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Chapter 1 – Reviewing linear equations Solutions to Exercise 1A 1 a x+3=6

3x =5 4

j

x=3

∴

3x = 20

b x−3=6

x=

∴

x=9

∴

3x =2 5 −3x = 10

k −

c 3−x =2 −x = −1 x=1

∴

l −

x+8=0

−x = −5

∴

∴ g 3x = 5

h −2x = 7

x=

b a

x =b a ∴ x = ab

7 x=− 2

d

i −3x = −7 ∴

x =a−b

c ax = b

5 3

∴

∴

x =a+b

b x+b=a

x=2

x=

−14 14 = 5 −5

2 a x−b=a

x=5

f 2x = 4

∴

5x = −2 7

x=

∴

e 2 − x = −3

∴

10 3

−5x = −14

x = −8

∴

x=−

∴

d x + 6 = −2 ∴

20 3

ax =c b ax = bc

e

7 x= 3

∴

x=

bc a

1 Cambridge Senior Maths AC/VCE

ISBN 978-1-107-52013-4

© Evans et al. 2016

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3 a

h t 1 1 + = 3 6 2 1 t = 3 3 t=1

2y − 4 = 6 2y = 10 y=5 b i

3t + 2 = 17

x +5=9 3 x =4 3

3t = 15 t=5 c

x = 12 2y + 5 = 2 j

2y = −3 y=−

3 − 5y = 12

3 2

−5y = 9

d

y=−

7x − 9 = 5

9 5

k

7x = 14

−3x − 7 = 14

x=2

−3x = 21

e

x = −7

2a − 4 = 7

l

2a = 11 a=

14 − 3y = 8

11 2

−3y = −6

f

y=2 3a + 6 = 14 3a = 8 a=

4 a 6x − 4 = 3x

8 3

3x = 4 ∴

g y − 11 = 6 8 y = 17 8 y = 136

x=

4 3

b x − 5 = 4 x + 10 −3x = 15 ∴

x=

15 = −5 −3 2

Cambridge Senior Maths AC/VCE

ISBN 978-1-107-52013-4

© Evans et al. 2016

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c 3x − 2 = 8 − 2x

g

5x = 10 x=2

∴

x x + = 10 2 3 5x = 10 6 5x = 60 x = 12

∴

5 a 2(y + 6) = 10 y+6= 5

h x+4=

∴ y = 5 − 6 = −1

x − = −4 2

b 2y + 6 = 3(y − 4)

−x = −8

2y + 6 = 3y − 12

∴

−y = −18 ∴ y = 18

i

c 2(x + 4) = 7x + 2

x=8

7x + 3 9x − 8 = 2 4 14 x + 6 = 9x − 8 5x = −14

2x + 8 = 7x + 2 −5x = −6 ∴

x=

3x 2

∴

6 5

j

d 5(y − 3) = 2(2y + 4) 5y − 15 = 4y + 8

x=−

14 5

2 4 2 (1 − 2x) − 2x = − + (2 − 3x) 3 5 3 10(1 − 2x) − 30x = −6 + 20(2 − 3x) 10 − 20 x − 30 x = −6 + 40 − 60 x

