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Contemporary Business Mathematics

Canadian 10th Edition by Hummelbrunner

Halliday Coombs ISBN 0133052311

9780133052312

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Chapter 4 Linear Systems

Exercise 4.1 A. 1. x + y = 9 x y = 7 (1)+(2) → In(1) Check: 2x = 16 x = 8 8+ y = 9 y = 1 (x, y) = ( 8, 1) (1) (2) In(1) LS= 8 1= 9 = RS In(2) LS=−8−(−1) =−7 = RS 2. x +5y = 0 x +2y = 6 (1) (2) (1)−(2)→ In(2)

Copyright © 2015 Pearson Canada Inc. Check: 3y =−6 y = 2 x 4 = 6 x =10 (x, y) = (10, 2) In(1) LS=10+5( 2) = 0 = RS In(2) LS=10+2( 2) = 6 = RS 3. 5x +2y = 74 7x 2y = 46 (1) (2) (1)+(2)→ 12x =120 x =10 In(1) 5(10)+2y = 74 2y = 24 y =12 (x, y) = (10,12) Check: In(1) LS= 5(10)+2(12) = 50+24 = 74 = RS In(2) LS= 7(10) 2(12) = 70 24 = 46 = RS

Copyright © 2015 Pearson Canada Inc. 128 CHAPTER 4 4. 2x +9y = 13 2x 3y = 23 (1) (2)→ 12y = 36 y =−3 (1) (2) In(1) 2x +9( 3) = 13 2x 27 = 13 2x =14 x = 7 (x, y) = (7, 3) Check: In(1) LS= 2(7)+9( 3) =14 27 = 13= RS In(2) LS= 2(7)−3(−3) =14+9 = 23= RS 5. y =3x +12 (1) x = y (2) Rearrange 3x y = 12 (3) x + y = 0 (4) (3)+(4)→ In(2) Check: 4x = 12 x = 3 3= y y = 3 (x, y) = (−3,3) In(1) LS = 3 RS = 3( 3)+12 = 9+12 = 3 In(2) LS = 3 RS = 3 6. 3x =10 2y 5y =3x 38 Rearrange (1) (2) 3x +2y =10 3x +5y = 38 (3) (4)

Copyright © 2015 Pearson Canada Inc. CHAPTER 4 129 (3)+(4)→ 7y = 28 y = 4 In(1) 3x =10 2( 4) 3x =18 x = 6 (x, y) = (6, 4) Check: In(1) LS = 3(6) =18 RS =10−2(−4) =10+8 =18 In(2) LS = 5( 4) = 20 RS = 3(6) 38 =18 38 = 20 B. 1. 4x + y = 13 x 5y = 19 To eliminate y (1) (2) (1)×5→ (3)+(2)→ 20x +5y = 65 x 5y = 19 21x = 84 x = 4 (3) (2) In(1) 4( 4)+ y = 13 16+ y = 13 y = 3 (x, y) = (−4,3) Check: In(1) LS= 4( 4)+3= 16+3= 13= RS In(2) LS= 4 5(3) = 4 15= 19 = RS 2. 6x +3y = 24 2x +9y = 8 To eliminate x (1) (2) (2)×3→ 6x + 27y = 24 (3) (1)×( 1) → 6x 3y = 24 (4)

Copyright © 2015 Pearson Canada Inc. 130 CHAPTER 4 (3)+(4)→ 24y = 48 y = 2 In(1) 6x +3( 2) = 24 6x 6 = 24 6x = 30 x = 5 (x, y) = (5, 2) Check: In(1) LS= 6(5)+3( 2) =30 6 = 24 = RS In(2) LS= 2(5)+9( 2) =10 18 = 8= RS 3. 7x 5y = 22 4x +3y =5 To eliminate y (1)×3→ (2)×5→ (3)+(4)→ In (2) Check: 21x 15y = 66 20x +15y = 25 41x = 41 x = 1 4+3y = 5 3y = 9 y = 3 (x, y) = ( 1,3) (1) (2) (3) (4) In(1) LS= 7(−1)−5(3) =−7−15 =−22 = RS In(2) LS= 4( 1)+3(3) = 4+9 =5 = RS 4. 8x +9y =129 6x +7y =99 To eliminate x (1)×3→ 24x +27y = 387 (1) (2) (3) (2)×( 4)→ 24x 28y = 396 (3)+(4)→ y = 9 y = 9 In (2) 6x +7(9) = 99 6x +63= 99 6x = 36 x = 6 (x, y) = (6,9) (4)

