CHAPTER Solutions Key 3 Linear Systems

[Pages:40]CHAPTER Solutions Key 3 Linear Systems

ARE YOU READY? PAGE 179

1. D

3. E

5. LCM = 32 ? 2 = 18

7. LCM = 23 ? 32 = 72

9. perpendicular

10. parallel;

5x 10y = 3

10y = 5x + 3

y = _21_x

_3__

10

and

y

=

_1_

2

x

6

12. neither;

2x 3y = 4

3y = 2x + 4

y

=

_2_

3x

+

_4_

3

13. 1.5(8) + 3(14) = 12 + 42 = 54

15. 4(0.25) 2(2)

= 1 4 =1 2 =1

17. 8x + 19 = 5 8x = 24 x= 3

19. 9x (x + 12) = 13

9x x 12 = 13

8x 12 = 13

8x = 1

x=

_1_

8

21. _14_x + _23_x = 8

3x + 8x = 96

11x = 96

x

=

_9_6_

11

2. C

4. B

6. LCM = 23 ? 7 = 56

8. LCM = 5 ? 33 = 135

11. perpendicular; x y=3 y=x 3

and x+y=4

y= x+4

and

3y x = 5

3y = x + 5

y

=

_1_

3x

+

_5_

3

14. 5(6) _3_( 4)

4

= 30 + 3

= 33

( ) 16.

_7_5_(_1_) 3 _1

3

= _7_5_

1

= 75

18. 5x + 4 = 25 2x 7x + 4 = 25 7x = 21 x=3

20. 3(4x 5) 1 = 20

12x + 15 1 = 20

12x + 14 = 20

12x = 6

x= x=

_6__ _11_2

2

22.

_25_x

+

_1_

6

=

4

12x + 5 = 120

12x = 125

x=

_1_2_5_

12

23.

x

+

_1_

2

=

_1_

5

10x + 5 = 2

10x = 7

x=

_7__

10

24.

_1_

2

=

3x

_31_x

3 = 18x 2x

3 = 16x

_3__

16

=

x

3-1 USING GRAPHS AND TABLES TO SOLVE LINEAR SYSTEMS, PAGES 182-189

CHECK IT OUT! 1a. ___x__+__2_y_=__1_0_

(4) + 2(3) 10

10 10 (4, 3) is a solution.

___3_x____y_=__9_ 3(4) (3) 9

9 9

b. ___6_x____7_y_=__1_ 6(5) 7(3) 1

9 1 (5, 3) is not a solution.

2a.

2y 4x

+ =

6 3

= +

x y

y = _12_x 3

x

y

2

2

0

3

2

4

y = 4x x 2 0 2

3 y 5 3 11

The solution to the system is (0, 3).

b.

x+ 2x

y=8 y=4

y= x+8

y = 2x 4

x

y

x

y

0

8

0

4

2

6

2

0

4

4

4

4

The solution to the system is (4, 4).

c.

y 3x

x=5 +y=1

y=x+5

x

y

y = 3x + 1

x

y

1

4

1

4

0

5

0

1

1

6

1

2

The solution to the system is ( 1, 4).

83

Holt McDougal Algebra 2

3a.

7x 3y

=

y= 21x

11 + 33

y = 7x + 11

y = 7x + 11

consistent, dependent;

infinite number of

solutions

b.

x+4= 5y = 5x

y +

35

y=x+4

y=x+7

inconsistent; no solution

4. Step 1 Write an equation for costs with each calling card. [then "Let x represent...." ending with "Card B" and its equation] Let x represent the number of minutes and y represent the total cost. Card A: y = 0.05x + 0.50 Card B: y = 0.08x + 0.20

Step 2 Solve the system by using a table of values.

y = 0.05x + 0.50

y = 0.08x + 0.20

x

y

x

y

5 0.75

5 0.60

10 1.00

10 1.00

15 1.25

15 1.40

So the cost is the same for each card at 10 min.

