ab ____

10b

= 48 × a × 1b

__________ 510 × 1b =

4a ___

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7 Complete the following to simplify.

a 12x ÷ 6 = □12 × x

_______ □6 b −12y ÷ 8 =

−□12 × y _________ □8

c 6pq ÷ 8p = □6 × 1p × q

___________ □8 × 1p

= ___ = ___ = ___

8 Simplify the following.

a 8a ÷ 4 b 25y ÷ 5 c 35x ÷ (−7)

d 2a ÷ 4 e 8a ÷ 2a f −6y

____ 3

g 12z

____ −3z h 20xy

____ 10x

i 16ab

_____ 12b

j −4pq

_____ 6p k

−20mn _______ −4m l

−16xy ______

20xy

WORKED EXAMPLE 5Simplify the following.

a 2 __

7 ×

5 __

9 b

4x ___

7 ×

3 ___

5y

Solve Think Apply

a 10

___ 63

2 __

7 ×

5 __

9 =

2 × 5 _____

7 × 9 Multiply the numerators

and multiply the

denominators.b 12x

____ 35y

4x

___ 7 ×

3 ___

5y = 4x × 3

______ 7 × 5y

9 Complete the following to simplify.

a 3 __

7 ×

k __

5 =

3 × □ ______ 7 × □ = ___ b

x __

5 ×

3 __ y =

x × □ ______ 5 × □ = ___

Simplify the following.

a 5 __

6 ×

m __

3 b

x __

2 ×

y __

5 c

m __

4 ×

3 __ n d

2a ___

7 ×

3 ___

4b

WORKED EXAMPLE 6Simplify these algebraic fractions.

a 16

___ 9 ×

27 ___

28 b

16a ____

9 ×

27 ____

28b c

6ab ____

5 ×

10 ___

9a

Solve Think Apply

a 416

___ 19 ×

327 ___ 728 =

12 ___

7

The HCF of 16 and 28 is 4. Divide the

numerator and the denominator by 4. The HCF

of 9 and 27 is 9. Divide the numerator and the

denominator by 9.

Divide the numerator and

denominator by the highest

common factor (HCF).b 416a

____ 19 ×

327 ____ 728b =

12a ____

7b

c 26 1a b

_____ 15 ×

210 ____ 39 1a =

4b ___

3

The HCF of 6 and 9 is 3. Divide the numerator

and the denominator by 3. The HCF of 5 and 10

is 5. Divide the numerator and the denominator

by 5. a is a common factor so divide the

numerator and the denominator by a.

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Complete the following.

a 10x

____ 9 ×

3 ____

16y =

510x ____ □9

× □3

____ 816y b

12ab _____

5 ×

10 ___

3a =

□121ab ______ 15

× 210

____ □31a

= ___ = ___


a 3x

___ 5 ×

2n ___

7 b

2a ___

9 ×

6b ___

5 c

4z __

3 ×

9w ___

8

d 8m

___ 5 ×

15n ____

16 e

3k ___

5 ×

10 ___

9m f

5a ___

3 ×

9 ___

2a

g 7 ___

2p × 3p

___ 14

h 2xy

___ 3 ×

6 ___

2y i

8ab ____

9 ×

15 ___

3a

j 6xy

___ 5 ×

25 ___

3y k

9c ____

10d × 25de _____

12c2 l a2

___ bc2 × bc

___ a

WORKED EXAMPLE 7Simplify these algebraic fractions.

a 3x

___ 7 ÷

8 ___

5y b 4a

___ 3b ÷

16 ___

9b

Solve Think Apply

a 3x

___ 7 ÷

8 ___

5y = 3x

___ 7 ×

5y ___

8

= 15xy

____ 56

To divide by 8 ___

5y invert it and

multiply, or multiply by its

reciprocal.

To divide by a fraction invert it

and multiply or multiply by its

reciprocal.

b 4a

___ 3b

÷ 16

___ 9b

= 14a

____ 131b ×

391b ____ 416

= 3a

___ 4

To divide by 16

___ 9b

invert it and

multiply.

Complete the following to simplify these algebraic fractions.

a 5x

___ 3 ÷

4y ___

7 =

5x ___

3 × ___ b

8m ___

5n ÷ 4m

___ 15

= 8m

___ 5n × ___

= ___ = ___

Simplify these algebraic fractions..

a 4x

___ 3 ÷

7x ___

2 b

11a ____

5 ÷

6 ____

10b c

15p ____

4q ÷ 3p

___ 8q

d 18xy

____ 7 ÷

6x ___

14 e

12a ____

9 ÷

4b ___

3c f

8 ___

5x ÷ 12y

____ 15

2B Expanding using the distributive lawExpanding using the distributive lawThe expression a × (b + c) can be written in expanded form as a × b + a × c. This is known as the

distributive law.

To expand a(b + c), each term inside the grouping symbols is multiplied by the term outside the grouping symbols.