Copyright © 2015 Pearson Canada Inc. CHAPTER 4 131 Check: In(1) LS= 8(6)+9(9) = 48+81=129 RS=129 In(2) LS= 6(6)+7(9) = 36+63= 99 RS= 99 5. 12y =5x =16 6x +10y 54= 0 Rearrange 5x +12y =16 6x +10y = 54 To eliminate x (3)×6→ 30x +72y = 96 (1) (2) (3) (4) (5) (4)×5→ (5)+(6)→ In (1) Check: 30x +50y = 270 122y = 366 y = 3 12(3) = 5x +16 36 16 = 5x 5x = 20 x = 4 (x, y) = (4,3) (6) In(1) LS=12(3) = 36 RS= 5(4)+16 = 20+16 = 36 In(2) LS= 6(4)+10(3) 54 = 24+30 54 = 0 RS= 0 6. 3x −8y +44 = 0 7x =12y 56 Rearrange 3x −8y =−44 7x 12y = 56 To eliminate y (3)×3→ 9x 24y = 132 (1) (2) (3) (4) (5) (4)×( 2)→ 14x +24y =112 (6)

Copyright © 2015 Pearson Canada Inc. 132 CHAPTER 4 (5)+(6)→ 5x = 20 x = 4 In (1) 3(4) 8y +44 = 0 12 8y +44 = 0 8y = 56 y = 7 (x, y) = (4,7) Check: In(1) LS = 3(4) 8(7)+44 =12 56+44 = 0 RS= 0 In(2) LS = 7(4) = 28 RS =12(7) 56 = 84 56 = 28 C. 1. 0.4x +1.5y =16.8 1.1x 0.9y = 6.0 To eliminate decimals (1) (2) (1)×10→ (2)×10→ To eliminate y (3)×3→ (4)×5→ 4x +15y =168 11x 9y = 60 12x +45y = 504 55x 45y = 300 (3) (4) Add: 67x = 804 x =12 In(3) 4(12)+15y =168 48+15y =168 15y =120 y = 8 (x, y) = (12,8) 2. 6.5

(1)

(2)

decimals

(3)

(4)

(4)×( 7)→ 35x 63y = 1484

x +3.5y

128

2.5x +4.5y =106

To eliminate

(1)×9 → 13x +7y = 256

(2)×2→ 5x +9y = 212

To eliminate y (3)×

→ 117x +63y = 2304

2.4

(1)

0.6

(2)

eliminate decimals

8y =38 (3)

×5→ 19x +3y = 36 (4) To eliminate y (3)×3→ 36x +24y =114 (4)×( 8)→ 152x 24y = 288 Add: 116x = 174 x =1.5 In(3) 12(1.5)+8y = 38 18+8y = 38 8y = 20 y = 2.5 (x, y) = (1.5,2.5) 4. 2.25x +0.75y = 2.25 (1) 1.25x +1.75y = 2.05 (2)

eliminate decimals

x +1.6y = 7.60

3.8x +

y = 7.20

(1)×

→

x +

(2)

(3)

(1)×100→ 225x +75y = 225

(4)

(2)×1005→125x +175y = 205

To simplify

(5)

(3)÷75→ 3x + y = 3

(6)

(4)÷(5) → 25x +35y = 41

To eliminate y

(5)×35→ 105x +35y =105

(6)×( 1) → 25x 35y = 41

Add: 80x = 64 x = 0.8

In(5) 3(0.8)+ y = 3

2.4+ y = 3 y = 0.6

(x, y) = (0.8,0.6)

5. 3x 2y = 13

4 3 6

4x + 3y = 123

5 4 10

To eliminate fractions

(1)×12→ 9x 8y = 26

(2)×20→16x +15y = 246

To eliminate y

(3)×15→135x 120y = 390

(4)×8→ 128x +120y =1968

Add: 263x =1578

x = 6

In(3) 9(6) 8y = 26 54 8y = 26 8y = 80 y =10

(x, y) = (6,10)