THINK AND DISCUSS

1. Graph the two equations on the same grid. If the line intersect once, there is 1 solution; if they are parallel, there is no solution; if they coincide, there are infinitely many solutions.

2. A solution to a system is represented on a graph by a point of intersection, and parallel lines never intersect.

3.

%XAMPLE 'RAPH

%XACTLY/NE 3OLUTION

yx yx

y

)NFINITELY-ANY 3OLUTIONS

yx yx

y

.O3OLUTION

yx yx

y

x

x

x

3LOPES y INTERCEPTS

$IFFERENT %ITHER

3AME 3AME

3AME $IFFERENT

EXERCISES

GUIDED PRACTICE

1. inconsistent

2. ___2_x____y_=__3_ 2(3) (3) 3

___y__+__x_=__6_ (3) + (3) 6

3 3

6 6

Because the point is a solution of both equations,

(3, 3) is a solution of the system.

3. _____y___4_x__=___7_

_____5_x_+__y__=___6_

( 3) 4(1) 7

5(1) + ( 3) 6

7 7

2 6

Because the point is not a solution of both

equations, (1, 3) is not a solution of the system.

4. _____5_y___5_x__=__1_0_

___3_x__+__1_0_=__2_y__

5(2) 5( 2) 10

3( 2) + 10 2(2)

20 10

4 4

Because the point is not a solution of both

equations, ( 2, 2) is not a solution of the system.

5. _y_=__3____x___

_____6_x_+__2_y__=__2_

4 3 ( 1)

6( 1) + 2(4) 2

4 4

2 2

Because the point is a solution of both equations,

( 1, 4) is a solution of the system.

6.

y3x+

x= 5y

5 =

1

y= x+5

x

y

1

6

0

5

3

2

y

=

_3_

5x

+

_1_

5

x

y

1

_2_

5

0

_1_

5

3

2

The solution to the system is (3, 2).

7.

3y x

+ 6x y=

= 3 7

y = 2x + 1

y=x+7

x

y

x

y

0

1

0

7

1

3

1

6

2

5

2

5

The solution to the system is ( 2, 5).

8.

y8x

x= + 4y

0 =

24

y=x

y = 2x 6

x

y

x

y

0

0

0

6

1

1

1

4

2

2

2

2

The solution to the system is ( 2, 2).

9. x4x

y= 1 2y = 2

y=x+1

y = 2x 1

x

y

x

y

0

1

0

1

1

2

1

1

2

3

2

3

The solution to the system is (2, 3).

84

Holt McDougal Algebra 2

10.y = 7x + 13

11. 2x

4y = 28x 12

y = 7x 3

inconsistent; no solution

3y = 15 3y = 2x + 15

y = _32_x + 5

12. 8y

24x = 64 8y = 24x + 64 y = 3x + 8

3y 2x = 15 3y = 2x + 15

y = _23_x + 5

consistent, dependent; infinite number of solutions 13. 2x + 2y = 10

2y = 2x 10 y= x 5

9y + 45x = 72 9y = 45x + 72 y = 5x + 8

consistent, independent; one solution

4x + 4y = 16 4y = 4x 16 y= x 4

inconsistent; no solution

14. Step 1 Write an equation for draining rate for

each tank. [then "Let x represent...." ending with

the two equations]

Let x represent the number of minutes, and let

y represent the depth of the water.

Tank A: y = 1x + 7

Tank B: y = 0.5x + 5

Step 2 Solve the system by using a table of values.

y= x+7

y = 0.5x + 5

x

y

x

y

1

6

1

4.5

2

5

2

4

3

4

3

3.5

4

3

4

3

5

2

5

2.5

So the two tanks have the same amount of water at 4 min.

PRACTICE AND PROBLEM SOLVING

15. __x__+__y_=__0_ 2+2 0

_____7_y____1_4_x_=__4_2_ 7(2) 14( 2) 42

0 0

42 42

Because the point is a solution of both equations,

( 2, 2) is a solution of the system.