11

12

13

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WORKED EXAMPLE 1Expand and simplify:

a 4(2a + 7) b 5(3m − 4)

Solve Think Apply

a 8a + 28 4(2a + 7) = 4 × 2a + 4 × 7 a(b + c) = ab + ac

a(b − c) = a(b + (−c))

= a × b + a × (−c)

= ab + (−ac)

= ab − ac

b 15m − 20 5(3m − 4) = 5(3m + (−4))

= 5 × 3m + 5 × (−4)

= 15m + (−20)

= 15m − 20

EXERCISE 2B1 Complete to expand the following.

a 5(x + 3) = 5 × ___ + 5 × ___ = ____

b 7(2y − 5) = ___ × 2y − ___ × 5 = ____

c 3a(4a − 7) = 3a × ___ − 3a × ___ = ____

2 Expand and simplify the following.

a 2(x + 3) b 2(a − 2) c 3(b − 5) d 9(xy + z)

e 3(2m + 1) f 6(3a − 7) g 7(1 − 3d) h a(a + 5)

i x(3 − x) j 3x(4x − 1) k w(w − 1) l 6a(4a − 7b)

WORKED EXAMPLE 2Expand and simplify the following.

a −3(x + 7) b −4(5 − 2y)

Solve Think Apply

a −3x − 21 −3(x + 7) = (−3) × x + (−3) × 7

= −3x − 21

Multiply each term inside the

parentheses by the negative

number at the front:

−a(b + c) = −ab − ac

−a(b − c) = −ab + ac

b −20 + 8y −4(5 − 2y) = (−4) × 5 − (−4)(2y)

= −20 − (−8y)

= −20 + 8y

3 Complete to expand the following.

a −2(x + 5) = (−2) × x + (−2) × 5 b −6(3y − 5) = (−6) × ___ − (−6) × ___

= ___ + ___ = ___ − ___

= ___ − ___ = ___ + ___


a −4(x + 3) b −5(4 − x) c −2(4x + 1)

d −4(2 − x) e −2(3p + 4) f −3(m + 2n)

g −x(x + y) h −a(2b + 3d) i −6(x + 2y + z)



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WORKED EXAMPLE 3Expand the following.

a −(7 + y) b −(y − 2)

Solve Think Apply

a −7 − y −(7 + y) = −1 × (7 + y)

= −7 − yA negative sign at the front of the

parentheses is an abbreviation for −1;

that is, −(a + b) = −1 × (a + b).

The eff ect of multiplying by −1 is to

change the sign of each term inside the

parentheses.

b −y + 2 −(y − 2) = −1 × (y − 2)

= −y + 2

5 Expand the following.

a −(x + 2) b −(x − 4) c −(5 + x)

d −(7p + 6) e −(4x + 2) f −(3 − 5x)

g −(4y + 8x) h −(8p − 7q) i −(4x2 − 3x)

6 Expand the following.

a −2a(3 − b) b −3y(x + 4) c −4a(a − 4)

d −4m(p + m) e −x(x + 6) f −3c(4c − d)

g −3a(2a + b) h −5x(x − 2y) i −7p(2p + 5q)


a 3 + 2(2 − 5x) b 4 − 3(x − 2)

Solve/Think Apply

a 3 + 2(2 − 5x) = 3 + 4 − 10x

= 7 − 10x

When expanding brackets, multiply every term inside

the brackets by the term outside the brackets.

Expand and collect like terms.b 4 − 3(x − 2) = 4 − 3x + 6

= 10 − 3x


a 4 + 3(5x + 2) = 4 + 15x + ___ b 7 + 2(3a − 5) = 7 + 6a − ___

= ___ + ___ = ___+ ___

= ___ − ___

c 4 − 3(a + 2) = 4 − 3a − ___ d 10 − 4(p − 3) = 10 − 4p + ___

= ___ − ___ = ___ − ___


a 7 + 2(3x + 5) b 4 − 3(1 + 2x) c 18 + 2(5x + 8)

d 7x − (3 − 4x) e 5x + 3(2x + 1) f 7 − 5(1 − 3x)

g 2x − (3x + 11) h 7 − 2(x − 3) i 5 − 6(3x − 1)

j 2(3x − 5) + 7x k 3(x − 2) + 4 l 5(x − 3) − 3x



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WORKED EXAMPLE 5Expand and simplify these expressions.

a 5(x + 2) + 2(5 − x) b 3(2 − x) − 4(x − 3)

Solve/Think Apply

a 5(x + 2) + 2(5 − x) = 5x + 10 + 10 − 2x

= 3x + 20

Expand and collect like terms.

b 3(2 − x) − 4(x − 3) = 6 − 3x − 4x + 12

= −7x + 18


a 3(a + 4) + 2(6 − a) = 3a + 12 + ___ − ___

= ___ + ___

b 2(y − 3) − 4(y − 2) = 2y − ___ − 4y + ___

= ___ + ___

Expand and simplify these expressions.

a 3(x + 4) + 2(x − 1) b 4(y + 1) + 2(y + 3)

c 3(p + 1) + 4(p − 5) d 2(1 − x) + 4(x − 2)

e d(d + 2) + 3(d − 3) f 2x(x + 1) + 3(x + 2)

g 2x(x + 2) − 4(4 − x) h 3x(x + 4) + 5(x + 2)

i 6(x − 3) − (5x + 4) j −(x + 6) − 7(x + 3)

k −(3 − x) + 2(4 − x) l 3(5 − 2x) − (4 − x)

2C Adding and subtracting algebraic fractions Adding and subtracting algebraic fractionsWORKED EXAMPLE 1Simplify the following.

a 9t

___ 20

+ 7t

___ 15

b 5k

___ 6 − 2k

Solve Think Apply

a 9t

___ 20

+ 7t

___ 15

= 27t

___ 60

+ 28t

___ 60

= 55t

___ 60

= 11t

___ 12

Lowest common denominator = 60:

9t

___ 20

= 27t

___ 60

and 7t

___ 15

= 28t

___ 60

55t

___ 60

= 55t ÷ 5

_______ 60 ÷ 5

= 11t

___ 12

Change the fractions to

equivalent fractions with a

common denominator.

