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Chapter 5
Chapter 5 Maintaining Mathematical Proficiency (p. 233)
y+4=x
1.
−2y + 12 = −3x
4.
−2y + 12 − 12 = −3x − 12
y+4−4=x−4
−2y = −3x − 12
y=x−4
−3x − 12
−2
3
y = —x + 6
2
−2y
−2
—=—
m = 1, b = −4
2
−4
y
−2
2
4
x
3
m = —, b = 6
2
−2
6
y
4
−6
2
6x − y = −1
2.
−6
−2
2 x
6x − 6x − y = −1 − 6x
−2
−y = −6x − 1
−y −6x − 1
−1
−1
y = 6x + 1
—=—
m = 6, b = 1
5. m + 4 >
−4
9
−4
m>5
The solution is m > 5.
y
8
5
0
4
2
4
6
8
6. 24 ≤ −6t
−4
−2
2
4 x
24 −6t
−6 −6
−4 ≥ t
—≥—
The solution is t ≤ −4.
4x + 5y = 20
3.
4x − 4x + 5y = 20 − 4x
5y = −4x + 20
−4x + 20
5
4
y = −—x + 4
5
5y
5
−8
−4
7. 2a − 5 ≤
+5
—=—
4
m = −—, b = 4
5
−6
0
13
+5
2a ≤ 18
2a 18
2
2
a≤9
—≤—
The solution is a ≤ 9.
y
9
0
2
−2
−2
4
8
12
16
8. −5z + 1 < −14
2
4
−1
x
−1
−5z < −15
−2
−5z −15
—>—
−5
−5
z>3
−4
The solution is z > 3.
3
0
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2
4
6
8
Algebra 1
Worked-Out Solutions
227
Chapter 5
9. 4k − 16 <
−k
k+2
3k − 16 <
+ 16
3x − 2y = 2
2x − y = 2
3x − 3x − 2y = 2 − 3x
2x − 2x − y = 2 − 2x
3.
−k
2y = −3x + 2
2
+ 16
3k <
−y = −2x + 2
−y
−1
−3x + 2
−2
3
y = —x − 1
2
−2y
−2
3k 18
3
3
k<6
—<—
−2x + 2
−1
—=—
—=—
18
y = 2x − 2
4
The solution is k < 6.
−6
0
2
4
6
7w + 12 ≥
10.
− 2w
8
2w − 3
− 2w
5w + 12 ≥
− 12
−3
Intersection
X=2
Y=2
−4
The point of intersection is (2, 2).
Section 5.1
− 12
5.1 Explorations (p. 235)
5w ≥ −15
1. a. An equation that represents the costs is C = 15x + 600.
5w −15
—≥—
5
5
w ≥ −3
b. An equation that represents the revenue is R = 75x.
c. A system of linear equations for this problem is
C = 15x + 600
The solution is w ≥ −3.
R = 75x.
−3
−6
−4
−2
0
2. a.
11. The lines intersect at the point (a, b).
Mathematical Practices (p. 234)
1.
6
x (nights)
0
1
2
3
4
5
6
C (dollars) 600
615
630
645
660
675
690
R (dollars)
0
75
150
225
300
375
450
x (nights)
7
8
9
10
11
C (dollars) 705
720
735
750
765
R (dollars) 525
600
675
750
825
2
−6
6
Intersection
X=0
Y=-3
−6
The point of intersection is (0, −3).
2.
b. Your family must rent the bedroom for 10 nights before
breaking even.
y
c.
4
y = 15x + 600
6
Intersection
X=1.5
Y=-.5
−4
The point of intersection is (1.5, −0.5).
Cost/Revenue
800
−6
600
400
y = 75x
200
0
0
2
4
6
8
10
12 x
Days
d. The point of intersection is (10, 750). This point
represents where both functions have the same x- and
y-values. This is the same as the break-even point.
3. Graph both equations and find the point of intersection.
To check your solution, substitute the x-coordinate for x
in both of the equations and verify that both results are the
y-coordinate of the point of intersection.
228
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Worked-Out Solutions
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Chapter 5
4. a.
x
−3
−2
−1
0
1
y = −4.3x − 1.3
11.6
7.3
3
−1.3
−5.6
y = 1.7x + 4.7
−0.4
1.3
3
4.7
6.4
5.1 Monitoring Progress (pp. 236–238)
1. Equation 1
The solution is (−1, 3).
Sample answer: A table was chosen because the slopes
and intercepts are decimals and would be difficult to graph
accurately.
Check
6
8
2. Equation 1
y
y = −3x + 8
6
y=x
(2, 2)
−2
2
y = 3x + 1
?
4 = 3(1) + 1
?
4=3+1
y = −x + 5
?
4 = −(1) + 5
?
4 = −1 + 5
4=4✓
4=4✓
y
3.
2
The solution is (2, 2).
(3, 1)
Sample answer: The graphing method was chosen
because both equations are in slope-intercept form with
slopes and y-intercepts that are whole numbers.
−2
y = −x + 4
Check Equation 1
6
Intersection
X=2
Y=2
−4
y
y = −x − 1
4.
y = −x + 4
?
1 = −3 + 4
1=1✓
1=1✓
1
y = 2x + 3 4
2
(−1.5, 0.5)
−4
−6
Sample answer: The graphing method was chosen
because the equations are in slope-intercept form and the
slopes and y-intercepts are whole numbers.
4
6
Intersection
X=-1.5
Y=.5
−4
−2
x
−2
The solution is (−1.5, 0.5).
−6
−4
y
2
(−4, 1)
2 x
−2
Check
Equation 2
y=x−2
?
1=3−2
The solution is (3, 1).
y = 3x + 5
−6
6 x
4
2
−4
4
−6
c.
y=x−2
6 x
4
Check
Equation 2
Because the ordered pair (1, 4) is a solution of each equation,
it is a solution of the linear system.
4
2
−5 ≠ 5 ✗
The ordered pair (1, −2) is a solution of the first equation,
but it is not a solution of the second equation. So, (1, −2)
is not a solution of the linear system.
Intersection
X=-1
Y=3
−4
b.
−x + 2y = 5
?
−(1) + 2(−2) = 5
?
−1 − 4 = 5
0=0✓
4
−6
Equation 2
2x + y = 0
?
2(1) + (−2) = 0
?
2−2=0
3
y = −2 x − 5
Check Equation 1
Equation 2
3
y = —12 x + 3
y = −—2 x − 5
?
1 = —12 (−4) + 3
? 3
1 = −—2 (−4) − 5
?