16. ______2_y____6_x_=__8_ 2( 5) 6( 3) 8

___4_y__=__8_x_+___4___ 4( 5) 8( 3) + 4

8 8

20 20

Because the point is a solution of both equations,

( 3, 5) is a solution of the system.

17. _y_=__2_ 2 2

__y_+__8__=__6_x__ (2) + 8 6(3)

10 18

Because the point is not a solution of both

equations, (3, 2) is not a solution of the system.

18. _y_=__8_x__+__2__ 1 8(6) + 2

1 50

Because the point is not a solution of both

equations, (6, 1) is not a solution of the system.

19.

2 x

+ +

y y

= =

x 4

y=x 2

y= x+4

x

y

x

y

0

2

0

4

1

1

1

3

3

1

3

1

The solution to the system is (3, 1).

20. y41=y0x_12_x2+x5y=1=410

x

y

y = 2x 2

x

y

0

1

0

2

2

2

2

2

4

3

4

6

The solution to the system is (2, 2).

21.12x2x

+ 4y y=

= 6

4

y = 3x 1

y = 2x 6

x

y

x

y

0

1

0

6

1

4

1

4

2

7

2

2

The solution to the system is (1, 4).

22.

y3x=

10 3y

x = 0

y = x + 10

y=x

x

y

x

y

5 15

5

5

0

10

0

0

5

5

5

5

The solution to the system is (5, 5).

23. 27y = 24x + 42

y

=

_8_

9x

_1_4_

9

9y = 8x 14

24.

y_23_x=+_23_x9

= +

45 9

y = _89_x

_1_4_

9

4y 6x = 36

consistent, dependent; infinite number of solutions

4y = 6x + 36

y

=

_3_

2x

+

9

consistent, dependent;

infinite number of

solutions

85

Holt McDougal Algebra 2

25. 7y + 42x = 56 7y = 42x + 56 y = 6x + 8

26. 3y = 2x

y = _32_x

25x 5y = 100 5y = 25x 100

y = 5x 20 consistent, independent; one solution

4x + 6y = 3

6y = 4x + 3

y

=

_32_x

+

_1_

2

inconsistent; no

solution

27. Let x be the number of systems sold, and y be the

total money earned.

Jamail: y = 100x + 2400

Wanda: y = 120x + 2200

y = 100x + 2400

y = 120x + 2200

x

y

2 2600

4 2800

6 3000

8 3200

10 3400

x

y

2 2440

4 2680

6 2920

8 3160

10 3400

So they have to sell 10 systems to earn the same amount.

28. ___y____x_=__2_ (2) (4) 2

___2_x__+__y_=__8_ 2(4) + (2) 8

2 2

10 8

Because the point is

not a solution of both

equations, (4, 2) is not a y = 2x + 8

solution of the system. y = 2(2) + 8

So y = x + 2 and

y= 4+8

y = 2x + 8.

y=4

x + 2 = 2x + 8

3x + 2 = 8

3x = 6

x=2

Because the point is a solution of both equations,

(2, 4) is a solution of the system.

29. _____3_x_+__y__=___1_ 3( 1) + (2) 1

___8_x__+__6__=___y_ 8( 1) + 6 (2)

1 1

2 2

Because the point is a solution of both equations,

( 1, 2) is a solution of the system.

30. ___x__+__y_=__9_

(7) + (2) 9

9 9

Because the point is

not a solution of both

equations, (7, 2) is not a

solution of the system.

So y = x + 9 and

y

=

_41_x

x+

9

1. =

_14_x

1

4x + 36 = x 4

5x + 36 = 4

5x = 40

x=8

__4_y__+__4_=__x_ 4(2) + 4 7

12 7

y= x+9 y = (8) + 9 y=1

Because the point is a solution of both equations, (8, 1) is a solution of the system.