Add or subtract the

numerators and simplify if

possible.b

5k ___

6 − 2k =

5k ___

6 −

12k ____

6

= − 7k

___ 6

Lowest common denominator = 6:

2k = 12k

____ 6

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EXERCISE 2C1 Complete the following to simplify the expression.

7y

___ 8 −

3y ___

4 =

7y ___

8 −

□ __ 8 = ___

2 Simplify the following expressions.

a 11x

____ 20

+ 8x

___ 20

b 2y

___ 3 +

3y ___

4 c

7a ___

8 −

2a ___

3 d

14t ___

9 −

3t __

5

e 4x

___ 5 +

3x ___

2 f

11p ____

10 −

p __

4 g

5a ___

7 −

3a ___

14 h

5z __

6 +

z __

3

i 3y

___ 7 + y j 2p +

2p ___

5 k m −

2m ___

3 l 2k −

3k ___

8

2D Indices IndicesINVESTIGATION 2.1

From Investigation 2.1 we see that the laws for the use of indices are:

am × an = am + n

am ÷ an = am − n

(am)n = amn

(ab)n = anbn


a p7 × p8 b t 16 ÷ t 4 c (m5)4

Solve Think Apply

a p7 × p8 = p15 p7 × p8 = p7 + 8 Use the index laws to simplify.

b t 16 ÷ t 4 = t 12 t 16 ÷ t 4 = t 16 − 4

c (m5)4 = m20 (m5)4 = m5 × 4

EXERCISE 2D1 Complete the following.

a a7 × a5 = a7 + □ b m15 ÷ m7 = m15 − □ c (k3)5 = k3 × □

= ___ = ___ = ___

2 Simplify the following.

a k10 × k6 b p12 ÷ p7 c (t3)5

d m11 × m10 e x25 ÷ x5 f (y7)5



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a 3x7 × 4x8 b x2y3 × x5y c 7x4y3 × 3x2y8

Solve Think Apply

a 12x15 3x7 × 4x8 = (3 × 4) × (x7 × x8)

= 12 × x15

Group the coeffi cients

and like pronumerals.

Multiply the coeffi cients

and use the index law for

multiplication to simplify

the pronumerals.

b x7y4 x2y3 × x5y = (x2 × x5) × (y3 × y)

= x7 × y4

c 21x6y11 7x4y3 × 3x2y8 = (7 × 3) × (x4 × x2) × (y3 × y8)

= 21 × x6 × y11


a 4p5 × 3p7 = (4 × 3) × (p5 × p7) = ___

b k3m4 × km7 = (k3 × k) × (m4 × m7) = ___

c 4x2y5 × 5x3y2 = (4 × 5) × (x2 × ___) × (y5 × ___) = ___

4 Simplify the following expressions.

a 5a6 × 3a8 b 4y9 × 3y c 6m11 × 2m4

d p8q3 × p4q2 e a7b8 × ab4 f xy3 × x7y

g 3m3n2 × 4m7n3 h 5a4b × 3a6b5 i 4p7q10 × 10p4q11

j 3k7 × 1

_ 4 k3 k

2

_ 5 x7y4 × 4xy l 5a3b4c2 × 2a2bc3


a (2x7)4 b (m2n3)5 c (5x6y5)3

Solve Think Apply

a (2x7)4 = 24 × (x7)4

= 16x28

(2x7)4 = 2x7 × 2x7 × 2x7 × 2x7

= (2 × 2 × 2 × 2) × (x7 × x7 × x7 × x7)

= 24 × (x7)4

(ab)n = anbn

b (m2n3)5 = (m2)5 × (n3)5

= m10n15

(m2n3)5 = m2n3 × m2n3 × m2n3 × m2n3 × m2n3

= (m2 × m2 × m2 × m2 × m2)

× (n3 × n3 × n3 × n3 × n3)

= (m2)5 × (n2)5

c (5x6y5)3 = 53 × (x6)3 × (y5)3

= 125x18y15

(5x6y5)3 = 5x6y5 × 5x6y5 × 5x6y5

= (5 × 5 × 5) × (x6 × x6 × x6)

× (y5 × y5 × y5)

= 53 × (x6)3 × (y5)3


a (2x3)5 = 2□ × (x3)□ b (a7b4)8 = (a7)□ × (b4)□

= ___ = ___



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6 Simplify:

a (7x12)2 b (2z10)4 c (5m6)3

d (a7b4)5 e (x4y5)10 f (pq5)3

g (10x3y5)4 h (3a7b9)2 i (2k5m4)3


a 6m7

____ 3m2 b

p7q9

____ p3q2 c

8a6b5

______ 10a3b3

Solve Think Apply

a 2m5

6m7

____ 3m2 =

6 __

3 ×

m7

___ m2

= 2 × m5

Group the coeffi cients and like

pronumerals.