6. 9x + 5y = 47

5 4 10

2x + 3y = 5

9 8 36

To eliminate fractions

(1)×20→ 36x +25y =94

(2)×72→16x +27y =10

To eliminate x

(3)×4→ 144x +100y =376

(4)×( 9) → 144x 243y = 90

134 CHAPTER 4

(1) (2)

(3)

(4) (1) (2) (3) (4)

Add: 143y = 286 y = 2

In(3) 36x 50 = 94 36x =144 x = 4 (x, y) = (4, 2)

7. x + 2y = 7

3 5 15

3x 7y = 1 2 3

(1)

(2)

(3)

(4)

To eliminate y

(4)×3→ 35x +42y = 49 27x 42y = 18

Add: 62x =31 x = 1 2

9 y

(x, y)

To eliminate fractions

3 4

 2 4 

(1)×84→ 21x +36y = 8

(1) (2) (3) (4)

To eliminate fractions

(1)×15→ 5x +6y = 7

(2)×60→ 9x 14y = 6

(3)×7 →

 1

3   

In(3) 5 1 +6y = 7 2  5 +6y = 7 2 5+12y =14

y =

8. x + 3y = 2 4 7 21 2x + 3y = 7 3 2 36

(2)×36 → 24x +54y = 7

→

x +108y = 24

48x 108y =14 Add: 15x = 10 2 x = 3 In(3) 21 2 +36y = 8 3  14+36y = 8 36y = 6 y = 1 6 (x, y) =  2 , 1   3 6    Exercise 4.2 A. 1. A(−4,−3) B(0, 4) C(3, 4) D(2,0) E(4,3) F(0,3) G( 4,4) H( 5,0) 2. (a)

×3

(4)×( 2)→

Copyright © 2015 Pearson Canada Inc. ; CHAPTER 4 137 (b) 3. (a) (b) (c) (d) 4. (a) 4x +5y =11 5y = 4x +11 y = −4x + 11 5 5 Slope, 4 m = y-intercept, b = 11 5 5 (b) 2y 5x =10 2y =5x +10 y = 5x + 10 2 2 Slope, m = 5 ; 2 y-intercept, b =5 (c) 1 1 y = 2x 2 1 y = 2x 1 2 y = 4x +2 Slope, m = 4; y-intercept, b = 2

(d)

(e)

Slope, m = 0;(line to x-axis); y-intercept, b = 2

3y +6 = 0 3y = 6 y = 2

) 0.15

= 0 15x +30y 12 = 0 30y = 15x +12 y = 15x + 12 30 30 1 2 y =− 2 5 Slope, m 1 ; y-intercept, b = 2 (g) 2 5 2 1 x = 0 2 2 = 1 x 2 x = 4

2x y =3 2x − y =9 y = 2x +9 y = 2x 9 Slope, m = 2; y-intercept, b = 9 (f

x +0.3y 0.12

2)(y

1) xy = 2 xy 2y + x 2 xy = 2 −2y =−x +4 y = x 2 2

-intercept,

Slope is undefined no y-intercept (h) (x

Slope, m = 1 ; 2 y

b = 2

8. 9. For y = 2x 3 slope, m = 2 y-intercept, b =−3

10. For y = 3x +9 slope, m = 3 y-intercept, b =9

C. 1. For y =3x +20 x 0 20 40

20 80 140 or m =3 b = 20

2. For y 2 x +40 5

0 50 100

CHAPTER 4 141

= =

2 5

y 40 20 0 or m

b = 40

For 3x +4y =1200 x 0 200 400 y 300 150 0 or m 3 4 b =300

For 3y 12x 2400 = 0 x 0 100 200 300 y 800 1200 1600 2000 or m = 4 b =800

Copyright © 2015 Pearson Canada Inc. CHAPTER 4 145 7. 5x 2y = 20 x 4 2 6 y 0 5 5 8. 3y =−5x x 0 3 3 y 0 5 5 B. 1. For y 4x = 0 x 0 5000 10 000 y 0 20 000 40 000 For y 2x 10000 = 0 x 0 5000 10 000 y 10 000 20 000 30 000

For 4x +2y = 200 x 0 20 40 50 y 100 60 20 0 For x +2y =80 x 0 40 80 y 40 20 0

For 3x +3y = 2400 x 0 400 800 y 800 400 0

Business Math News Box

SOLUTIONS:

1. Let x represent Mega Brands’ 2011 revenues

111.5% x = 420.3 million

x = 420.3 / 1.115 = 376.9

Mega Brands’ 2011 revenue was $377.0 million.