31. ___3_x__+__4_y_=____9_ 3(0) + 4(6) 9

24 9

(0, 6) is not a solution.

4y = 3x 9

y=

So y

_34_x

=

_34__9x__34_=x

4

_9_

4

2x

_9_

4

and

+ 6.

y

=

2x

+ 6. y=

2x

+

6

3x 9 = 8x + 24

y = 2( 3) + 6

11x 9 = 24 11x = 33 x= 3

y=0

The solution is ( 3, 0).

32a.Roberto: r = 15x + 12 Alexandra: a = 18x + 8

y

b.

from

the

graph:

_4_

3

h

c. from the graph: 32 mi

x

33a.Lynn: l = 200x + 10,000 Miguel: m = 50x + 5000

y

x

b. from the graph: 20 min

c. from the graph: 6000 ft

34a. Plan A: y = 0.4x + 15 Plan B: y = 0.25x + 30

y

b. from the graph: 100 min

x

c. 2 hours = 120 min Plan A: y = 0.4(120) + 15 = 63 Plan B: y = 0.25(120) + 30 = 60 He should use plan B. It is $3.00 cheaper than plan A.

86

Holt McDougal Algebra 2

35.

y y

= =

x 2x

+ 6 3

consistent, independent

x + 6 = 2x 3

3x + 6 = 3

3x = 9

x=3

The solution of the

system is (3, 3).

y= x+6 y = (3) + 6 y=3

36.

x y

= =

2 3

37.y y

= =

3x 3x

+

1 3

consistent, independent inconsistent; no solution

The solution to the

system is (2, 3).

38. (2, 3)

39. ( 0.25, 4)

40. (6.444, 131)

41. (2.831, 30.403)

42a. Truck:

city: _4_0_8_ = 24 mi/gal

17

hwy: _4_7_6_ = 28 mi/gal

17

b. Truck:

_4_7_6_ = 7_1_4 h

60

15

Car:

_4_9_0_ = 8_1 h

60

6

Temhepttyrutcaknkwiallfthearve7_1a4_nh. 15

Car:

city: _3_6_4_ = 26 mi/gal

14

hwy: _4_9_0_ = 35 mi/gal

14

c. _4_7_6_ = 8_1 h

x

6

( )_81 6

x

=

476

_ x

=

58 2

7

mi/h

Tath5e8t_r2umcki/hm.ust travel 7

The car will empty tank

ahfatevre8a_1nh.

6

43. 2y = x + 4

y = _21_x + 2

infinite number of solutions any multiple of the

original equation. For example,

2y = x + 4

4y = 2x + 8...etc.

no solution any equation with the same slope as

y

_1_

2

=

but a

_12_x

different 6

y-intercept.

For

example,

one solution any equation with a different slope.

For example,

y=

_3_

4x

+

2

44. Possible answer: (3.5, 0.25)

45. Consistent, independent; one solution. The solution is the y-intercept since it is the same point for each line.

46. Possible answer: One hot-air balloon starts at 120 ft and rises quickly. The other balloon starts at 200 ft and rises slowly. After 4 min, the balloons are at the same height, 280 ft, as represented by the point of intersection.

TEST PREP

47. D

48. G

49. B

50.

y

x+y = 4x

=

8

So y = x + 8 and y = 4x.

x + 8 = 4x

8 = 3x

_8_

3

=

x

The solution of the

( ) system is _8_, _3_2_ . 33

y = 4x

( ) y

=

4

_8_

3

y

=

_3_2_

3

Three times the value of y is 32.

CHALLENGE AND EXTEND

51. 55x + 100 = 20x + 600

55x + 100 = 600

35x = 500

x x

= =

_5_0_0_ _13_05_0_

7

( ) y = 20x + 600

y y y

= = =

2_2_00_0__10__70_+0_ _6_27_0_0_

7

+ 600

_4_2_0_0_

7

( ) The solution of the system is _1_0_0_, _6_2_0_0_ . 77

52. 5y = 20x + 135

y = 4x + 27

y = 50x + 32.