5y − 4y = 18 + 8

10 x = 24

∴ y = 23 ∴ e x − 6 = 2(x − 3) x − 6 = 2x − 6

k

−x = 0 ∴ f

x=0

x=

12 5

4y − 5 2y − 1 =y − 6 2 (12y − 15) − (2y − 1) = 6y 12y − 15 − 2y + 1 = 6y

y+2 =4 3 y + 2 = 12

4y = 14 ∴ y=

7 2

∴ y = 10

3 Cambridge Senior Maths AC/VCE

ISBN 978-1-107-52013-4

© Evans et al. 2016

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6 a ax + b = 0

h

ax = −b x=−

∴

ax + c =d b ax + c = bd

b a

ax = bd − c

cx = e − d ∴

x=

bd − c a

7 a 0.2x + 6 = 2.4

e−d c

0.2x = −3.6

c a(x + b) = c c x+b= a c ∴ x= −b a d

x=

∴

b cx + d = e

∴

x = −18

b 0.6(2.8 − x) = 48.6 2.8 − x = 81 −x = 78.2

ax + b = cx ∴ ax − cx = −b x(c − a) = b ∴

e

x=

c

b c−a

∴

f

g

x = 16.75

d 0.5x − 4 = 10

x(a + b) = ab x=

2x + 12 = 6.5 7 2x + 12 = 45.5 x + 6 = 22.75

x x + =1 a b bx + ax = ab

∴

x = −78.2

0.5x = 14

ab a+b

∴

a b + =1 x x ∴ x =a+b

e

1 (x − 10) = 6 4 x − 10 = 24 ∴

ax − b = cx − d

x = 28

x = 34

ax − cx = b − d

f 6.4x + 2 = 3.2 − 4x

x(a − c) = b − d

10.4x = 1.2

∴

x=

b−d a−c

∴

x=

1.2 3 = 10.4 26

4 Cambridge Senior Maths AC/VCE

ISBN 978-1-107-52013-4

© Evans et al. 2016

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8

b − cx a − cx +2=0 + b a b(b − cx) + a(a − cx) + 2ab = 0

b2 − bcx + a2 − acx + 2ab = 0

b2 + a2 + 2ab = acx + bcx (a + b)2 = cx(a + b) ∴

so long as a + b , 0

x=

a+b c

9

a b a+b + = x+a x−b x+c a(x − b) + b(x + a) a + b = x+c (x + a)(x − b) ax − ab + bx + ab a + b = x+c (x + a)(x − b) ax + bx a+b = (x + a)(x − b) x + c x 1 = (x + a)(x − b) x + c x(x + c) = (x + a)( x − b)

x2 + cx = x2 + ax − bx − ab

cx − ax + bx = −ab x(a − b − c) = ab

ab a−b−c so long as a + b , 0 ∴

x=

5 Cambridge Senior Maths AC/VCE

ISBN 978-1-107-52013-4

© Evans et al. 2016

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Solutions to Exercise 1B 1 a x+2=6

4

x=4

∴

b 3x = 10 x=

∴

x 1 + =3 3 3 x+1=9 ∴

10 3

x = 8 kg

5 L = w + 0.5; A = Lw

c 3x + 6 = 22

P = 2(L + w)

3x = 16

= 2(2w + 0.5)

16 3

= 4w + 1 4w + 1 = 4.8

d 3x − 5 = 15

4w = 3.8

x=

∴

3x = 20 x=

∴

∴ w = 0.95 A = 0.95(0.95 + 0.5)

20 3

= 1.3775 m2

e 6(x + 3) = 56 56 28 = 3 6 19 x= 3

x+3= ∴ f

x+5 = 23 4 x + 5 = 92 ∴

6 (n − 1) + n + (n + 1) = 150 3n = 150 ∴ n = 50 Sequence = 49, 50 & 51, assuming n is the middle number. 7 n + (n + 2) + (n + 4) + (n + 6) = 80

x = 87

4n + 12 = 80 4n = 68

2 A + 3A + 2A = 48

∴ n = 17 17, 19, 21 and 23 are the odd numbers.

6A = 48 ∴ A=8 A gets $8, B $24 and C $16

8 6(x − 3000) = x + 3000 3 y = 2x; x + y = 42 = 3x

5x = 21000

42 3 x = 14, y = 28

x= ∴

6x − 18000 = x + 3000 ∴

x = 4200 L

6 Cambridge Senior Maths AC/VCE

ISBN 978-1-107-52013-4

© Evans et al. 2016

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140(p − 3) = 120 p

9

13 t=

140 p − 420 = 120 p 20 p = 420

60 ×

p = 21

∴

x x + 60 × = 45 4 6

15 x + 10 x = 45 25 x = 45

x x 48 + 10 = 6 10 60 5 x + 3 x = 24

45 25 9 = 5 = 1.8

x=

8x = 24 x = 3 km 11 Profit = x for crate 1 and 0.5x for crate 2, where x = amount of dozen eggs in each crate. x+3 x+ = 15 2 2x + x + 3 = 30 3x = 27

12 3

60

 30 

=6 60 9 x + =6 4 2 x 15 = 4 2 15 ∴ x= = 7.5km/hr 2

+x

Total = 2 × 1.8 = 3.6 km (there and back) Total = 4 × 0.9 = 3.6 km there and back twice f = b + 24

14

( f + 2) + (b + 2) = 40

∴ x=9 Crate 1 has 9 dozen, crate 2 has 12 dozen.  45 

x x 45 + = 4 6 60

b + 26 + b + 2 = 40 2b = 12 ∴ b=6 The boy is 6, the father 30.