1 = −2 + 3
?
1=6−5
1=1✓
1=1✓
The solution is (−4, 1).
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Algebra 1
Worked-Out Solutions
229
Chapter 5
2. The one that is different is “Solve each equation for y.” When
2x + y = 5
3x − 2y = 4
2x − 2x + y = 5 − 2x
3x − 3x − 2y = 4 − 3x
5.
y = −2x + 5
−2y = −3x + 4
−3x + 4
−2
3
y = —x − 2
2
−2y
−2
y
—=—
3
y = 2x − 2
2
(2, 1)
−2
4
2
you solve both equations for y, you get y = 2x + 2 and
y = 4x + 6. The other three ask for the solution of the
system, which is (−2, −2).
−4x + 2y = 4
4x − y = −6
−4x + 4x + 2y = 4 + 4x
4x − 4x − y = −6 − 4x
2y = 4x + 4
−y = −4x − 6
—=—
4x + 4
2
—=—
y = 2x + 2
y = 4x + 6
2y
2
6 x
y = −2x + 5
−2
−4
Check Equation 1
2x + y = 5
?
2(2) + 1 = 5
?
4+1=5
2
3x − 2y = 4
?
3(2) − 2(1) = 4
?
6−2=4
5=5✓
−6
18
Number of
science
exercises
+
6
y=6+x
y + x = 18
y + x − x = 18 − x
y = −x + 18
y + x = 18
?
12 + 6 = 18
18 = 18 ✓
4x − y = −6
?
4(−2) − (−2) = −6
?
−8 + 2 = −6
3. Equation 1
x+y=8
?
2+6=8
8=8✓
y=x+6
−6 = −6 ✓
Equation 2
3x − y = 0
?
3(2) − 6 = 0
?
6−6=0
0=0✓
8
Because the ordered pair (2, 6) is a solution of each equation,
it is a solution of the linear system.
4
Check Equation 1
Equation 2
−4x + 2y = 4
?
−4(−2) + 2(−2) = 4
?
8−4=4
Monitoring Progress and Modeling with Mathematics
16
0
Check Equation 1
4=4✓
y
12 (6, 12)
2 x
−2
−4
Variables Let x be the number of science exercises, and let y
be the number of math exercises.
System y + x = 18
−2
4=4✓
Number
Number
of math + of science =
exercises
exercises
Number
of math =
exercises
−4
(−2, −2)
y = 2x + 2
The solution is (2, 1).
6. Words
−4x − 6
−1
y
y = 4x + 6
Equation 2
−y
−1
y = −x + 18
0
4
8
12
16
x
Equation 2
y=6+x
?
12 = 6 + 6
12 = 12 ✓
The solution is (6, 12). So, you have 6 exercises in science
and 12 exercises in math.
4. Equation 1
x−y=6
?
8−2=6
6=6✓
Equation 2
2x − 10y = 4
?
2(8) − 10(2) = 4
?
16 − 20 = 4
−4 ≠ 4 ✗
The ordered pair (8, 2) is a solution of the first equation, but
it is not a solution of the second equation. So, (8, 2) is not a
solution of the linear system.
5.1 Exercises (pp. 239 –240)
Vocabulary and Core Concept Check
1. yes; Because 5y − 2x = 18 and 6x = −4y − 10 are both
linear equations in the same variables, they form a system of
linear equations.
230
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Chapter 5
5. Equation 1
10. The lines appear to intersect at (3, 2).
Equation 2
y = −7x − 4
?
3 = −7(−1) − 4
?
3=7−4
y = 8x + 5
?
3 = 8(−1) + 5
?
3 = −8 + 5
3=3✓
3 ≠ −3 ✗
Check Equation 1
−4 = −4 ✓
The solution is (3, 2).
11. The lines appear to intersect at (−4, 5).
Equation 2
y = 2x + 6
?
−2 = 2(−4) + 6
?
−2 = −8 + 6
y = −3x − 14
?
−2 = −3(−4) − 14
?
−2 = 12 − 14
−2 = −2 ✓
−2 = −2 ✓
Check Equation 1
18 = 18 ✓
Check Equation 1
2x − 4y = −8
?
2(−2) − 4(1) = −8
−4 − 4 = −8 ✓
4x + y = 14
?
4(5) + (−6) = 14
?
20 − 6 = 14 ✓
12 = 12 ✓
14 = 14 ✓
Because the ordered pair (5, −6) is a solution of each
equation, it is a solution of the linear system.
9. The lines appear to intersect at (1, −3).
Check Equation 1
x−y=4
?
1 − (−3) = 4
?
1+3=4
Equation 2
4x + y = 1
?
4(1) + (−3) = 1
?
4−3=1
4=4✓
2x + 4y = 8
?
2(0) + 4(2) = 8
?
0+8=8
−2 = −2 ✓
8=8✓
The solution is (0, 2).
y
13.
Equation 2
Equation 2
2x − y = −2
?
2(0) − 2 = −2
?
0 − 2 = −2
−7 = −7 ✓
6x + 3y = 12
?
6(5) + 3(−6) = 12
?
30 − 18 = 12
24 = 24 ✓
The solution is (−4, 5).
Because the ordered pair (−2, 1) is a solution of each
equation, it is a solution of the linear system.
8. Equation 1
−x + 4y = 24
?
−(−4) + 4(5) = 24
?
4 + 20 = 24
12. The lines appear to intersect at (0, 2).
Equation 2
6x + 5y = −7
?
6(−2) + 5(1) = −7
?
−12 + 5 = −7
Equation 2
6y + 3x = 18
?
6(5) + 3(−4) = 18
?
30 − 12 = 18
Because the ordered pair (−4, −2) is a solution of each
equation, it is a solution of the linear system.
7. Equation 1
y − 2x = −4
?
2 − 2(3) = −4
?
2 − 6 = −4
5=5✓
The ordered pair (−1, 3) is a solution of the first equation,
but it is not a solution of the second equation. So, (−1, 3) is
not a solution of the linear system.
6. Equation 1
Equation 2
x+y=5
?
3+2=5
y = −x + 7
6
4
(3, 4)
2
y=x+1
2
4
Check Equation 1
6 x
Equation 2
y = −x + 7
?
4 = −3 + 7
y=x+1
?
4=3+1
4=4✓
4=4✓
The solution is (3, 4).
1=1✓
The solution is (1, −3).
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Algebra 1
Worked-Out Solutions
231
Chapter 5
14.