So y = 4x + 27 and

50x + 32 = 4x + 27

46x + 32 = 27

46x = 5

x

=

_5__

46

( ) y = 4x + 27

y y y y

= = = =

4 _5__ 6__1_24_21__1__0630__4++6

23

+ 27

27

_6_2_1_

23

( ) The solution of the system is _5__, 6__1_1_ . 46 23

53. 18y = 9x + 126

y=

_1_

2x

+

7

14y = y=

infinite number of solutions

7x + 98

_1_

2x

+

7

54. 0.25x y = 2.25 y = 0.75x + 3.75 So y = 0.75x + 3.75 and y = 0.25x 2.25. 0.25x 2.25 = 0.75x + 3.75 0.5x 2.25 = 3.75 0.5x = 6 x = 12

y = 0.25x 2.25 y = 0.25( 12) 2.25 y = 3 2.25 y = 5.25 The solution of the system is ( 12,

5.25).

55. The solution is meaningless in the real world. Time and cost cannot be negative. The costs for the 2 products will never be equal.

87

Holt McDougal Algebra 2

56a. Brad: y = 12x + 70 Cliff: y = 15x + 100 12x + 70 = 15x + 100 3x + 70 = 100 3x = 30 x = 10 days

b. Brad: y = 12(10) + 70 y = 50 lb

Cliff: y = 15(10) + 100 y = 50 lb

No; they will both run out of food before that time.

c. Brad: y = 12(4) + 70 + 100 y = 122 lb

Cliff: y = 15(4) + 100 + 100 y = 140 lb

On day 10 (6 days later) they have used up: Brad: 12(6) = 72 lb

122 72 = 50 lb Cliff: 15(6) = 90 lb

140 90 = 50 lb

The answer would make sense. Each farmer would

have 50 lb of food on day 10.

SPIRAL REVIEW

57. __4__ ? __1_2_

12 12

= _4__1_2_

12

= _4__4___3_

12

= _8__3_

12

= _2__3_

3

58. __1__ ? __5_

25 5

= __5_

10

59. ___6_

12

60. _7__1_4_ ? __5_

5 5

= __2___3_

4 3

= _7__7_0_

5

= __2_

2

61. _52_x

5x x

1

=

_1_

2

+

3x

2 = 1 + 6x

2 = 1

x=3

x= 3

62. 6(7n + 2) = (34 + 11n)3 42n + 12 = 102 + 33n 9n + 12 = 102 9n = 90 n = 10

63. _4_5_ = _9_0_0_

2x

45x = 1800

x = 40 h

It would take 40 h to package 900 lb of cheese.

3-2 USING ALGEBRAIC METHODS TO SOLVE LINEAR SYSTEMS, PAGES 190-197

CHECK IT OUT!

1a.

y = 2x 3x + 2y

1 = 26

Step 1 Solve one equation for one variable. The first equation is already solved for y: y = 2x 1.

Step 2 Substitute the expression into the other equation. 3x + 2y = 26

3x + 2(2x 1) = 26 3x + 4x 2 = 26 7x 2 = 26 7x = 28 x=4

Step 3 Substitute the x-value into one of the original equations to solve for y. y = 2x 1

y = 2(4) 1 y=7 The solution is the ordered pair (4, 7).

2a. Step 1 Find the value

of one variable.

4x + 7y = 25

12x7y=19

8x

=6

x x

= =

_6_ _83_

4

Step 2 Substitute the

x-value into one of the

original equations to

solve for y.

( )4 _3_ 4

+ 7y =

25

3 + 7y = 25

7y = 28

y= 4

( ) The solution is the

ordered pair _3_, 4 .

4

b.

5x + y=

6y = 2x +

9 2

Step 1 Solve one

equation for one

variable. The first

equation is already

solved for y: y = ?2x + 2.