Divide the coeffi cients and use

the index law for division to

simplify the pronumerals.b p4q7

p7q9

____ p3q2 =

p7

___ p3 ×

q9

__ q2

= p4 × q7

c 4a3b2

_____ 5

8a6b5

______ 10a3b3 =

8 ___

10 ×

a6

__ a3 ×

b5

__ b3

= 4 __

5 × a3 × b2


a 15k6

____ 3k2 =

15 ___

3 ×

k6

__ k2 = ___ b

x11y8

____ x4y2 =

x11

___ x4 ×

y8

__ y2 = ___

c 9m6n7

______ 12m4n6 =

9 ___

12 ×

m6

___ m4 ×

n7

__ n6 = ___

8 Simplify:

a 10k8

____ 5k3 b

12w9

_____ 4w2 c

15z6

____ 20z3

d p6q10

____ p5q4 e

a12b9

____ a10b4 f

m8n8

____ m4n5

g 6k3m7

_____ 2km2 h

6w4v6

_____ 8w2v2 i

18b10c8

______ 27b3c8

WORKED EXAMPLE 5

Simplify 9x4y3

_____ 5x2y

× x7y2

____ 3xy2 .

Solve Think Apply

9x4y3

_____ 5x2y

× x7y2

____ 3xy2 =

3x8y2

_____ 5

9x4y3

_____ 5x2y

× x7y2

____ 3xy2 =

9x11y5

______ 15x3y3

= 9 ___

15 ×

x11

___ x3 ×

y5

__ y3

= 3 __

5 × x8 × y2

Multiply and divide the

coeffi cients, then multiply and

divide the pronumerals using

the index laws.



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a a4b6

____ a2b3 ×

a10b4

____ a8b5 =

a14b10

_____ a□b□ b x5y7 ×

x6y3

____ x8y9 =

x□y□

____ x8y9 c

(p7q4)5

______ p10q15 =

p35q□

_____ p10q15

= ___ = ___ = ___


a a3b4 × a7b2

__________ a5b3 b

m7n8

___________ m2n4 × m3n3 c

x7y3

____ x2y2 ×

x5y4

____ x4y2 d

(a2b3)4

______ a5b2

e 4x8y10

___________ 5x3y × 2x2y2 f

a5b4 × a2b5

__________ a3b9 g

x5y4

____ x2y2 ×

x6y5

____ x7y4 h

3m5n2 × 2m7n4

_____________ 4m3n × 5m6n3


a 2x3(x2 + 4) b 3a2(4a7 − 5a)

Solve Think Apply

a 2x5 + 8x3 2x3(x2 + 4) = 2x3 × x2 + 2x3 × 4 Use a(b + c) = ab + ac

and the index laws.

b 12a9 − 15a3 3a2(4a7 − 5a) = 3a2 × 4a7 − 3a2 × 5a


a 4k3(3k + 2) = 4k3 × 3k + 4k3 × 2 b m6(4m3 − m2) = m6 × 4m3 − m6 × m2

= ___ = ___

Expand and simplify the following.

a y4(y2 + 3) b n7(5 + n2) c p3(2p2 − 4)

d 2b4(b2 − 7) e 3k(2k2 + 4m) f 4w3(2w2 − 3z)

g a6(a3 + a2) h 2t7(3t4 − t2) i 3w4(4w3 − 2w6)

WORKED EXAMPLE 7Expand and simplify 2a4(a + 3) + 3a4(a − 5).

Solve/Think Apply

2a4(a + 3) + 3a4(a − 5) = 2a4 × a + 2a4 × 3 + 3a4 × a − 3a4 × 5

= 2a5 + 6a4 + 3a5 − 15a4

= 2a5 + 3a5 + 6a4 − 15a4

= 5a5 − 9a4

Expand using the distributive law

and index law. Collect like terms.


3a2(a + 5) + a2(4a − 1) = 3a2 × a + 3a2 × 5 + a2 × 4a − a2 × 1

= 3a□ + 15a□ + ___ − a2 = ___


a 3x2(6x − 3) + 2x2(7x + 5) b 4x3(2x3 + 7) − x3(3x3 − 4)

c 2p4(5p3 − p) + 3p4(2p3 − 4p) d 3x5(2x4 − 5x2) − x5(3x4 − 4x2)

e 5y2(2y5 + 3y3) − 4y2(3y5 + 2y3) f 7x3(2x2 − 3x) + 5x2(x3 − 3x2)

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2E Equations EquationsWORKED EXAMPLE 1Solve these equations.

a 6x = 41 b x __

3 = −4 c x + 7 = 3 d x −4 = −1

Solve Think Apply

a x = 6 5

_ 6

6x ___

6 =

41 ___

6

x = 6 5

_ 6

Add or subtract the same number to

or from both sides of the equation.