8
y
9x + 3y = −3
2x − y = −4
9x − 9x + 3y = −3 − 9x
2x − 2x − y = −4 − 2x
3y = −9x − 3
−y = −2x − 4
—=—
−9x − 3
3
—=—
y = −3x − 1
y = 2x + 4
17.
y = 2x − 8
(4, 0)
−4
3y
3
12 x
8
−4
y = −x + 4
−y
−1
−2x − 4
−1
y
Check Equation 1
Equation 2
y = −x + 4
?
0 = −4 + 4
4
y = 2x − 8
?
0 = 2(4) − 8
?
0=8−8
0=0✓
y = 2x + 4
(−1, 2) 2
−4
4 x
2
0=0✓
−2
y = −3x − 1
The solution is (4, 0).
Check Equation 1
y
15.
2
y = 3x + 5
(−9, −1)
−12
−4
x
9x + 3y = −3
2x − y = −4
?
9(−1) + 3(2) = −3
?
−9 + 6 = −3
?
2(−1) − 2 = −4
?
−2 − 2 = −4
−3 = −3 ✓
1
y = 3x + 2
Check Equation 1
Equation 2
y = —13x + 2
y = —23 x + 5
?
−1 = —13(−9) + 2
?
−1 = −3 + 2
?
−1 = —23 (−9) + 5
?
−1 = −6 + 5
−1 = −1 ✓
−1 = −1 ✓
The solution is (−9, −1).
y
18.
4x − 4y = 20
4x − 4x − 4y = 20 − 4x
−4y = −4x + 20
−4y −4x + 20
−4
−4
y=x−5
—=—
y
−2
x
2
−2
1
y = − 2 x + 11
−4
8
y=x−5
(0, −5)
3
4
−4 = −4 ✓
The solution is (−1, 2).
−4
16.
Equation 2
y = 4x − 4
y = −5
−8
(12, 5)
8
12
x
Check Equation 1
Equation 2
4x − 4y = 20
Check Equation 1
Equation 2
1
y = —34 x − 4
y = −—2 x + 11
?
5 = —34(12) − 4
?
5=9−4
? 1
5 = −—2 (12) + 11
?
5 = −6 + 11
5=5✓
5=5✓
?
4(0) − 4(−5) = 20
?
0 + 20 = 20
y = −5
−5 = −5 ✓
20 = 20 ✓
The solution is (0, −5).
The solution is (12, 5).
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Chapter 5
19.
x − 4y = −4
−3x − 4y = 12
x − x − 4y = −4 − x
−3x + 3x − 4y = 12 + 3x
−4y = −x − 4
−4y = 3x + 12
−x − 4
−4
1
y = —x + 1
4
−4y
−4
3x + 12
−4
3
y = −—x − 3
4
−4y
−4
—=—
not −3.
x − 3y = 6
2x − 3y = 3
x − x − 3y = 6 − x
2x − 2x − 3y = 3 − 2x
−3y = −x + 6
−3y = −2x + 3
−3y
−3
−2x + 3
−3
2
y=—x−1
3
−x + 6
−3
1
y=—x−2
3
−3y
−3
—=—
—=—
y
4
1
y = 4x + 1
—=—
21. The graph of 2x − 3y = 3 should have a y-intercept of −1,
y
2
2
y = 3x − 1
2
−2
x
(−4, 0)
−6
−4
−2
2 x
3
y = −4 x − 3
−4
1
Check Equation 1
Equation 2
x − 4y = −4
−3x − 4y = 12
?
−4 − 4(0) = −4
?
−4 − 0 = −4
?
−3(−4) − 4(0) = 12
?
12 − 0 = 12
−4 = −4 ✓
Check Equation 1
12 = 12 ✓
2x − 3y = 3
?
2(−3) − 3(−3) = 3
?
−6 + 9 = 3
6=6✓
3y + 4x = 3
x + 3y = −6
3y + 4x − 4x = 3 − 4x
x − x + 3y = −6 − x
3y = −4x + 3
−4x + 3
3
4
y = −—x + 1
3
3y
3
—=—
y
Equation 2
x − 3y = 6
?
−3 − 3(−3) = 6
?
−3 + 9 = 6
The solution is (−4, 0).
20.
y = 3x − 2
−4
3y = −x − 6
−x − 6
3
1
y = −—x − 2
3
3y
3
—=—
3=3✓
The solution of the linear system x − 3y = 6 and 2x − 3y = 3
is (−3, −3).
22. The solution of the system should be the ordered pair for
the point of intersection, not just the x-value where the
lines intersect.
y = 2x − 1
6
y
4
y = −3 x + 1
2
4
4
x
6
(2, 3)
2
y=x+1
1
y = −3 x − 2
−4
2
(3, −3)
4
6 x
−6
Check Equation 1
Check Equation 1
Equation 2
3y + 4x = 3
x + 3y = −6
?
3(−3) + 4(3) = 3
?
−9 + 12 = 3
?
3 + 3(−3) = −6
?
3 − 9 = −6
3=3✓
The solution is (3, −3).
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−6 = −6 ✓
y = 2x − 1
?
3 = 2(2) − 1
?
3=4−1
Equation 2
y=x+1
?
3=2+1
3=3✓
3=3✓
The solution of the linear system y = 2x − 1 and y = x + 1
is (2, 3).
Algebra 1
Worked-Out Solutions
233
Chapter 5
4x − y = 1.5
2x + y = 1.5
4x − 4x − y = 1.5 − 4x
2x − 2x + y = 1.5 − 2x
26.
0.2x + 0.4y = 4
23.
0.2x − 0.2x + 0.4y = 4 − 0.2x
0.4y = −0.2x + 4
−y = −4x + 1.5
−0.2x + 4
0.4
—=—
0.4y
0.4
−4x + 1.5
−1
−y
−1
—=—
y = −2x + 1.5
y = 4x − 1.5
y = −0.5x + 10
4
−0.6x + 0.6y = −3
−0.6x + 0.6x + 0.6y = −3 + 0.6x
−6
0.6y = 0.6x − 3
Intersection
X=.5
Y=.5
−4
0.6y
0.6x − 3
—=—
0.6
0.6
The solution is (0.5, 0.5).
y=x−5
8
27. Words
Time on
elliptical
8
Intersection
Y=5
0 X=10
0
12
−1.6x − 3.2y = −24
−1.6x + 1.6x − 3.2y = −24 + 1.6x
+ 6
⋅
Time on
bike
= 300
8x + 6y = 300
1.6x − 24
−3.2
—=—
y = −0.5x + 7.5
x + y = 40
8x + 6y = 300
x − x + y = 40 − x
8x − 8x + 6y = 300 − 8x
y = −x + 40
2.6x + 2.6y = 26
2.6x − 2.6x + 2.6y = 26 − 2.6x
6y = −8x + 300
−8x + 300
6
4
y = − — x + 50
3
6y
6
—=—
Exercise with an Elliptical
Trainer and Stationary Bike
2.6y = −2.6x + 26
Calories burned
−2.6x + 26
2.6
2.6y
2.6
Time on
elliptical
= 40
x + y = 40
System
−3.2y = 1.6x − 24
−3.2y
−3.2
⋅
Time on
bike
+
Variables Let x be how much time (in minutes) you
spend on the elliptical trainer, and let y be
how much time (in minutes) you spend on the
stationary bike.