7 Cambridge Senior Maths AC/VCE

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Solutions to Exercise 1C 1 a

Subsitute in (2).

y = 2x + 1 = 3x + 2

2(4x + 6) − 3x = 4

−x = 1, ∴ x = −1

5x + 12 = 4

∴ y = 2(−1) + 1 = −1 b

5x = −8

y = 5x − 4 = 3 x + 6

x=−

2x = 10, ∴ x = 5

! 8 Substitute in (1). y − 4 × − = 6. 5 50 y= 3 8 2 Therefore x = − and y = − . 5 5

∴ y = 5(5) − 4 = 21 y = 2 − 3x = 5x + 10

c

−8x = 8, ∴ x = −1 ∴ y = 2 − 3(−1) = 5 d y − 4 = 3x (1) y − 5x + 6 = 0 (2) From (1) y = 3x + 4 Subsitute in (2). 3x + 4 − 5x + 6 = 0 −2x + 10 = 0

2 a x+y=6 x − y = 10

2x = 16 ∴ x = 8; y = 6 − 8 = −2 b y− x = 5 y+ x = 3 2y =8 ∴ y = 4; x = 3 − 4 = −1

x=5 Substitute in (1). y − 4 = 15. Therefore x = 5 and y = 19. e y − 4x = 3 (1) 2y − 5x + 6 = 0 (2) From (1) y = 4x + 3 Subsitute in (2).

8 5

c

x − 2y = 6 −(x + 6y = 10) −8y = −4 2 1 ∴ y= , x =6+ =7 2 2

2(4x + 3) − 5x + 6 = 0 3x + 12 = 0

3 a 2x − 3y = 7 9x + 3y = 15 11 x = 22 ∴ px = 2 4 − 3y = 7, ∴ y = −1

x = −4 Substitute in (1). y + 16 = 3. Therefore x = −4 and y = −13. f y − 4x = 6 (1) 2y − 3x = 4 (2) From (1) y = 4x + 6

b

4x − 10y = 20 −(4 x + 3y = 7) 8

Cambridge Senior Maths AC/VCE

ISBN 978-1-107-52013-4

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−13y = 13 ∴ y = −1 4x − 3 = 7, ∴ x = 2.5 c 4m − 2n = 2 m + 2n = 8 5m = 10 ∴ m=2 8 − 2n = 2, ∴ n = 3 d 14 x − 12y = 40 9x + 12y = 6 23 x = 46 ∴ x=2 14 − 6y = 20, ∴ y = −1 e 6s − 2t = 2 5s + 2t = 20 11 s = 22 ∴ s=2 6 − t = 1, ∴ t = 5 f

15 x − 4y = 6 −(18x − 4y = 10) −3x = −4 4 ∴ x=   3 4 9 − 2y = 5 3

−2y = −7, ∴ y =

35 p − 10q = 45 p=1 ∴ q = −1 i 2x − 4y = −12 6x + 4y = 4 8x = −8 ∴ x = −1 5 2y − 3 − 2 = 0, ∴ y = 2 4 a 3x + y = 6 (1) 6x + 2y = 7 (2) Multiply (1) by 2. 6x + 2y = 12 (3) Subtract (2) from (3) 0 = 5. There are no solutions. The graphs of the two straight lines are parallel. b 3x + y = 6 (1) 6x + 2y = 12 (2) Multiply (1) by 2. 6x + 2y = 12 (3) Subtract (2) from (3) 0 = 0. There are infinitely many solutions. The graphs of the two straight lines coincide.

16 x − 12y = 4 −15 x + 12y = 6 x = 10 ∴ 4y − 5(10) = 2 ∴ y = 13

g

4p + 10q = −6 39 p = 39

c 3x + y = 6 (1) 6x − 2y = 7 (2) Multiply (1) by 2. 6x + 2y = 12 (3) Add (2) and (3) 12 x = 19. 5 19 and y = . There is only one x= 4 12 solution.

7 2

h 2p + 5q = −3 7p − 2q = 9

9 Cambridge Senior Maths AC/VCE

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The graphs intersect at the ! 19 5 , point 12 4 d 3x − y = 6 (1) 6x + 2y = 7 (2) Multiply (1) by 2. 6x − 2y = 12 (3)

Add (2) and (3) 12 x = 19. 5 19 and y = − . There is only x= 12 4 one solution. The graphs intersect at the ! 19 5 ,− point 12 4

10 Cambridge Senior Maths AC/VCE

ISBN 978-1-107-52013-4

© Evans et al. 2016

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Solutions to Exercise 1D a 4B + 4W = 4 × 15 + 4 × 27

1 x + y = 138

= 60 + 108 = $168

x − y = 88 2x = 226 ∴ x = 113

b 3B = 3 × 15 = $45 c B = $15

y = 138 − 113 = 25

5 x + y = 45

2 x + y = 36

x − 7 = 11 2x = 56 ∴ x = 28; y = 17

x−y=9 2x = 45 ∴ x = 22.5 y = 36 − 22.5 = 13.5

6 m + 4 = 3(c + 4) . . . (1) m − 2 = 5(c − 4) . . . (2)

3 6S + 4C = 58

From (1), m = 3c + 8.

