Section 4.2 Solving Systems of Linear Equations by the Substitution Method
39.
0.7 0.1 0.6
0.8 0.3 0.8
xy
xy


Multiply both sides of both equations by 10.
76
83 8
xy
xy


Solve the first equation for y.
76
xy

40.
1.25 0.01 4.5
0.5 0.02 1
xy
xy


Multiply both sides of both equations by 100.
125 1 450
50 2 100
xy
xy


Solve the first equation for y.
125 1 450
xy

Back-substitute.
125 450 125 4 450
500 450 50
yx 

The solution set is
4,50 .
41. a. Substitute
0.001 3x
for p in the first equation.
0.002 6
0.001 3 0.002 6
0.001 3 0.002 6
0.002 0.001 3 0.002 0.002 6
p
px
xx
xx
xx xx
 
 
 


b. If unleaded premium gasoline is sold for $4 per
gallon, there will be a demand for 1000 gallons
per day and 1000 gallons will be supplied per
day.
42. a. Substitute
0.5 5x
for y in the first equation.
0.5 105
0.5 5 0.5 105
y
yx
xx
 
 

$0.50
b. If cookies are sold for $0.50 per package, there
will be a demand for 110 million packages and
bakers will supply 110 million packages per
Chapter 4 Systems of Linear Equations
49. does not make sense; Explanations will vary.
Sample explanation: Solving for x in the second
equation will allow us to avoid fractions.
53. true
54. false; Changes to make the statement true will vary.
A sample change is: The graphing method allows us
to visualize solutions.
56.
3
26
311
xyz
xyz
xyz



First substitute 3yzfor x in the second
equation.
26
xyz

Now substitute 3yz in the third equation of the
given system.
311
xyz

Substitute
12 3z
in the last equation.
94 2 11
9 4 12 3 2 11
yz
zz


33351xyz
Thus, x = 1, y = −3, and z = 5.
57.
3
52 7
ymx
xy


257
57
22
yx
yx


The system will be inconsistent if the graphs of the
two equations have the same slope and different y-
intercepts. The y-intercepts are different. Therefore,
the system will be inconsistent if
5
m
.
4612
4(0) 6 12
612
xy
y
y


Section 4.3 Solving Systems of Linear Equations by the Addition Method
Draw a line through (3,0), (0, 2), and (−3, 4).
59.
412533
442539
xx
xx
 
  
60. The integers in the given set are −73, 0, and
33.
1
61.
32 48
32(12)48
32448
xy
x
x



62.
14 168
14 168
14 14
12
y
y
y



4.3 Check Points
1.
5
9
xy
xy


Now solve for x.
214
x
2
y

The solution set is
7, 2 .
2.
422
34 26
xy
xy


2
y
The solution set is
6, 2 .
Chapter 4 Systems of Linear Equations
3.
45 3
23 7
xy
xy


Multiply each term of the second equation by –2
and add the equations to eliminate x.
45 3
xy

4.
293
483
xy
yx


Rewrite each equation in the form .
A
xByC
23 9
xy

Back-substitution of
60
17
to find y would cause
cumbersome arithmetic.
Instead, use the system that is in the form
A
xByC
to eliminate x and find y.
23 9
xy

5.
24
36 13
xy
xy


Multiply the first equation by −3.
36 12
xy
 
The true statement indicates that the system has
infinitely many solutions.
The solution set is
,57xy x y
or
, 3 15 21 .xy x y
4.3 Exercise Set
1.
3
11
xy
xy


Add the equations to eliminate the y-terms.
Section 4.3 Solving Systems of Linear Equations by the Addition Method
2.
6
2
xy
xy


Add the equations to eliminate the y-terms.
6xy
4
y
The solution set is
2, 4 .
3.
23 6xy
23 6xy
412
3
x
x
4.
32 14xy
32 10xy
624
4
x
x
Back-substitute into either of the original equations
to solve for y.
5. 27xy
318xy
525
y
6. 2 2xy
23 6xy 
28
4
y
y

Back-substitute into either of the original equations
to solve for x.
2 2
xy

to solve for x.
5114
515
3
x
x
x

The solution set is
3,1 .
Chapter 4 Systems of Linear Equations
9.
3 7
25 1
xy
xy


Multiply each term of the first equation by 5 and
add the equations to eliminate y.
10. 311
2513
xy
xy


Multiply first equation by 5 and add.
15 5 55
2513
xy
xy


x
11. 34
45 2
xy
xy


Multiply each term of the first equation by −4 and
add the equations to eliminate x.
Back-substitute into either of the original equations
to solve for x.
34
32 4
xy
x


to solve for x.
21
22 1
41
3
xy
x
x
x



The solution set is
3, 2 .
to solve for y.
213
27 13
14 13
xy
y
y



Section 4.3 Solving Systems of Linear Equations by the Addition Method
14. 25 1
37
xy
xy