The solution is (10, 5).
24.
6
—=—
y = −x + 10
8
y
4
y = − 3 x + 50
40
y = −x + 40
20
0
(30, 10)
0
20
40
x
60
Time (minutes)
Intersection
Y=5
0 X=5
0
Check Equation 1
12
The solution is (5, 5).
25.
−7x + 6y = 0
0.5x + y = 2
−7x + 7x + 6y = 0 + 7x
0.5x − 0.5x + y = 2 − 0.5x
6y = 7x
7x
6
7
y = —x
6
y = −0.5x + 2
6y
6
—=—
The solution is (1.2, 1.4).
234
40 = 40 ✓
8x + 6y = 300
?
8(30) + 6(10) = 300
?
240 + 60 = 300
300 = 300 ✓
The solution is (30, 10). So, you should spend 30 minutes on
the elliptical trainer and 10 minutes on the stationary bike.
4
−6
Equation 2
x + y = 40
?
30 + 10 = 40
6
Intersection
X=1.2
Y=1.4
−4
Algebra 1
Worked-Out Solutions
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Chapter 5
28. Words
Number of
small candles
4
⋅
Number of
large candles
+
Number of
small candles + 6
⋅
Check Equation 1
= 28
Number of
large candles = 144
Variables Let x be how many small candles you sell, and
let y be how many large candles you sell.
4x + 6y = 144
4x + 6y = 144
x − x + y = 28 − x
4x − 4x + 6y = 144 − 4x
y = −x + 28
6y = −4x + 144
6y −4x + 144
—=—
6
6
2
y = −— x + 24
3
Selling Candles at a Craft Fair
Money earned (dollars)
y = 6x + 6
?
18 = 6(2) + 6
?
18 = 12 + 6
18 = 18 ✓
18 = 18 ✓
y = −x + 28
y
400
30.
Balance (dollars)
x + y = 28
y
32
y = 18x − 18
?
18 = 18(2) − 18
?
18 = 36 − 18
The solution is (2, 18). This means that when x = 2, the
area of the rectangle will be 18 square centimeters, and
the perimeter will be 18 centimeters.
x + y = 28
System
300
y = 25x + 250
200
100
0
24
0
8
16
24
32
x
Number of candles
Check Equation 1
Equation 2
x + y = 28
?
12 + 16 = 28
28 = 28 ✓
31. a.
⋅
y = 6 (3x − 3)
y = 2(6) + 2 (3x − 3)
y = 6 (3x) − 6(3)
y = 12 + 2 (3x) − 2(3)
y = 18x − 18
y = 12 + 6x − 6
−x
2 = 2x − 4
+4
6 2x
2
2
3=x
—=—
The solution is x = 3.
b.
6
y
(3, 5)
y=x+2
4
y = 3x − 4
y = 6x + 6
2
4
6 x
−2
y
(2, 18)
18
Check Equation 1
y = 6x + 6
12
6
y = 18x − 18
0
+4
6 = 2x
The solution is (12, 16). So, you sell 12 small candles and
16 large candles.
P = 2ℓ+ 2w
x + 2 = 3x − 4
−x
4x + 6y = 144
?
4(12) + 6(16) = 144
?
48 + 96 = 144
144 = 144 ✓
29. A =ℓ w
8 x
Sample answers: A linear equation that could represent my
account balance is y = 12.5x + 325. The slope is 12.5, and
the y-intercept is 325. So, you must deposit $325 initially
and then deposit an additional $12.50 each month for 6
months.
(12, 16)
0
6
Time (months)
y = − 3 x + 24
8
0
4
2
2
16
Equation 2
0
1
2
3
Equation 2
y=x+2
?
5=3+2
y = 3x − 4
?
5 = 3(3) − 4
?
5=9−4
5=5✓
x
5=5✓
The solution is (3, 5).
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Algebra 1
Worked-Out Solutions
235
Chapter 5
c. The x-value of the solution in part (b) is the solution of the
equation in part (a). The expressions with x on the right
sides of the equations in part (b) make up each side of
the equation in part (a). So, when x = 3, each side of the
equation in part (a) is equal to the y-value of the solution
in part (b).
1
3
—4 x + —4 y = 5
36.
3
3
1
b. Each of the answers in part (a) is the x-value of a solution
of a system of two linear equations formed by each pair of
equations.
33. a. Let x be the time (in hours) you and your friend have
been hiking, and let y be the distance (in miles) from the
trailhead. A system of linear equations that represents this
situation is
y = 5x
y = 3x + 3.
Distance (miles)
y
16
y = 5x
12
⋅
(
y = −3x + 20
The rewritten literal equation is y = −3x + 20.
Section 5.2
5.2 Explorations (p. 241)
1. a. Method 1
Method 2
Equation 1
x + y = −7
Equation 1
x + y = −7
x + y − y = −7 − y
x − x + y = −7 − x
x = −y − 7
Equation 2
−5x + y = 5
y = −x − 7
Equation 2
−5x + y = 5
−5(−y − 7) + y = 5
−5x + (−x − 7) = 5
−5(−y) − 5(−7) + y = 5
−5x − x − 7 = 5
5y + 35 + y = 5
−6x − 7 = 5
6y + 35 = 5
−35
0
1
2
3
b. no; According to the graph, after 1 hour of hiking, you
will be 5 miles from the trailhead, and your friend will
be 6 miles from the trailhead. So, you will still be 1 mile
apart. The graphs intersect at (1.5, 7.5). So, after hiking
for 1.5 hours, you will meet at a location that is 7.5 miles
from the trailhead.