5S + 2C = 35, ∴ 10S + 4C = 70 10S + 4C = 70

Substitute into (2): 3c + 8 − 4 = 5(c − 4)

−(6S + 4C) = 58 4S = 12 ∴ S = $3

3c + 4 = 5c − 20 −2c = −24, ∴ c = 12

2C = 35 − 35, ∴ C = $10

∴ m − 4 = 5(12 − 4)

a 10S + 4C = 10 × 3 + 4 × 10

m = 44

= 30 + 40 = $70 h = 5p

7

b 4S = 4 × 3 = $12

h + p = 20

c S = $3

∴ 5p + p = 30 ∴

4 7B + 4W = 213 B + W = 42, ∴ 4B + 4W = 168 7B + 4W = 213 −(4B + 4W = 168) 3B = 45 ∴ B = 15 15 + W = 42, ∴ W = $27

p = 5; h = 25

8 Let one child have x marbles and the other y marbles.

11 Cambridge Senior Maths AC/VCE

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x + y = 110 x = y − 20 2 ∴ x = 2y − 40 ∴ 2y − 40 + y = 110 3y = 150 ∴ y = 50; x = 60 They started with 50 and 60 marbles, and finished with 30 each. 9 Let x be the number of adult tickets and y be the number of child tickets. x + y = 150 (1) 4x + 1.5y = 560 (2) Multiply (1) by 1.5. 1.5x + 1.5y = 225 (1’) Subtract (1’) from (2) 2.5x = 335 x = 134 Substitute in (1). y = 16 There were 134 adult tickets and 16 child tickets sold. 10 Let a be the numerator and b be the denominator. a + b = 17 (1) a+3 =1 (2). b From (2), a + 3 = b (1’) Substitute in (1) a + a + 3 = 17 2a = 14 a = 7 and hence b = 10. 7 The fraction is 10 11 Let the digits be m and n. m+n=8 (1) 10n + m − (n + 10m) = 36

9n − 9m = 36 n − m=4 (2) Add (1) and (2) 2n = 12 implies n = 6. Hence m = 2. The initial number is 26 and the second number is 62. 12 Let x be the number of adult tickets and y be the numbr of child tickets. x + y = 960 (1) 30 x + 12y = 19 080 (2) Multiply (1) by 12. 12x + 12y = 11 520 (1’) Subtract (1’) from (2). 18 x = 7560 x = 420. There were 420 adults and 540 children. 13 0.1x + 0.07y = 1400 . . . (1) 0.07 x + 0.1y = 1490 . . . (2) From (1), x = (14 000 − 0.7y) From (2): 0.07(14 000 − 0.7y) + 0.1y = 1490 ∴

980 − 0.049 y + 0.1y = 1490 0.051y = 510

∴

y=

510 .051

= 10 000 From (1): 0.1x + 0.07 × 10 000 = 1400 0.1x = 1400 − 700 = 700 ∴ x = 7000 So x + y = $17 000 invested.

12 Cambridge Senior Maths AC/VCE

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14

100s + 20t = 10 000 . . . (1) 3     2t 100  s + 20 = 6000 3 3 2  40t  50s + ∴ = 6000 . . . (2) 3 3 From (1): 100s 20t = 10 000 − 3 5s ∴ t = 500 − . . . (3) 3 Substitute into (2): 5s  50s   40 500 − + = 6000 3 3 3   5s = 54 000 150 s + 120 500 − 3 150 s + 60 000 − 200 s = 54 000 −50s = −6000 s = 120

∴ Substitute into (3): 5 t = 500 − × 120 3

Substitute in (1): 30x + 28 000 = 24(6000 − x) + 35 200 30 x + 28 000 = 144 000 − 24 x + 35 200 54x = 151 200 ∴

x = 2800 ; y = 3200

17 Tea A = $10; B = $11, C = $12 per kg B = C; C + B + A = 100 10A + 11B + 12C = 1120 10A + 23B = 1120 ∴

A = 100 − 2B

10(100 − 2B) + 23B = 1120 3B = 1120 − 1000 ∴ B = 40 A = 20kg, B = C = 40 kg

= 500 − 200

∴ t = 300 He sold 120 shirts and 300 ties. 15 Outback = x, BushWalker = y; x = 1.2y 200 x + 350y = 177 000 200(1.2y) + 350y = 177 000 240y + 350y = 177 000 177 000 = 300 590 x = 1.2 × 300