Multiply the second equation by 5 and add.
25 1
xy

15. 314 6
5710
xy
xy


Multiply each term of the second equation by 2 and
add the equations to eliminate y.
314 6
xy

The solution set is
2,0 .
16. 54 19
32 7
xy
xy


Multiply the second equation by 2 and add.
Back-substitute into either of the original equations
to solve for y.
32 7
33 2 7
xy
y


17x 17
x = 1
Back-substitute into either of the original equations
to solve for y.
23 4
xy

35 140
4
y
y


Back-substitute into either of the original equations
to solve for x.
Chapter 4 Systems of Linear Equations
19. 32 1
27 9
xy
xy

 
Multiply the first equation by 2 and the second
equation by 3.
64 2xy
A
20. 53 27
7213
xy
xy


Multiply the first equation by 2.
Multiply the second equation by 3.
10 6 54xy
21. 327
5213
xy
xy


Rewrite each equation in the form .
A
xByC
32 7
52 13
xy
xy


Back-substitute into either of the original equations
to solve for y.
327
33 2 7
xy
y


9 25xy
92 4xy 
321
7
y
y


Back-substitute into either of the original equations
612 6
xy
 
Rewrite the first equation in the form .
A
xByC
23 4xy
Multiply the first equation by 3 and add to
eliminate x.
69 12xy
612 6xy
Section 4.3 Solving Systems of Linear Equations by the Addition Method
24. 548
37 14
xy
xy


Rewrite the first equation in the form .
A
xByC
54 8
37 14
xy
xy


25. 23
44 1
xy
xy


Multiply the first equation by 4 and add to
eliminate y.
84 12xy
A
6
The solution set is
11 7
,.
12 6






26.
322
45 21
xy
xy


Multiply the first equation by 5.
15 5 110xy
45 21xy
The solution set is
89 151
,.
19 19






27.
452
23 4
xy
xy


Rewrite the first equation in the form
16 23
23
16
x
x

The solution set is
23 3
,.
16 8






Chapter 4 Systems of Linear Equations
28.
341
431
xy
xy


Rewrite the first equation in the form
,
A
xByC
and multiply the second equation
by −3.
12 16 4
A
xy

34 1
431
xy
xy


To eliminate y, multiply the first of these equations
by 3 and the second by 4.
9 12 3
xy

29.
31
32
xy
xy


Multiply the first equation by −1.
31
xy

30.
49 2
49 2
0 4, false
xy
xy

 

31. 32xy
39 6xy
Multiply the first equation by −3.
39 6xy 
39 6xy
42 2
42 2
0 0, true
xy
xy

 
The true statement indicates that the system has
14 6 8xy
14 6 7xy
0 = 1, false
The false statement indicates that the system is
Multiply the second equation by −2.
612 15
612 12
0 3, false
xy
xy

 
35.
52
31
xy
xy


Multiply the second equation by −1.
52xy
31xy
36.
25 1
21
xy
xy


Multiply the first equation by −1.
251
2 1
xy
xy
 


37.
53
26 10
xy
xy


The true statement indicates that the system has
infinitely many solutions.
The solution set is
,53xy x y
or
,26 10.xy x y
The solution set is
,4 368xy x y
or
,36 27.xy x y
39.
43 0
3310
xy
xy


Rewrite both equations.
3
The solution set is
1,1 .
3






Chapter 4 Systems of Linear Equations
40.
22 3 0
73232
xy
xy


Rewrite each equation in the form .
A
xByC
Now solve the system formed by the two rewritten
equations.
46 0
76 11
xy
xy


41.
11
1
57
xy
xy


Multiply the second equation by the LCD, 35, to
clear fractions.
35 35 1
57
75 35
xy
xy





42.
3
1
97
xy
xy


Now solve the system
3
79 63
xy
xy


Multiply the first equation by −7 and add the result
to the second equation.
43.
5
21
5
xy
xy


Multiply both equations by 5 to clear fractions.
45 5xy
25 5
xy

60
0
x
x
Section 4.3 Solving Systems of Linear Equations by the Addition Method
44.
3
3
1
24
xy
xy


To clear fractions, multiply the first equation by 3
and the second equation by 4 to obtain the system.
39
xy

45.
32 8
2
xy
xy


The substitution method is a good choice because
the second equation is already solved for x.
Substitute −2y for x in the first equation
46.
210
3
xy
yx

The substitution method is a good choice because
Back-substitute −10 for x into the second equation
of the system.
3
310 30
yx
y
 
The solution set is
10, 30 .
47.
32 3
xy

33
1
x
x
The solution set is
1, 3 .
48.
27 17
xy

2717
27117
2717
xy
x
x



49.
32 6
3
xy
y

The substitution method is a good choice because
the second equation is already solved for y.
Substitute 3 for y in the first equation.
32 6
xy

50.
23 7
2
xy
x

The substitution method is a good choice because
the second equation is already solved for x.
Substitute 2 for x in the first equation.
23 7
xy