Multiply or divide both sides by the

same number.b x = −12

x __

3 × 3 = −4 × 3

x = −12

c x = −4 x + 7 − 7 = 3 − 7

x = −4

d x = 3 x − 4 + 4 = −1 + 4

x = 3

EXERCISE 2E1 Solve the following equations.

a 5p = −30 b 7k = 31 c x __

7 = 4 d

p __

6 = −5

e p + 7 = 16 f x + 8 = 16 g x − 3 = 9 h y − 4 = 26

i k − 24 = −25 j x + 11 = 0 k x − 17 = 0 l x __

5 = 0

WORKED EXAMPLE 2Solve the following equations.

a 8x − 7 = −31 b 19 − 3x = 7

Solve Think Apply

a 8x − 7 = −31

8x = −24

x = −3

8x − 7 + 7 = −31 + 7

8x = −24

8x

___ 8 = −

24 ___

8

x = −3


or from both sides of the equation,

then divide both sides by the same

number.

b 19 − 3x = 7

−3x = −12

x = 4

19 − 3x − 19 = 7 − 19

−3x = −12

−3x

____ −3 =

−12 ____ −3

x = 4



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2 Solve these equations.

a 3x − 10 = 2 b 4p + 7 = 3 c 6x − 25 = −5

d 9a + 10 = −8 e 4x − 5 = 9 f 17 − 2x = −3

g 5 − 4x = 21 h 12 − 7x = 15 i 18 = 17 + 5m

j 5t + 19 = 4 k −3p − 11 = −2 l −2k + 7 = −13


a x __

5 + 7 = 3 b 4 −

x __

3 = 2

Solve Think Apply

a x __

5 + 7 = 3

x __

5 = −4

x = −20

x __

5 + 7 − 7 = 3 − 7

x __

5 = −4

x __

5 × 5 = −4 × 5

x = −20


or from both sides of the equation,

then multiply both sides by the same

number.

b 4 − x __

3 = 2

− x __

3 = −2

x = 6

4 − x __

3 − 4 = 2 − 4

− x __

3 = −2

− x __

3 × −3 = −2 × −3

x = 6


a m

__ 3 + 1 = 6 b

k __

2 − 3 = 7 c

p __

5 + 6 = 2

d y ___

10 − 3 = −7 e 5 −

x __

2 = 2 f 9 −

m __

4 = −1

g 1 − t __

6 = 5 h 2 −

q __

7 = −1 i

w __

5 − 9 = 6

j k __

3 + 5 = 1 k

m __

4 − 3 = −5 l 5 −

z __

3 = 1

WORKED EXAMPLE 4

Solve 2x

___ 3 = 7.

Solve Think Apply

2x

___ 3 = 7

2x = 21

x = 10 1

_ 2

2x

___ 3 × 3 = 7 × 3

2x = 21

2x

___ 2 =

21 ___

2

x = 10 1

_ 2

Multiply both sides by the same number


number.

4 Solve the following equations.

a 2x

___ 3 = 5 b

3x ___

4 = 2 c

2x ___

5 = −4

d 5x

___ 4 = 3 e

3x ___

2 = −5 f

5x ___

6 = 14



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WORKED EXAMPLE 5

Solve 2x

___ 3 =

5 __

4 .

Solve Think Apply

2x

___ 3 =

5 __

4

2x = 15

___ 4

x = 15

__

8 or 1 7

_ 8

2x

___ 3 × 3 =

5 __

4 × 3

2x = 15

___ 4

2x

___ 2 =

15 ___

4 ×

1 __

2

x = 15

__

8 or 1 7

_ 8

Multiply both sides of the

equation by the same number.


number.

(Dividing by 2 is the same as

multiplying by 1

_ 2 .)


a 2x

___ 5 =

3 __

4 b

3m ___

4 =

1 __

2 c

2p ___

3 =

4 __

7

d 3k

___ 5 = −

2 __

3 e

5t __

9 = −

5 __

4 f

7y ___

5 = −

3 __

5

WORKED EXAMPLE 6Solve 3(p − 2) = 21.

Solve Think Apply

3(p − 2) = 21

3p − 6 = 21

3p = 27

p = 9

Expanded 3(p − 2) = 3p − 6

3p − 6 + 6 = 11 + 6

3p = 27

p = 9

Expand and solve the resulting

equation.


a 5(n + 1) = 65 b 3(p − 7) = 21 c 2(x + 11) = 25

d 4(y − 3) = 7 e 3(2m − 5) = 24 f 2(4x + 3) = 19


a 5x + 3 = 3x − 1 b 14 − 3x = 2 − x

Solve Think Apply

a 5x + 3 = 3x − 1

2x + 3 = −1

2x = −4

x = −2

5x + 3 − 3x = 3x − 1 − 3x

2x + 3 = −1

2x + 3 − 3 = −1 − 3

2x = −4

Eliminate the variable on the

right-hand (or left-hand) side

by adding it to or subtracting it

from both sides of the equation.

Complete the solution as for

previous examples.b 14 − 3x = 2 − x

14 − 2x = 2

−2x = −12

x = 6

14 − 3x + x = 2 − x + x

14 − 2x = 2

14 − 2x − 14 = 2 − 14

−2x = −12



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a 6x + 2 = 2x + 14 b 3x + 8 = 13 − 2x

c 4x − 5 = 3x − 7 d 10 − 2x = 4 − 5x

e 2x − 10 = x + 4 f 5x − 7 = 13 + 7x

g 3x − 6 = 9 − 2x h 4x − 5 = 5x − 2

i 3a + 2 = a − 5 j 2k − 3 = 12 − 3k

k −4m + 1 = 3 − 5m l −2t − 3 = 7 − 4t

WORKED EXAMPLE 8Solve

4 __ n = 22.