∴ y= ∴

= 360 16 Mydney = x jeans; Selbourne = y jeans 30 x + 28 000 = 24y + 35 200 . . . (1) x + y = 6000 . . . (2) From (2): y = 6000 − x 13 Cambridge Senior Maths AC/VCE

ISBN 978-1-107-52013-4

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Solutions to Exercise 1E i 4x − 4 ≤ 2

1 a x+38 x > 8 + 5, ∴ x > 13

2 a 4x + 3 < 11 4x < 8, ∴ x < 2

c 2x ≥ 6

2x 6 ≥ , ∴ x≥3 2 2

d

–2

x ≤4 3  x ≤ 12, ∴ x ≤ 12 3 3

e −x ≥ 6

–2

f −2x < −6 −x < −3 0< x−3 3 < x, ∴ x > 3

2

d

–1

0

1

2

1 (x + 1) − x > 1 2 x 1 + −x>1 2 2 x 1 − > 2 2 −x > 1, ∴ x < −1

–2

3−x >5

–1

0

1

2

1 (x + 3) ≥ 1 6 x + 3 ≥ 6, ∴ x ≥ 3

−x > 2

–2

0> x+2 −2 > x, ∴ x < −2 h −

1

2x < −2, ∴ x < −1

−6 ≥ x, ∴ x ≤ −6

g 6 − 2x > 10

0

b 3x + 5 < x + 3

c

0≥6+x

–1

3x ≤6 4 −x ≤ 8

e

–1

0

1

2

3

4

2

3

4

2 (2x − 5) < 2 3 2x − 5 < 3 2x < 8, ∴ x < 4

0≤ x+8 −8 ≤ x, ∴ x ≥ −8

–2

–1

0

1

14 Cambridge Senior Maths AC/VCE

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© Evans et al. 2016

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f

3x − 1 2x + 3 < −2 − 2 4 (3 x − 1) − (4 x + 6) < −8 −x − 7 < −8

6x − 4 > −3 6x > 1, ∴ x > 6

−x < −1, ∴ x > 1 –2

g

–1

0

1

2

0 3

2x > −1, ∴ x > −

2 > x, ∴ x < 2

2

1

2

7 2

c 100 + 20 x > 0

3

1 − 7x ≥ 10 −2 7x − 1 ≥ 10 2 7x − 1 ≥ 20

20 x > −100, ∴ x > −5

4

4 Let p be the number of sheets of paper. 3p < 20 20 p< 3 p ∈ Z, ∴ p = 6

7x ≥ 21, ∴ x ≥ 3 –2

–1

0

1 2

100 > 50x

2x < 7, ∴ x <

h

3

b 100 − 50 x > 0

2x − 1 < 6

0

2

3 a 2x + 1 > 0

4x − 3 − (2x − 2) < 6

–1

1

4

4x − 3 3x − 3 − −1 3 3 (5x − 2) − (2 − x) > −3

5

66 + 72 + x ≥ 75 3 138 + x ≥ 225 ∴ x ≥ 87 Lowest mark:87

15 Cambridge Senior Maths AC/VCE

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© Evans et al. 2016

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Solutions to Exercise 1F d T = dp + cq

1 a c = ab = 6 × 3 = 18

e T = 60a + b

b r = p+q 3 a E = IR

= 12 + −3 = 9 c

= 5 × 3 = 15

c = ab c ∴ b= a 18 = =3 6

d

b C = pd = 3.14 × 10 = 31.4 T  c P=R V 150 = 60 × = 1000 9

r = p+q ∴ q=r−p = −15 − 3 = −18

e c=

√

=

√

f

E R 240 = = 12 20

d I=

a

9=3

c=

√

e A = πrl

a

= 3.14 × 5 × 20 = 314

∴ a = c2 = 92 = 81

f S = 90(2n − 4) = 90(6 × 2 − 4) = 720

u v 10 = =5 2

g p=

h

4 a P V = c, ∴ V =

u v ∴ u = pv p=

= 2 × 10 = 20

b F = ma, ∴ a =

F m

c I = Prt, ∴ P =

I rt

d

2 a S =a+b+c

c P

w = H + Cr ∴ Cr = w − H

b P = xy

∴ r= c C = 5p

w−H C

16 Cambridge Senior Maths AC/VCE

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6 a l = 4a + 3w

e S = P(1 + rt) S = 1 + rt P S −P S ∴ rt = − 1 = P P S −P ∴t= rP ∴

f V=

b H = 2b + h c A = 3 × (h × w)...