2. ($4 million – $234 000) / $234 000 = 16.094017 = 1609.4%

3. 118% × $108 million = $127.4 million

Mega Brands’ fourth quarter net sales for 2012 was $127.4 million.

4. 2012 annual sales: $420.3 million

Growth rate in sales revenue in 2012: 11.5%

Assuming an 11.5% growth in sales in 2013 and 2014: $420.3(1.115)(1.115) = $522.5 million

Yes, Mega Brands is on track to meet (exceed) its goal of delivering $500 million in annual sales by 2014.

Exercise 4.4

A. 1. Let the number of units of Product A be x and the number of units of Product B be y

2. Let the number of units of Product 1 be x and the number of units of Product 2 be y.

3. Let the number of tax returns be x and the total cost be $y.

4. Let monthly sales be $x and the total salary be $y.

B. 1. Let x represent the number of employees at the larger location, y, the number of employees at the smaller location.

∴ x + y = 24

2x =3y +3

(1) (2)

(1)×3→

From(2) → 3x +3y = 72 2x 3y =3

Add: 5x = 75 x = 15

The number of employees is 15 at the larger location, and 9 at the smaller one.

Sum =15+9 = 24

2×15 =30 = (3×9)+3

2. Let x represent the number of orders for the 1st special, y those for the 2nd special.

There are 12 orders for the 1st special, 18 for the 2nd special

3. Let the number of jars of Brand X be x and the number of jars of the No-Name brand be y

Sales were 90 jars of Brand X and 50 jars of No-Name brand.

Check:

7x 4y

3 x + 2 y = 21 4 3 (2)×12→ (1)×2→ 9x +8y = 252 14x 8y = 24 (1) (2) Add: 23x = 276 x =12 In(1) 84 4y =12 4y = 72 y =18

=12

Check: 7×12 4×18 = 84 72 =12 3 2 ×12+ ×18 = 9+12 = 21 4 3

x + y =140 2.25x +1.75y = 290 (1) (2) (2)×4 → (1)×7 → 9x +7y =1160 7x +7y = 980 Subtract: 2x =180 x =90 y =50

Check

Totalsold

140 Value: 90

50×1.75

87.50

= 90+50 =

×2.25 = $202.50

= $290.00

4. Let Nancy’s sales be $x and Andrea’s sales be $y.

x + y =940

(2)

x =3y 140 (1)

(1)×3→

From (2)→ 3x +3y = 2820 x 3y = 140

4x = 2680 x = 670 y = 270

Nancy's sales were $670; Andrea's sales were $270.

Total = 670+270 =$940

3×270 140 =810 140 =$670

5. Let Kaya’s investment be $x, and Fred’s investment be $y.

∴x + y =55000 y = 2 x +2500 3

(2)×3→ 2x +3y = 7500

(1)×2→ 2x +2y =110 000

Add: 5y =117 500

y = 23 500

x = 31500

(1) (2)

Kaya's investment is $31 500 and Fred's investment is $23 500.

Check:

Total = 31500+23 500 = $55 000

2 ×31 500+2500 = 21 000+2500 = $23 500

6. Let the number of chairs produced by the first shift be x and the number of chairs produced by the second shift be y.

∴x + y = 2320

y = 4 x 60 3

(2)×3→ 4x +3y = 180

(1)×4→ 4x +4y = 9280

Add: 7y = 9100

y = 1300

x =1020

(1)

(2)

Canada Inc. CHAPTER 4 151

The first shift produced 1020 chairs and the second shift 1300 chairs.

Check: Total =1020+1300 = 2320

4 ×1020 60 =1360 60 =1300 3

7. Let the number of Type A lights be x and the number of Type B lights be y.

∴ x + y = 60 40x +50y = 2580 (1)

(2)÷10→ 4x +5y = 258

(1)×( 4)→ 4x 4y = 240

Add: y =18 x = 42

The number of Type A is 42, and the number of Type B is 18.