Solve Think Apply

4 __ n = 22

4 = 22n

4 ___

22 = n

n = 2

__ 11

4 __ n × n = 22 × n

4 = 22n

4 ___

22 =

22n ____

22

∴ n = 2

__ 11


equation by the same number,

then divide both sides by the

same number.


a 75

___ n = 3 b 48

___ m = 6 c 6 __ k = −2 d

39 ___ t = −3

e 3 __ p = 4 f

7 __ a = 2 g

12 ___ w = 7 h

5 __ z = −6

WORKED EXAMPLE 9Solve this equation:

w __

2 +

w __

5 = −14

Solve Think Apply

w

__ 2 +

w __

5 = −14

5w + 2w = −140

7w = −140

w = −20

Lowest common denominator = 10

( w __ 2 +

w __

5 ) × 10 = −14 × 10

w

__ 2 × 10 +

w __

5 × 10 = −140

5w + 2w = −140

7w = −140

7w

___ 7 =

−140 _____

7

w = −20


equation by the lowest common

denominator of the fractions.


a k __

3 +

k __

2 = 10 b

z __

3 +

z __

5 = 4 c

2x ___

3 +

3x ___

4 = 1

d y __

2 −

y __

6 = 3 e

3x ___

5 −

x __

2 = −4 f

4k ___

3 −

3k ___

4 = 2



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WORKED EXAMPLE 10Solve

x __

3 + 4 = 2x − 11.

Solve Think Apply

x __

3 + 4 = 2x − 11

x + 12 = 6x − 33

−5x + 12 = −33

−5x = −45

x = 9

Lowest common denominator = 3

( x __

3 + 4) × 3 = (2x − 11) × 3

x + 12 = 6x − 33

x + 12 − 6x = 6x − 33 − 6x

−5x + 12 = −33

−5x + 12 − 12 = 33 − 12

−5x = −45


equation by the lowest common

denominator of the fractions.

Solve the following equations.

a x __

3 + 2 = x − 4 b x + 3 =

x __

2 + 1 c 2x − 5 =

3x ___

4 + 1

d 2x

___ 3 − 1 = 3x − 15 e

x __

5 − 1 =

x __

3 + 2 f 11 −

4x ___

3 =

x __

2 − 1

WORKED EXAMPLE 11Solve these equations.

a x2 = 9 b y3 = 7

Solve Think Apply

a x = 3 or −3 x = √ __

9 or x = − √ __

9 Take the square root of both sides.

Note: 3 or −3 is often written as ±3.

b y = 3 √ __

7

= 1.9 (1 decimal place)

3 √ __

y3 = 3 √ __

7 Take the cube root of both sides.

Solve the following equations.

a y3 = 8 b x2 = 100 c y3 = 35

d x2 = 36 e y3 = 10 f x2 = 7


a √ __

x = 5 b 3 √ __

x = 4

Solve Think Apply

a x = 25 ( √ __

x )2 = 52 Square both sides.

b x = 64 ( 3 √ __

x )3 = 43 Cube both sides.

Solve these equations.

a √ __

x = 3 b 3 √ __

x = 2 c √ __

m = 7

d 3 √ __

k = 1 e √ ___

2p = 10 f 3 √ ___

2w = 5

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2F Practical equations Practical equationsA formula links two or more pronumerals according to a rule. The pronumeral that is expressed in terms of the

others is called the subject of the formula. If the pronumeral to be found is the subject of the formula then its

value is found directly by substitution. If the pronumeral to be found is not the subject then, after substitution, the

resulting equation is solved to fi nd its value.

WORKED EXAMPLE 1a Find the value of t when v = 117, u = 5, a = 8 and v = u + at.

b Find the value of N when R = 23, I = 4 and R = I __ N + 1.

Solve Think Apply

a v = u + at

117 = 5 + 8t

112 = 8t

t = 14

In the formula, replace v by 117,

u by 5 and a by 8. Solve the

resulting equation.

Substitute the given values

into the formula and solve the

resulting equation.

b R = I __ N + 1

23 = 4 __ N + 1

22 = 4 __ N

22N = 4

N = 4 ___

22

= 2 ___

11

Replace R by 23 and I by 4. Solve

the resulting equation.

EXERCISE 2F1 If v = u + at, fi nd the value of:

a t when v = 102, u = 18, a = 7 b a when v = 54, u = 12, t = 14

2 Use the formula d = 1

_ 2 ct to fi nd the value of:

a c when d = 100, t = 8 b t when d = 320, c = 16

3 If M = 5k

___ 18

fi nd k when M = 15.

4 If V 2 = gR fi nd R when V = 12, g = 10.

5 If B = m

__ h2 fi nd m when B = 23, h = 1.63.

6 Use the formula I = Prn

____ 100

to fi nd the value of:

a P when I = 19 500, r = 3, n = 25 b n when I = 2100, P = 5000, r = 6

c r when I = 2560, P = 8000, n = 8.



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7 Given s = ut + 1

_ 2 at2, fi nd u when s = 360, t = 8 and a = 10.

8 If S = n __

2 (a + l) fi nd a when S = 560, n = 20 and l = 53.

9 If S = n __

2 {2a + (n − 1)d} fi nd the value of a given that S = 610, d = 3 and n = 20.

Use the formula A = 180 − 360

____ n to fi nd n when A = 120.

Use the formula a = 2Rn

_____ n + 1

to fi nd the value of:

a R when a = 12, n = 24 b R when a = 19.5, n = 39

Use the formula I = E _____ R + r to fi nd the value of:

a E when I = 8, R = 15, r = 3 b R when I = 2, E = 24, r = 3

Given R = I __ N + 1, fi nd the value of N when:

a R = 3, I = 7 b R = 11, I = 5

Use t = √ __

l __ g to fi nd l when t = 1.3, g = 10.