Maintaining Mathematical Proficiency
10x + 5y = 5x + 20
10x − 10x + 5y = 5x − 10x + 20
5y = −5x + 20
5y −5x + 20
—=—
5
5
y = −x + 4
+7
−35
Equation 1
x + y = −7
x + (−5) = −7
+5
+5
x = −2
+7
−6x = 12
12
−6x
—=—
−6
−6
x = −2
6y = −30
6y −30
—=—
6
6
y = −5
4 x
Time (hours)
34.
)
3
y = 4 −—4 x + 4(5)
4
0
⋅ ( −— x + 5 )
3
4
4 —14 y = 4
y = 3x + 3
8
3
1
—4 y = −—4 x + 5
32. a. If the teacher buys 20 binders, companies B and C charge
the same. For 35 binders, companies A and B charge
the same. For 50 binders, companies A and C charge the
same. These are the x-values of the points of intersection
in the graph.
3
—4 x − —4 x + —4 y = 5 − —4 x
Equation 1
x + y = −7
−2 + y = −7
+2
+2
y = −5
The solution (−2, −5) is the same using both methods.
For this problem, Method 2 is preferred because both
equations can be solved for y easily.
The rewritten literal equation is y = −x + 4.
9x + 18 = 6y − 3x
35.
9x + 3x + 18 = 6y − 3x + 3x
12x + 18 = 6y
12x + 18 6y
—=—
6
6
2x + 3 = y
The rewritten literal equation is y = 2x + 3.
236
Algebra 1
Worked-Out Solutions
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Chapter 5
b. Method 1
c. Method 1
Method 2
Method 2
Equation 1
x − 6y = −11
Equation 1
x − 6y = −11
Equation 2
3x − 5y = −18
x − 6y + 6y = −11 + 6y
x − x − 6y = −11 − x
3x − 5y + 5y = −18 + 5y
4x − 4x + y = −1 − 4x
−6y = −x − 11
3x = 5y − 18
3x 5y − 18
— =—
3
3
5
x = —y − 6
3
y = −4x − 1
x = 6y − 11
−6y −x − 11
—=—
−6
−6
11
1
y = —x + —
6
6
Equation 2
Equation 2
3x + 2y = 7
3x + 2y = 7
11
1
3(6y − 11) + 2y = 7
3x + 2 — x + — = 7
6
6
1
11
3(6y) − 3(11) + 2y = 7
3x + 2 — x + 2 — = 7
6
6
11
1
18y − 33 + 2y = 7
3x + — x + — = 7
3
3
11
10
20y − 33 = 7
—x + — = 7
3
3
11
11
+33 +33
−— −—
3
3
20y = 40
(
)
( ) ( )
20y
20
10
10
3
3
3 10
3 10
— ∙ —x = — ∙ —
10 3
10 3
x=1
40
20
—=—
—x = —
y=2
Equation 1
Equation 1
x − 6y = −11
x − 6y = −11
x − 6(2) = −11
1 − 6y = −11
x − 12 = −11
+12
+12
x=1
−1
−1
−6y = −12
−6y −12
—=—
−6
−6
y=2
The solution (1, 2) is the same using both methods.
For this system, Method 1 is preferred because the first
equation can be solved for x easily.
Equation 1
4x + y = −1
5
4 — y − 6 + y = −1
3
5
4 —y − 4(6) + y = −1
3
20
— y − 24 + y = −1
3
23
— y − 24 = −1
3
+24 +24
(
Equation 1
4x + y = −1
Equation 2
3x − 5y = −18
)
( )
3x − 5(−4x − 1) = −18
3x − 5(−4x) − 5(−1) = −18
3x + 20x + 5 = −18
23x + 5 = −18
−5
23
3
3 23
3
— — y = — 23
23 3
23
y=3
— y = 23
⋅
23x = −23
23x −23
23
23
x = −1
⋅
—=—
Equation 1
4x + y = −1
Equation 1
4x + y = −1
4x + 3 = −1
4(−1) + y = −1
−3
−5
−3
−4 + y = −1
4x = −4
4x −4
—=—
4
4
x = −1
+4
+4
y=3
The solution (−1, 3) is the same using both methods.
For this system, Method 2 is preferred because the first
equation can be solved for y easily.
2. a. Sample answer: randInt (−5, 5, 2)
{−1 3}
b. Sample answer: 2x − y = 2(−1) − 3
= −2 − 3
= −5
One linear equation that has (−1, 3) as a solution is
2x − y = −5.
x + 5y = −1 + 5(3)
= −1 + 15
= 14
Another linear equation that has (−1, 3) as a solution is
x + 5y = 14. So, a system of linear equations that has
(−1, 3) as its solution is
2x − y = −5.
x + 5y = 14
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Algebra 1
Worked-Out Solutions
237
Chapter 5
c. Sample answer: I would solve my system by Method 1.
First, I would solve Equation 2 for x, because x has a
coefficient of 1.
Equation 2
x + 5y = 14
y has a coefficient of 1. So, it will take fewer steps to
isolate y and I will get a relatively simple expression with
no fractions or decimals.
Equation 1
2x − y = −5
x + 5y − 5y = 14 − 5y
b. Sample answer: Start by solving Equation 2 for y, because
2(−5y + 14) − y = −5
Equation 2
Equation 1
2x + y = −2
x − 2y = −6
2(−5y) + 2(14) − y = −5
2x − 2x + y = −2 − 2x
x − 2(−2x − 2) = −6
Equation 2
x + 5y = 14
−10y + 28 − y = −5
−11y + 28 = −5
y = −2x − 2
x − 2(−2x) − 2(−2) = −6
x + 5(3) = 14
−28 −28
x = − 5y + 14
x + 15 = 14
−15
−11y = −33
−11y −33
— =—
−11 −11
y=3
−15
x = −1
The solution of the system is (−1, 3).
2x + y = −2
equations for one of the variables. Substitute the expression
for this variable into the other equation to find the value of
the other variable. Then substitute this value into one of the
original equations to find the value of the other variable.
x has a coefficient of 1. So, it will take fewer steps to
isolate x and I will get a relatively simple expression with
no fractions or decimals.
Equation 2
x + 2y = −7
2x − y = −9
x + 2y −2y = −7 − 2y
2(−2y − 7) − y = −9
x = −2y − 7
2(−2y) − 2(7) − y = −9
−4y − 14 − y = −9
x + 2y = −7
−5y − 14 = −9
x + 2(−1) = −7
+14
x − 2 = −7
+2
+2
x = −5
x + 2y = −7
?