Check:

Total = 42+18 = 60

Value: 42×40 = $1680 18×50 = 900 $2580

8. Let the number of units of Product A be x and the number of units of Product B be y

∴ x + y = 60

4x +3y = 200 (1)×4→

The number of units of Product A is 20, and of Product B is 40.

Check:

Total number = 20+40 = 60

Number of hours:

ProductA:20×4 = 80

ProductB:40×3= 120

Totalhours = 200

(2)

y = 40 x = 20 (1) (2)

Subtract: 4x +

y = 240

= 200

9. Let the number of quarters be x and the number of loonies be y. ∴

x +100y =8575

Substitute (2) x = 3 (72)+1=54+1=55 4

Marysia has 55 quarters and 72 loonies.

Value:

55×0.25 =$13.75

72×1.00 = 72.00 $85.75

10. Let the number of $12-tickets be x and the number of $15-tickets be y.

The club bought 30 $12-tickets and 21 $15-tickets.

Check:

Value =30(12)+21(15) = 360+315=$675

x = 3 y +1 4 (1) (2) (1)÷25 → (2)×4 → (3)×4 → x +4y = 343 4x = 3y +4 4x +16y =1372 (3) (4) (5) Rearrange (4) 4x 3y = 4 (6)

→ 19y =1368 y = 72

(5) (6)

y = 4 x 3 5 Substitute (2) in (1). 12x +15 4 x 3 = 675 (1) (2)  5    12x +12x 45= 675 24x = 720 x =30

x +

y = 675

Substitute in (2). y = 4 (30) 3 5

= 21

Review Exercise

1. (a) For 7x +3y = 6

3y =−7x +6 y = 7x +2 3

Slope, m 7 ; y-intercept, 3 b = 2

(b) For 10y =5x y = 5x 10 y = 1 x 2

Slope, m = 1 ; y-intercept, 2 b = 0

Slope, m = 3 ; y-intercept, 2 b = 4

(d) For 1.8x +0.3y 3= 0 18x +3y 30 = 0

3y = 18x +30 y = 6x +10

Slope, m = 6; y-intercept, b =10

(e) For 1 x = 2 3 x = 6 Line is parallel to the y-axis.

Slope, m is undefined. There is no y-intercept.

Canada Inc.

=− 154 CHAPTER 4

(f) For 11x 33y =99 33y = 11x +99 y = 11x + 99 33 y = 1 x 3 3

Slope, m = 1 ;3 y-intercept, b = 3

(g) For xy (x +4)(y 1) =8 xy xy 4y + x +4 =8 4y = x +4 y = 1 x 1 4

Slope, m = 1 ; y-intercept, 4

(h) For 2.5y −12.5 = 0 25y 125 = 0 y =5

b = 1

Line is parallel to the x-axis; slope, m = 0; y-intercept, b =5

2. (a) 2x y = 6 x 3 0 5 y 0 6 4

(b) 3x +4y = 0 x 0 4 4 y 0 3 3

5x +2y =10 x 0 2 4 y 5 0 −5

y = 3

5y = 3x +15 x 0 5 2.5 y 3 0 1.5

) 5x 4y = 0 x 0 4 4 y 0 5 5

(c)

(d)

(e)

158 CHAPTER 4

(b) x +4y = 8 and 3x +4y = 0

x 0 4 4

y 2 3 1

x 0 4 4 y 0 3 3

x 0 3 3 y 0 5 5

(d) 2x +6y =8 and x = 2

x 4 2 1 y 0 2 1

(e) y =3x 2 and y =3

x 0 5/3 1 y 2 3 5

(h) y =−2 and 5x +3y =15 x 3 0 4.2 y 0 5 2

x 0

y 0 8 4

(f ) y = 2x and x = 4

4 2

(g) x = 2 and 3x +4y =12 x 0 2 4 y 3 4.5 0

Copyright © 2015 Pearson Canada Inc. 160 CHAPTER 4 4. (a) 3x +2y = 1 5x +3y = 2 To eliminate y (1)×3→ 9x +6y = 3 (1) (2) (2)×(−2)→ −10x −6y = 4 Add: x =1 x = 1 In(1) 3( 1)+ 2y = 1 2y = 2 y =1 (x, y) = (−1,1) (b) 4x 5y = 25 3x +2y =13 To eliminate y (1) (2) (1)×2→ (2)×5→ 8x 10y = 50 15x +10y = 65 (c) Add: 23x =115 x = 5 In(2) 15+2y =13 2y = 2 y = 1 (x, y) = (5, 1) y = 10x 3y = 29 x Rearrange (1) (2) 10x + y = 0 x +3y = 29 (3) (4)