Use r = 3 √ __

V

__ k to fi nd V when r = 3, k = 4.2.

Use the formula E = 1

_ 2 mv2 to fi nd the value of v when E = 135 and m = 2.8.

INVESTIGATION 2.1Complete the following.

1 y3 × y4 = y × y × y × ___ × ___ × ___ × ___ = y□

2 k8 ÷ k5 = k × k × k × k × k × k × k × k

__________________________ k × □ × □ × □ × □ = k□

3 (p2)3 = p2 × p2 × p2 = p□

4 (km)3 = km × km × km = k × k × k × ___ × ___ × ___= k□m□

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REVIEW 2 FURTHER ALGEBRAIC SKILLS

Language and terminologyHere is a list of terms used in this chapter. Explain each term in a sentence.

algebraic fraction, expand, expression, equation, formula, index, index law, indices, like terms, pronumeral,

substitute, value, variable

Having completed this chapter you should be able to:• add and subtract like algebraic terms

• multiply and divide algebraic terms

• add and subtract algebraic fractions

• establish and apply index laws in algebra

• solve linear equations.

2 REVIEW TEST1 8x − x =

A 8 B 8x C 7x D 7

2 5x − 7 + 4x − 3 =A x − 10 B 9x − 10 C 9x − 4 D x − 4

3 −5d − 3c + 2d − 4c + 6 =A 3d − 7c + 6 B 6 C −3d + 7c + 6 D −3d − 7c + 6

4 5x × −3y =A 2xy B −2xy C −15xy D 15xy

5 8xy

___ 6x

=

A 8y B 3 ___

4y C 4xy

___ 3 D

4y ___

3

6 9ab

____ 10

× 20

___ 6a2 =

A 3b

___ a B 3b

___ 2a C

2b ___

3a D 2b

___ a

7 5x

___ 4 ÷

3y ___

7 =

A 15xy

____ 28

B 35x

____ 12y C

12y ____

35x D 28

____ 15xy

8 6(3m − 4) =A 18m − 4 B 9m − 4 C 9m − 10 D 18m − 24

9 −3(2a − 5) =A −6a − 15 B −6a + 15 C −5a − 5 D −5a − 8

3 + 4(5 − 2x) =A 23 − 8x B 35 − 14x C 23 −2x D 35 − 2x

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Chapter 2 Further algebraic skills 43

5 − 3(2a − 1) =A 4a − 1 B 4a − 2 C 8 − 6a D 2 − 6a

2x(x − 3) + x(x − 7) =A x2 − 13x B x2 + x C 3x2 + 13x D 3x2 − 13x

4t

__ 5 +

t __

2 =

A 5t

__ 7 B

13t ___

20 C

13t ___

10 D

4t2

___ 7

4p6 × 3p7 =A 12p42 B 7p42 C 12p67 D 12p13

(2m7n4)3 =A 2m21n12 B 8m21n12 C 8m10n7 D 2m10n7

x4y3 × x3y7

_________ x7y5 =

A y5 B xy5 C y2 D xy2

2x5(4x2 − 3x) =A 8x7 − 6x6 B 8x10 − 6x6 C 8x7 − 6x5 D 8x10 − 6x5

2a2(3a − 1) − 3a2(1 − 2a) =A −5a2 B −a2 C 12a3 − 5a2 D 12a3 − a2

The solution of 5p + 3 = 7 is:

A p = 2 B p = − 4

_ 5 C p =

4

_ 5 D p = 1

1

_ 4

The solution of 7x − 4 = 2x + 6 is:

A x = −2 B x = 2 C x = 1 1

_ 9 D x =

9

__ 10

The solution of 3x

___ 4 = 7 is:

A x = 9 1

_ 3 B x =

3

__ 28 C x = 5

1

_ 4 D x =

4

__ 21

The solution of 5 __ n = 3 is:

A n = 15 B n = 2 C n = 3

_ 5 D n = 1

2

_ 3

The solution of t ___

15 − 50 = 175 is:

A t = 15 B t = 125 C t = 1875 D t = 3375

The solution of w

__ 3 +

w __

2 = −10 is:

A w = −25 B w = −8 1

_ 3 C w = −12 D w = −24

Given the formula V = IR − E, the value of I when V = 13, R = 4 and E = 7 is:

A 1 1

_ 2 B

1

_ 4 C 5 D

2

_ 3

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

Question 1–7 8–12 13 14–18 19–24 25

Section A B C D E F

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2A REVIEW SET1 Simplify:

a 7ab − 5ab b 7k + k c 5p − 3 + 7p − 1 d x2 + 2x + 2x2 + x

2 Simplify:

a 5 × −3b b 15pq

_____ 10p c

8ab ____

15 ×

5 ___

4a d

2x ___

3 ÷

3y ___

5

3 Expand and simplify:

a 5m(2m − 3) b −(x + 7) c 2(3x − 5) + 8x d 2(4x − 1) + 3(2x + 1)

4 Simplify:

a 2x

___ 3 +

x __

4 b

k __

6 + 2k

5 Simplify:

a 5m3n2 × 7m4n b 2k7m10

______ 8k4m7

c (a2b3)5

d 4p6q5 × 6p2q8

____________ pq × 3p3q3

e 5a3(2a2 + 7) + 2a3(3a2 − 4)

6 Solve:

a 8x = 57 b x __

5 = −2 c x − 5 = −3

d x + 8 = 5 e 4x − 5 = 3 f 5 − 3x = −4

g 3(m + 5) = 17 h 15 + 4x = 3 + x i x __

2 + 1 = 3x − 9

j z __

5 +

z __

2 = 1 k x2 = 16 l 3 √

__ y = 5

7 Use the formula E = 1

_ 2 mv2 to fi nd the value of m when E = 390 and v = 20.