Step 2 Substitute the

expression into the

other equation. 5x + 6y = ?9

5x + 6( 2x + 2) = 9 5x 12x + 12 = 9 7x + 12 = 9 7x = 21 x=3

Step 3 Substitute the

x-value into one of the

original equations to

solve for y. y = 2x + 2

y = 2(3) + 2 y= 4

The solution is the

ordered pair (3, 4).

b. Step 1 To eliminate y,

multiply both sides of

the first equation by 5

and both sides of the

second equation by 3.

5(5x 3y) = 5(42)

3(8x + 5y) = 3(28)

Add to eliminate y.

25x 15y = 210

_2_4_x_+__1_5_y_=___8_4_

49x

= 294

x=6

Step 2 Substitute the

x-value into one of the

original equations to

solve for y. 5(6) 3y = 42

30 3y = 42

3y = 12

y= 4

The solution is the

ordered pair (6, 4).

88

Holt McDougal Algebra 2

3a. 56x + 8y = 32 b. 6x + 3y = 12

_5_6_x_ + _8_y_ = _3_2_

88

8

7x + y = 4

7x + y = 4

2x + y = 6 y = 2x 6

6x + 3( 2x 6) = 12 6x 6x 18 = 12

Since = , the

18 = 12

system is consistent, Since 18 12, the

dependent, and has

system is inconsistent and

an infinite number

has no solution.

of solutions.

4. Let x represent amount of Sumatra beans.

Let y represent amount of Kona beans.

x + y = 50

From y = 50 x

5x + 13y = 10(50)

y = 50 18.75

5x + 13y = 500

y = 31.25 lb

5x + 13(50 x) = 500

5x + 650 13x = 500

8x = 150

x = 18.75 lb

There is 18.75 lb of Sumatra beans and 31.25 lb of

Kona beans.

THINK AND DISCUSS

1. Possible answer: elimination, because the y-terms have opposite coefficients.

2.

'RAPHING yx

yx

xy

y

x

3OLVING,INEAR3YSTEMS

3UBSTITUTION yx xy xx  xx x xy

%LIMINATION yx yx

y

yx

EXERCISES

GUIDED PRACTICE

1. elimination

2. Step 1 Solve one equation for one variable. The second equation is already solved for y: y = x + 7. Step 2 Substitute the expression into the other equation. x + y = 17 x + (x + 7) = 17 2x + 7 = 17 2x = 10 x=5 Step 3 Substitute the x-value into one of the original equations to solve for y. y=x+7 y=5+7 y = 12 The solution is the ordered pair (5, 12).

3. Step 1 Solve one equation for one variable. The first equation is already solved for y: y = x 19. Step 2 Substitute the expression into the other equation. 2x ? y = 27 2x (x 19) = 27 2x x + 19 = 27 x=8 Step 3 Substitute the x-value into one of the original equations to solve for y. y = x ? 19 y = 8 19 y = 11 The solution is the ordered pair (8, 11).

4. Step 1 Solve one

5. Step 1 Solve one

equation for one

equation for one

variable. 2x ? y = 2

variable. The second

y = 2x 2

equation is already

Step 2 Substitute the

solved for x:

expression into the

x = 3y 5.

other equation.

Step 2 Substitute the

3x 2y = 11

expression into the

3x 2(2x 2) = 11

other equation.

3x 4x + 4 = 11

y = 3x + 5

x=7

y = 3( 3y 5) + 5

x= 7

y = 9y 15 + 5

Step 3 Substitute the

10y = 10

x-value into one of the

y= 1

original equations to

Step 3 Substitute the

solve for y.

y-value into one of the

y = 2x ? 2

original equations to

y = 2( 7) 2

solve for x.

y = 14 2

y = 3x + 5

y = 16

1 = 3x + 5

The solution is the

6 = 3x

ordered pair ( 7, 16).

2=x

The solution is the

ordered pair ( 2, 1).