51.
21
23
yx
yx


The substitution method is a good choice, because
both equations are already solved for y. Substitute
A
52.
24
21
yx
yx


The substitution method is a good choice, because
both equations are already solved for y.
Substitute
24
x
for y in the second equation.
2421
xx
 
The addition method is a good choice since the
equations can easily be simplified to give equations
of the form .
A
xByC
24 6
xy

36 90
36 9
xy
xy


Solve the resulting system.
The solution set is
,2 2 6
xy x y

or
,3 2 30.
xy x y

54.
241
xy x

Section 4.3 Solving Systems of Linear Equations by the Addition Method
Now solve the system formed by the two rewritten
equations.
22 1
xy
 
2


55.
32
29 24
yx
xy

The substitution method is a good choice because
56.
45
58 20
yx
xy


The addition method is a good choice because the
Multiply the second equation by −1, and the result
to the rewritten equation.
54 0
xy

57.
514
44
xy
xy


Add the equations.
514
xy

5
x
Back-substitute into the first equation of the original
system.
32 11
3(5) 2 11
xy
y


Chapter 4 Systems of Linear Equations
59.
43 0
7
xy
xy


Multiply the second equation by 3 and then add the
equations.
60.
36 15
2
xy
xy


Multiply the second equation by 6 and then add the
equations.
36 15
xy

61.
341
55
3
1
48
xy
xy


Multiply the first equation by 5 and the second
Multiply the first equation by 3 and the second
equation by 4.
912 15
812 32
xy
xy


62.
2
32 3
24
33
xy
xy


2
x
Back-substitute 2 for x in the equation and solve for
y.
23 4
22 3 4
xy
y


Section 4.3 Solving Systems of Linear Equations by the Addition Method
63.
517 17
61551
xy
xy
 
 
Simplify both equations.
51717
557 77
xy
xy
 
 
57 5
65 6
xy
xy


Multiply the first equation by −6 and the second
equation by 5, and solve by addition.
30 42 30
30 25 30
17 0
xy
xy
y
 

64.
65 3
3451
xxy x
xy y y

 
Simplify both equations.
65 3
xxy x

Solve the rewritten system.
2515
34 5
xy
xy


Multiply the first equation by -3 and the second
equation by 2, and solve by addition.
615 45
xy
 
65. 0.4 2.2
0.5 1.2 0.3
xy
xy


Multiply the first equation by 1.2 and solve by
addition.
0.48 1.2 2.64
0.50 1.2 0.30
xy
xy


66. 1.25 1.5 2
3.5 1.75 10.5
xy
xy


Multiply the first equation by –2.8 and solve by
addition.
Chapter 4 Systems of Linear Equations
Back-substitute 2 for y to find x.
1.25 1.5 2
xy

67. 8
23
211
23
xy
xy

Simplify the first equation.
8
23
xy
When simplified, the equations are the same. This
means that the system is dependent and there are an
infinite number of solutions.
68. 8
24
35
24
xy
xy

Simplify the first equation.
Simplify the second equation.
35
21
21
xy
xy
yx



When simplified, the both equations have the same
slope, but different y-intercepts. This means that
the system is inconsistent and there are no solutions.
The solution set is
.
69. 0.45 0.8
xy

2.7 0.8
3.5
y
y

The solution is (6, 3.5). This means that in week 6,
b. In 2015 both will receive 50%.
c. 2112
288
xy
xy

 
Add the equations.
Section 4.3 Solving Systems of Linear Equations by the Addition Method
Back-substitute 50 for y and solve for x.
2112
xy

d. reasonably well
71. – 75. Answers will vary.
76. makes sense
77. does not make sense; Explanations will vary.
Sample explanation: The addition method does not
involve graphing.
81. false; Changes to make the statement true will vary.
A sample change is: If (2, 2) satisfies
22Ax y
and 210,xBythen A and B can be
found by substitution.
Find A.
22
Ax y

83. false; Changes to make the statement true will vary.
A sample change is: After these multiplications, the
2
xxba
xb a
xab
xab

 

 
Back-substitute ab for x to find y.
2
2
yxb
yabb


2
A
37
35 2 7
15 2 7
28
xBy
B
B
B




Chapter 4 Systems of Linear Equations
88. 29,700 150 5000 1100
29,700 950 5000
950 24,700
26
xx
x
x
x



The solution set is {26}.
89. a. 28
6
xy
xy


Mid-Chapter Check Point – Chapter 4
1.
32 6
2 4
xy
xy


x
y
-4 -2 2 4
-4
-2
2
4
0
The lines intersect at (2, 0).
The solution set is
2,0 .
3. 21
6312
yx
xy


x
y
-4 -2 2 4
-4
-2
2
4
0
373571578yx 
The solution set is
5,8 .
5.
65 7
37 13
xy
xy


Multiply the second equation by −2 and add to
eliminate x.
6 5 7
614 26
19 19
1
xy
xy
y
y

 


Back-substitute −1 for y and solve for x.
65 7
xy