2B REVIEW SET1 Simplify:

a 4x − x b 12x2 − 8x2 c 4x − 5y − 7x + y d 4y + x2 − 2x2 + y

2 Simplify:

a −7m × −4n b 12pq ÷ 18q c 6km

____ 5 ×

4 ___

3m d

5t __

3 ÷

10t ___

9


a −6(5p − 2) b −(m − 7) c 7 − 4(2x + 3) d 2(2 − 3x) + 4(x − 1)

4 Simplify:

a 2m

___ 5 +

3m ___

4 b x −

3x ___

4

5 Simplify:

a 4k2m5 × 1

__ 4 km2 b

4w13

____ 8w8 c (2p3)4

d 15p7q8

____________ 3p4q2 × 5p3q4 e 4p2(3p − 5q2) + 2p2(7p + 3q2)



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Chapter 2 Further algebraic skills 45

6 Solve:

a 7p = 64 b 3x

___ 2 = 5 c

x __

3 + 12 = 7 d 8 − x = 15

e 3x − 7 = 5 f 4 − 7x = −2 g 2(p − 3) = 7 h 12 + 7x = 3x + 4

i y3 = 8 j √ __

w = 6 k 4 __ x = 3 l 3 √

__ x = 5

7 Use the formula A = 1

_ 2 bh to fi nd b given that A = 72 and h = 8.

2C REVIEW SET1 Simplify:

a 8p − p b 6x + 7x c 2y − 5x − 5y − 2x d 5 − x2 + 3x2 + 2

2 Simplify:

a 4xy × 3yz b 10m ÷ 15mn c 4 ____

5ab ×

15a ____

16 d

4t __

3 ÷

8t __

9


a 4z(3 − z) b −(5 − 2x) c 4 + 2(x − 3) d 4(x + 1) − 5(2x + 1)

4 Simplify:

a 5m

___ 6 −

2m ___

3 b 3k +

3k ___

4

5 Simplify:

a v6w5 × 2

_________ 5v3w2

b m7n10

_____ m3n4

c (3yz3)2

d 6t 7v3

_____ 10t4v

× 5t3v4

____ tv5

e 3x2(4x3 − 2) + x2(3x3 + 9)

6 Solve:

a 5x = 16 b 4x

___ 3 = −2 c 4 − x = −9 d

x __

2 + 5 = 3

e 4x − 11 = 8 f 3 − 5x = −9 g 5(a + 1) = 35 h y __

3 − 1 = y + 5

i 3z

__ 2 −

5z __

6 = 8 j

4 __ x = 2 k y2 = 3 l √

_ t = 14

7 Use the formula T = a + (n − 1)d to fi nd n given that T = 41, a = 3 and d = 2.

2D REVIEW SET1 Simplify:

a 3x − 2x b 4x2 + 8x + x2

c 3x − 4y + 4x − y d a2 − 5a2b2 + 2a2b2 − a2

2 Simplify:

a −7pq × 3qr b 16a × 20ab c 6 ___

5x × 20xy

____ 9 d

8k ___

3 ÷

4 ___

6k


a −5w(w − 4) b −(2x − 11) c 8 + 3(3x − 2) d 2(p + 1) − 3(p − 2)



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4 Simplify:

a 14p

____ 5 −

3p ___

2 b

2x ___

7 +

5x ___

9

5 Simplify:

a 2x4y5

_____ 5 ×

5x2y3

_____ 2 b

15a7b10

______ 20a6b

c (10u5v4)4

d 4a6b2 × 3a7b9

____________ 6a3b6 × 4a10b5 e 5y2(2y4 − 5) + 3y2(3y4 − 4)

6 Solve:

a 2x = 29 b y __

5 =

4 __

3 c m − 7 = −7

d 5t + 2 = −3 e w

__ 7 − 1 = 5 f 5v + 3 = 9 − 7v

g 4(2a + 1) = 11 h 5p

___ 3 −

p __

6 =

3 __

4 i

4 __ x = 3

j t 2 = 1 k 3 √ __

y = 3 l x + 7

_____ 4 = 3

7 Use the formula B = m

__ h2 to fi nd m if B = 31.25 and h = 1.6.

2 EXAMINAT ION QUEST ION (15 MARKS)a Simplify:

i 3(5x − 2y) − 2(4x − y) (2 marks)

ii 5y3 × 7y5 (1 mark)

iii (p2q3)4 (1 mark)

iv 8p6q7

______ 12p5q2 (1 mark)

b Solve:

i 3(m + 6) = 22 (2 marks)

ii w

___ 10

− 11 = 13 (2 marks)

iii 18 = 54

___ n (2 marks)

iv 5y + 3 = 3y − 7 (2 marks)

c Use the formula a = 2Rn

_____ n + 1

to fi nd the value of R when a = 95 and n = 19. (2 marks)



Documents

Chapter 2 Further algebraic skills