−5 + 2(−1) = −7
?
−5 − 2 = −7
+14
+4
Check Equation 1
−7 = −7 ✓
The solution of the system is (−5, −1).
−6 = −6 ✓
−2 = −2 ✓
c. Sample answer: Start by solving Equation 2 for y, because
y has a coefficient of 1. So, it will take fewer steps to
isolate y and I will get a relatively simple expression with
no fractions or decimals.
Equation 2
Equation 1
−2x + y = −6
−3x + 2y = −10
−2x + 2x + y = −6 + 2x
y = 2x − 6
−3x + 2(2x − 6) = −10
−3x + 2(2x) − 2(6) = −10
−3x + 4x − 12 = −10
−2(2) + y = −6
−9 = −9 ✓
2x + y = −2
?
2(−2) + 2 = −2
?
−4 + 2 = −2
The solution of the system is (−2, 2).
5
−5y
— =—
−5 −5
2x − y = −9
?
2(−5) − (−1) = −9
?
−10 + 1 = −9
Equation 2
x − 2y = −6
?
−2 − 2(2) = −6
?
−2 − 4 = −6
Equation 2
Equation 2
−4
5x = −10
5x −10
— =—
5
5
x = −2
+4
−5y = 5
y = −1
Check Equation 1
−4
−4 + y = −2
4. a. Sample answer: Start by solving Equation 1 for x, because
Equation 1
5x + 4 = −6
2(−2) + y = −2
y=2
3. To solve a system of linear equations, first solve one of the
Equation 1
x + 4x + 4 = −6
Equation 2
−2x + y = −6
x − 12 = −10
+12
−4 + y = −6
+4
+12
x=2
+4
y = −2
Check Equation 1
Equation 2
−3x + 2y = −10
?
−3(2) + 2(−2) = −10
?
−6 − 4 = −10
−2x + y = −6
?
−2(2) − 2 = −6
?
−4 − 2 = −6
−10 = −10 ✓
−6 = −6 ✓
The solution of the system is (2, –2).
238
Algebra 1
Worked-Out Solutions
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Chapter 5
d. Sample answer: Start by solving Equation 2 for x, because
x has a coefficient of 1. So, it will take fewer steps to
isolate x and I will get a relatively simple expression with
no fractions or decimals.
Equation 2
f. Sample answer: Start by solving Equation 1 for y, because
y has a coefficient of −1. So, I will get a relatively simple
expression with no fractions or decimals.
Equation 1
Equation 1
x – 3y = –3
3x + 2y = 13
x – 3y + 3y = –3 + 3y
x = 3y – 3
3(3y – 3) + 2y = 13
3x − y = −6
4x + 5(3x + 6) = 11
−y = −3x − 6
−y −3x − 6
—=—
−1
−1
y = 3x + 6
4x + 5(3x) + 5(6) = 11
9y – 9 + 2y = 13
x – 3y = –3
11y – 9 = 13
x – 3(2) = –3
+9
x – 6 = –3
+6
+9
11y = 22
11y 22
—=—
11
11
y=2
+6
x=3
Check Equation 1
−4 + 5y = 11
+4
3x − 2y = 9
3(−3y − 8) − 2y = 9
−x = 3y + 8
3(−3y) − 3(8) − 2y = 9
−x 3y + 8
−9y − 24 − 2y = 9
—=—
−1
−1
x = −3y − 8
−11y − 24 = 9
+24
3x + 6 = 9
−6
+24
−11y = 33
33
−11y
—=—
−11
−11
y = −3
3x − 2(−3) = 9
4x + 5y = 11
?
4(−1) + 5(3) = 11
?
−4 + 15 = 11
−6 = −6 ✓
11 = 11 ✓
The solution of the system is (–1, 3).
Equation 1
3x − 2y = 9
Equation 2
3x − y = −6
?
3(−1) − 3 = −6
?
−3 − 3 = −6
x has a coefficient of −1. So, I will get a relatively simple
expression with no fractions or decimals.
Equation 1
+4
Check Equation 1
–3 = –3 ✓
−x − 3y = 8
−30
19x = −19
19x −19
—=—
19
19
x = −1
4(−1) + 5y = 11
e. Sample answer: Start by solving Equation 2 for x, because
−x − 3y + 3y = 8 + 3y
−30
4x + 5y = 11
The solution of the system is (3, 2).
Equation 2
19x + 30 = 11
5y = 15
5y 15
—=—
5
5
y=3
x – 3y = –3
?
3 – 3(2) = –3
?
3 – 6 = –3
13 = 13 ✓
4x + 15x + 30 = 11
Equation 2
Equation 2
3x + 2y = 13
?
3(3) + 2(2) = 13
?
9 + 4 = 13
4x + 5y = 11
3x − 3x − y = −6 − 3x
3(3y) – 3(3) + 2y = 13
Equation 2
Equation 2
−6
5.2 Monitoring Progress (pp. 242–244)
1. Substitute −4x for y in Equation 1 and solve for x.
y = 3x + 14
−4x = 3x + 14
−3x
−3x
−7x = 14
14
−7x
−7
−7
x = −2
—=—
Substitute −2 for x in Equation 2 and solve for y.
y = −4x
3x = 3
3x 3
—=—
3
3
x=1
y = −4(−2)
y=8
Check Equation 1
Equation 2
3x − 2y = 9
?
3(1) − 2(−3) = 9
?
3+6=9
−x − 3y = 8
?
−1 − 3(−3) = 8
?
−1 + 9 = 8
9=9✓
8=8✓
Check y = 3x + 14
?
8 = 3(−2) + 14
?
8 = −6 + 14
y = −4x
?
8 = −4(−2)
8=8✓
8=8✓
The solution is (−2, 8).
The solution of the system is (1, –3).
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Algebra 1
Worked-Out Solutions
239
Chapter 5
1
2
2. Substitute — x − 1 for y in Equation 1 and solve for x.
3x + 2y = 0
1
3x + 2 —x − 1 = 0
2
1
3x + 2 —x − 2(1) = 0
2
3x + x −2 = 0
Check
)
(
( )
4x + y = −3
?
4(−1) + 1 = −3
?
−4 + 1 = −3
−1 = −1 ✓
−3 = −3 ✓
The solution is (−1, 1).
x + y = −2
4. Step 1
4x −2 = 0
+2
x = 6y − 7
?
−1 = 6(1) − 7
?