89

Holt McDougal Algebra 2

6. Step 1 Find the value of one variable. Add to eliminate y. 2x + y = 12 5x y = 33 3x = 21 x=7 Step 2 Substitute the x-value into one of the original equations to solve for y. 2(7) + y = 12 y= 2 The solution is the ordered pair (7, 2).

7. Step 1 Find the value of one variable. Add to eliminate x. 2x 5y = 5 2x + 8y = 58 3y = 63 y = 21 Step 2 Substitute the y-value into one of the original equations to solve for x. 2x 5( 21) = 5 2x = 110 x = 55 The solution is the ordered pair ( 55, 21)

8. Step 1 To eliminate y,

multiply both sides of

the second equation by

2. 2(5x 3y) = 2(88)

Add to eliminate y.

2x + 6y = 8

_1_0_x___6_y_=__1_7_6_

12x

= 168

x = 14

Step 2 Substitute the

x-value into one of the

original equations to

solve for y.

2(14) + 6y = 8

6y = 36

y= 6

The solution is the

ordered pair (14, 6).

9. Step 1 To eliminate x, multiply both sides of the first equation by 4.

( ) 4 _21_x + y = 4(4)

Add to eliminate x. 2x + 4y = 16

__2_x____2_y_=___6_ 2y = 10 y=5

Step 2 Substitute the y-value into one of the original equations to solve for x.

2x 2(5) = 6 2x = 4 x= 2

The solution is the ordered pair ( 2, 5).

10. 5x y = 3

11. x = 2y 8

y = 5x + 3

4(2y 8) = 8y 56

15x 3(5x + 3) = 9

8y ? 2 32 = 8y 56

15x 15x 9 = 9

32 = 56

9= 9

inconsistent, no solution

consistent, dependent,

infinite number of

solutions

( ) 12. 2x + 3y = 24

13. x

3y = 2x 24

y= 8x + 12

_2_ _2 3x

3x

8 8 = 60

6x

_1_ _13_y

3y

= =

2 x

2

y = 3x + 6

2(3x + 6) =

12

( ) 8x + _234_x 96 = 60

6x 6x 12 = 12 12 = 12

consistent, dependent,

8x 8x 96 = 60 infinite number of

96 = 60 solutions

inconsistent, no solution

14. Let x be the amount of 85% ethanol, and y be the

amount of 25% ethanol.

x + y = 20

0.85x + 0.25y = 20(0.5)

x + y = 20

0.85x + 0.25y = 10

x + y = 20

y = 20 x

0.85x + 0.25(20 x) = 10

0.85x + 5 0.25x = 10

0.6x = 5

( ) y = 20

y

=

11 _2_

3

8 _1_

3

Denise needs

8 _1_

x gal

= of

8 _1_

3 85%

ethanol

fuel

and

11 _2_

gal

of

25%

3 ethanol

fuel.

3

PRACTICE AND PROBLEM SOLVING

15. Step 1 Solve one 16. Step 1 Solve one equation

equation for one

for one variable.

variable. The first

12x + y = 21

equation is already y = 12x + 21

solved for x:

Step 2 Substitute the

x = ?4y.

expression into the

Step 2 Substitute

other equation.

the expression into

18x ? 3y = ?36

the other equation.

18x 3( 12x + 21) = 36

2x + 6y = ?3

18x + 36x 63 = 36

2( 4y) + 6y = 3

8y + 6y = 3

2y = 3

y

=

_3_

2

Step 3 Substitute

the y-value into

one of the original

equations to solve

for x.

x = ?4y

( ) x =

4

_3_

2

=6

The solution is the

( ) ordered pair 6, _3_ . 2

54x = 27

x

=

_1_

2

Step 3 Substitute the

x-value into one of the original equations to

solve for y.

y = ?12x + 21

( ) y =

12

_1_

2

+ 21

y = 6 + 21

= 15

The solution is the ordered

( ) pair _1_, 15 . 2

90

Holt McDougal Algebra 2

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