(d) 2y =3x +17 3x =11 5y

(e) 2x 3y =13 3x 2y =12

(2)

From(1)→ 2x 3y = 11

(2)×3→ 9x +3y = 39

Add: 7x = 28 x = 4

30x +3y = 0 x +3y = 29

29x = 29 x = 1 In

Rearrange y = 10( 1) y =10 (x, y) = ( 1,10) (1) (2)

y = 4 In (2) 3x =11 20 3x = 9 x = 3 (x, y) = ( 3,4)

To eliminate y (3)×3→

Subtract:

(1)

3x +2y =17 3x +

y =11 Add: 7y = 28

(1) (2)

To eliminate y

(3)×2→ 4x 6y = 26

x = 2 In

3 (

(2,

×(3)→ 9x +6y = 36 Add: 5x = 10

(1) 2(2) 3y =13

y =9

x, y) =

(1) (2)

(f) 2x =3y 11 y =13+3x

Copyright © 2015 Pearson Canada Inc. 162 CHAPTER 4 In (2) (g) 2a 3b 14 = 0 a + b 2 = 0 y =13+3( 4) y =13 12 y =1 (x, y) = ( 4,1) (1) (2) Rearrangeandmultiply(2)×2 2a 3b =14 2a +2b = 4 (h) Subtract: In (2) a + c = 10 8a +4c = 0 (1)×4→ Subtract: 5b =10 b = 2 a 2 2 = 0 a = 4 (a,b) = (4, 2) 4a +4c = 40 8a +4c = 0 4a = 40 a =10 (1) (2) In (1) 10+ c = 10 c = 20 (a,c) = (10, 20) (i) 3b 3c = 15 2b +4c =14 (1) (2) (1)×2→ 6b 6c = 30 (2)×3→ 6b +12c = 42 Add: 6c =12 c = 2 In (1) 3b 6 = 15 3b = 9 b = 3 (b,c) = (−3,2) (j) 48a 32b =128 16a +48b =32 (1) (2) (1)÷16→ (2)÷16→ 3a 2b =8 a +3b = 2 (3) (4)

Copyright © 2015 Pearson Canada Inc. CHAPTER 4 163 (4)×3→ (3)→ 3a +9b = 6 3a 2b = 8 Subtract: 11b = 2 b = −2 11 In (4) a +3 2 = 2 11   a = 2 6 11 (a,b) =  2 6 , 2   11 11   (k) 0.5m +0.3n =54 0.3m +0.7n = 74 (1) (2) (1)×10→ 5m +3n = 540 (2)×10→ 3m +7n = 740 (3)×3→ 15m +9n =1620 (4)×5→ 15m +35n = 3700 (3) (4) Subtract: 26n = 2080 n = 80 In (3) 5m +240=540 5m =300 m = 60 (m,n) = (60,80) (l) 3 m + 5 n = 3 4 8 4 5 n + 2 m = 7 6 3 9 To eliminate fractions (1)×8→ 6m +5n = 6 (2)×18→ 15n +12m =14 (3)×2→ 12m +10n =12 (4)→ 12m +15n =14 Subtract: 5n = 2 n = 2 5 (1) (2) (3) (4)

5. (a) Let the number of announcements be x and the total cost be $y.

(b) Let the number of units of Product A be x and the number of units of Product B be y.

6. (a) Let x represent the number of sports tires sold, y the number of all-season tires sold.

(b) Let the number of $2.50-tickets be x and that of $3.50-tickets be y.

6x +5

(1) 3 x 2 y = 0 4 3 (2) (1)×12→ 9x – 8y = 0 (3) To eliminate y (3)×5→ 45x – 40y = 0 (1)×8→ 48x + 40y = 744 Add: 93x = 744 In (3) x =8 9(8) 8y = 0 8y = 72 y =9 8 sports tires were sold, 9 all-season tires were sold.

y =93

x + y =

(1) 2.5x +3.5

(2)

450

y =1300

The number of $2.50-tickets is 275 and the number of $3.50-tickets is 175.