−1 = 6 − 7
x + y − y = −2 − y
+2
x = −y − 2
4x = 2
4x 2
—=—
4
4
1
x=—
2
−3x + y = 6
Step 2
−3(−y − 2) + y = 6
−3(−y) − 3(−2) + y = 6
1
Substitute — for x in Equation 2 and solve for y.
2
1
y = —x − 1
2
1 1
y=— — −1
2 2
1
y=—−1
4
3
y = −—
4
3y + 6 + y = 6
4y + 6 = 6
−6 −6
4y = 0
4y 0
—=—
4
4
y=0
()
3x + 2y = 0
Check
() ( )
1
3 ?
3 — + 2 −— = 0
2
4
3 3 ?
—−—=0
2 2
0=0✓
1 3
The solution is —, −— .
2 4
(
)
1
y = —x − 1
2
3 ? 1 1
−— = — — − 1
4 2 2
3 ? 1
−— = — −1
4 4
3
3
−— = −— ✓
4
4
Step 3
x = −2
Check
()
4(6y − 7) + y = −3
4(6y) − 4(7) + y = −3
24y − 28 + y = −3
x + y = −2
?
−2 + 0 = −2
−2 = −2 ✓
The solution is (−2, 0).
−x + y = −4
5. Step 1
−x + x + y = −4 + x
y=x−4
4x − y = 10
Step 2
4x − (x − 4) = 10
4x − x + 4 = 10
3x + 4 = 10
25y − 28 = −3
+28
−4
+28
25y = 25
25y 25
—=—
25
25
y=1
Substitute 1 for y in Equation 1 and solve for x.
x = 6y − 7
x = 6(1) − 7
x=6−7
−3x + y = 6
?
−3(−2) + 0 = 6
?
6+0=6
6=6✓
3. Substitute 6y − 7 for x in Equation 2 and solve for y.
4x + y = −3
x + y = −2
x + 0 = −2
−4
3x = 6
3x 6
—=—
3
3
x=2
Step 3 −x + y = −4
−2 + y = −4
+2
+2
y = −2
x = −1
240
Algebra 1
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Chapter 5
Check −x + y = −4
?
−2 − 2 = −4
4x − y = 10
?
4(2) − (−2) = 10
?
8 + 2 = 10
−4 = −4 ✓
x −2y = 7
Step 3
( )
9
x − 2 −— = 7
2
x+9=7
−9
10 = 10 ✓
2x − y = −5
6. Step 1
x − 2y = 7
Check
3x − 2y = 3
( )
2x − 2x − y = −5 − 2x
( )
9 ?
−2 − 2 −— = 7
2
?
−2 + 9 = 7
−y = −2x − 5
−y −2x − 5
−1
−1
y = 2x + 5
—=—
9 ?
3(−2) − 2 −— = 3
2
?
−6 + 9 = 3
7=7✓
3=3✓
)
(
9
The solution is −2, −— .
2
3x − y = 1
Step 2
−9
x = −2
The solution is (2, −2).
3x − (2x + 5) = 1
8. Words
3x −2x − 5 = 1
Number of
students in
drama club
x−5=1
+5
+5
Number of
students in
drama club
x=6
Step 3
3x − y = 1
3(6) − y = 1
−18
−y = −17
−y −17
—=—
−1
−1
y = 17
Check
System
2x − y = −5
?
2(6) − 17 = −5
?
12 − 17 = −5
3x − y = 1
?
3(6) − 17 = 1
?
18 − 17 = 1
x + y = 64
Equation 1
x = 10 + y
Equation 2
1=1✓
(10 + y) + y = 64
10 + 2y = 64
−10
x − 2y + 2y = 7 + 2y
x = 2y + 7
3x − 2y = 3
3(2y + 7) − 2y = 3
3(2y) + 3(7) − 2y = 3
6y + 21 − 2y = 3
4y + 21 = 3
−10
2y = 54
2y 54
—=—
2
2
y = 27
x − 2y = 7
−21
Number of
students in
yearbook club
x + y = 64
The solution is (6, 17).
Step 2
= 10 +
= 64
Substitute 10 + y for x in Equation 1 and solve for y.
−5 = −5 ✓
7. Step 1
Number of
students in
yearbook club
Variables Let x be the number of students in the drama
club, and let y be the number of students in the
yearbook club.
18 − y = 1
−18
+
Substitute 27 for y in Equation 2 and solve for x.
x = 10 + y
x = 10 + 27
x = 37
The solution is (37, 27). So, there are 37 students in the
drama club and 27 students in the yearbook club.
−21
4y = −18
18
4y
— = −—
4
4
9
y = −—
2
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Algebra 1
Worked-Out Solutions
241
Chapter 5
10. Substitute −3x for y in Equation 1 and solve for x.
5.2 Exercises (pp. 245–246)
Vocabulary and Core Concept Check
1. To solve a system of linear equations by substitution, solve
one of the equations for one of the variables. Then substitute
the expression for this variable into the other equation to find
the value of the other variable. Finally, substitute this value
into one of the original equations to find the value of the
other variable.
2. Sample answer: If one of the variables has a coefficient of
1 or −1, I solve for that variable in Step 1.
Monitoring Progress and Modeling with Mathematics
3. Sample answer: Solve Equation 2 for x because x has a
coefficient of 1.
6x − 9 = y
6x − 9 = −3x
−6x
−6x
−9 = −9x
−9 −9x
—=—
−9
−9
1=x
Substitute 1 for x in Equation 2 and solve for y.
y = −3x
y = −3(1)
y = −3
Check
4. Sample answer: Solve Equation 2 for y because y has a
coefficient of 1.
5. Sample answer: Solve Equation 2 for y because y has a
coefficient of −1.
6x − 9 = y
?
6(1) − 9 = −3
?
6 − 9 = −3
x and y each have a coefficient of 1.
7. Sample answer: Solve Equation 1 for x because x has a
coefficient of 1.
−3 = −3 ✓
11. Substitute 16 − 4y for x in Equation 2 and solve for y.
3x + 4y = 8
3(16 − 4y) + 4y = 8
3(16) − 3(4y) + 4y = 8
8. Sample answer: Solve Equation 2 for x because x has a
coefficient of 1.
48 − 12y + 4y = 8
48 − 8y = 8
−48
9. Substitute 17 − 4y for x in Equation 2 and solve for y.
−48
−8y = −40
−8y −40
—=—
−8
−8
y=5
y=x−2
y = (17 − 4y) − 2
y = 17 − 4y −2
y = −4y + 15
Substitute 5 for y in Equation 1 and solve for x.