+2y =175

=3y Substitute (2) in (1)

y +2y =175

y =175 y =35 x =105

The price of the jacket is $105.

(d) Let the cost of a case of white bordeaux be $x and that of a case of red bordeaux be $y. 3x +5y = 438

Self-Test

eliminate x (1)×2→ (2)×5→ 5x 7y = 2600 5x +5y = 2250

y =

x =

Subtract: 2 y = 350

175

275

(1) (2)

3x +5(2x 6) = 438 3x +10x 30 = 438 13x = 468 x =36 y = 2(36) 6 = 66 (1) (2)

Substitute (2) in (1)

A case of white bordeaux is $36; and a case of red bordeaux is $66.

(a)

y 11 = 3 Slope,

0; y-intercept, b

For4y +11= y

m =

= 11 3

166

CHAPTER

Slope, m = 6; y-intercept, b = 9

3y = x y

Slope, m

1 = 3 1 ; y-intercept, b = 0 = 3

(d) For 6y 18= 0 6y =18 y = 3

Line is parallel to the x-axis; slope, m = 0; y-intercept, b = 3

(e) For 13 1 x = 0 2 26 x = 0 x = 26

Line is parallel to the y-axis; slope, m isundefined ; there is no y-intercept

(f ) For ax + by = c

by =−ax + c

Slope, m

a c y = b b a ; y-intercept, b = c = b b

2. (a) y = x 2 x 0 2 2 y 2 0 4 x y = 4 x 0 4 2 y 4 0 2

Inc.

x x +

(b) For 2 x 1 y =1 3 9 6x y =9 y = 6x 9

(b)

3x = 2y and x = 2 x 0 2 2 y 0 3 3 3. (a) For − x =−55+ y x 0 25 55 y 55 30 0

For 3x +2y = 600 x 0 100 200 y 300 150 0

4. (a) 6x +5y =9

4x 3y = 25

To eliminate y

(1)×3→ 18x +5y = 27

(2)×5→ 20x 15y =125

(b) 12 – 7x = 4y

6 – 2y = 3x

Rearrange 12 = 7x +4y

6 =3x +2y

To eliminate y

(1)×2→ 12 = 6x +4y 12 = 7x +4y

Subtract: 0 = x

(1)

(2)

3 (x, y)

(4, 3)

Add: 38x =152 x = 4 In (1) 24+5y =9

y =−15

(1)

(2)

(3) (4)

0,3)

0.7

(1)

(2) (1)

2a +3b = 0 (2)

7a 2b = 2500 (1) (2) (3) (4) (3)

2→

0 21a

In (3)

200 (a,b

)

4 b − 3 c =− 17 3 5 3 5 b + 4 c = 5 6 9 9 (1) (2)

In (1) 12 = 4y y =3

, y) =(

a −0.2b = 250 To eliminate b from

and

→

(4)×3→

−

= 7500 Add: 25a = 7500

=300

600+3

= 0

) =(300, 200

(d)

(1)×15→ 20b 9c = 85 (3)

(2)×18→ 15b +8c =10 (4)

(3)×3→ 60b −27c =−255

(4)×4→ 60b +32c = 40

Subtract: 59c = 295

5. Let the amount invested at 4% be $x, and the amount invested at 6% be $y.

+ y =12000

The amount of interest earned at 4% is $0.04x; the amount of interest earned at 6% is $0.06y 0.04x +0.06y =560

To eliminate x from (1) and (2)

×100→

×4→ 4x +6y = 56 000 4x +4y = 48 000

Subtract: 2y = 8000

= 4000

(1) x +4000 =12 000

= 8000

Theamountinvestedat 4% is $8000; the amount invested at 6% is $4000.

6. Let Eyad’s share of the profit be $x, and Rahia’s share of the profit be $y.

+ y =12 700

x = 2 y +2200 5

To eliminate fractions

×5→ 5x = 2y +11 000

To eliminate y from (1) and (3) Rearrange(3)→

In (4) c

(

=5 15b +40 =10 15b = 30

= 2

=( 2,5)

(1)

(2)

(1)

(2)

(1) (2) (2)

(3)

(1)

Add:

x =

×2→ 5x 2y =11 000

x +2y = 25 400

7x = 36 400

5200