+4y
x = 16 − 4y
5y = 15
5y 15
—=—
5
5
y=3
x = 16 − 4(5)
x = 16 − 20
x = −4
Substitute 3 for y in Equation 1 and solve for x.
x = 17 − 4y
x = 17 − 4(3)
x = 17 − 12
Check x = 16 − 4y
?
−4 = 16 − 4(5)
?
−4 = 16 − 20
−4 = −4 ✓
x=5
Check x = 17 − 4y
?
5 = 17 − 4(3)
?
5 = 17 − 12
−3 = −3 ✓
The solution is (1, −3).
6. Sample answer: Solve Equation 2 for either x or y because
+4y
y = −3x
?
−3 = −3(1)
3x + 4y = 8
?
3(−4) + 4(5) = 8
?
−12 + 20 = 8
8=8✓
The solution is (−4, 5).
y=x−2
?
3=5−2
3=3✓
5=5✓
The solution is (5, 3).
242
Algebra 1
Worked-Out Solutions
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Chapter 5
12. Substitute 10x − 8 for y in Equation 1 and solve for x.
−5x + 3y = 51
14. Solve Equation 2 for x.
x − 9 = −1
+9
−5x + 3(10x − 8) = 51
x=8
−5x + 3(10x) −3(8) = 51
Substitute 8 for x in Equation 1 and solve for y.
−5x + 30x − 24 = 51
2x − y = 23
25x − 24 = 51
+24
2(8) − y = 23
+24
16 − y = 23
25x = 75
25x 75
—=—
25
25
x=3
−16
Substitute 3 for x in Equation 2 and solve for y.
y = 10x − 8
y = 10(3) − 8
−16
−y = 7
7
−y
—=—
−1 −1
y = −7
Check
y = 30 − 8
y = 22
Check
+9
−5x + 3y = 51
?
−5(3) + 3(22) = 51
?
−15 + 66 = 51
51 = 51 ✓
y = 10x − 8
?
22 = 10(3) − 8
?
22 = 30 − 8
22 = 22 ✓
2x − y = 23
?
2(8) − (−7) = 23
?
16 + 7 = 23
The solution is (8, −7).
15. Step 1
x + y = −3
x − x + y = −3 − x
y = −x − 3
5x + 2y = 9
5x + 2(−x − 3) = 9
Step 2
13. Solve Equation 1 for x.
2x = 12
2x 12
—=—
2
2
x=6
5x + 2(−x) − 2(3) = 9
5x − 2x − 6 = 9
3x − 6 = 9
+6 +6
Substitute 6 for x in Equation 2 and solve for y.
3x = 15
3x 15
—=—
3
3
x=5
x − 5y = −29
6 − 5y = −29
−6
Step 3 x + y = −3
5 + y = −3
−5y = −35
−5y −35
—=—
−5
−5
y=7
Check
−1 = −1 ✓
23 = 23 ✓
The solution is (3, 22).
−6
x − 9 = −1
?
8 − 9 = −1
2x = 12
?
2(6) = 12
12 = 12 ✓
−5
−5
y = −8
x − 5y = −29
?
6 − 5(7) = −29
?
6 − 35 = −29
−29 = −29 ✓
The solution is (6, 7).
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Check:
5x + 2y = 9
?
5(5) + 2(−8) = 9
?
25 − 16 = 9
x + y = −3
?
5 − 8 = −3
−3 = −3 ✓
9=9✓
The solution is (5, −8).
Algebra 1
Worked-Out Solutions
243
Chapter 5
16. Step 1
18. In Step 3, 6 should be substituted for x, not for y.
x − 2y = −4
3x + y = 9
Step 3
x − 2y + 2y = −4 + 2y
3(6) + y = 9
x = 2y − 4
18 + y = 9
11x − 7y = −14
Step 2
−18
11(2y − 4) − 7y = −14
y = −9
11(2y) − 11(4) − 7y = −14
22y − 44 − 7y = −14
+44
Acres of
= 3
corn
15y = 30
15y 30
—=—
15
15
y=2
Step 3
Acres of
Acres of
+ wheat = 180
corn
19. Words
15y − 44 = −14
+44
System
x − 2(2) = −4
x − 4 = −4
x + y = 180
Equation 1
x = 3y
Equation 2
x + y = 180
11x − 7y = −14
?
11(0) − 7(2) = −14
?
0 − 14 = −14
x − 2y = − 4
?
0 − 2(2) = −4
?
0 − 4 = −4
−14 = −14 ✓
−4 = −4 ✓
The solution is (0, 2).
3y + y = 180
4y = 180
4y 180
—= —
4
4
y = 45
Substitute 45 for y in Equation 2 and solve for x.
x = 3y
17. In Step 2, the expression for y should be substituted in the
other equation.
Step 1
Acres of
wheat
Substitute 3y for x in Equation 1 and solve for y.
+4
x=0
Check
⋅
Variables Let x be the number of acres of corn the farmer
should plant, and let y be the number of acres of
wheat the farmer should plant.
x − 2y = −4
+4
−18
5x − y = 4
5x − 5x − y = 4 − 5x
x = 3(45)
x = 135
The solution is (135, 45). So, the farmer should plant
135 acres of corn and 45 acres of wheat.
−y = −5x + 4
−y −5x + 4
—=—
−1
−1
y = 5x − 4
Step 2
8x + 2y = −12
8x + 2(5x − 4) = −12
8x + 2(5x) − 2(4) = −12
8x + 10x − 8 = −12
18x − 8 = −12
+8
Number of
1 person tubes
Number of
+ “cooler” tubes = 15
⋅
Number of
1 person tubes
+ 12.5
20
18x −4
—=—
18
18
2
x=−—
9
Algebra 1
Worked-Out Solutions
Number of
⋅ “cooler”
tubes = 270
Variables Let x be the number of 1 person tubes the group
rents, and let y be the number of “cooler” tubes
the group rents.
System
+8
18x = −4
244
20. Words
Step 1
x + y = 15
Equation 1
20x + 12.5y = 270
Equation 2
x + y = 15
x + y − y = 15 − y
x = 15 − y
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Chapter 5
20x + 12.5y = 270
Step 2
20(15 − y) + 12.5y = 270
20(15) − 20(y) + 12.5y = 270
300 − 20y + 12.5y = 270
300 − 7.5y = 270
−300
−300
−7.5y = −30
−30
−7.5y
—=—
−7.5
−7.5
y=4
Step 3
23. Sample answer:
x − y = −4 − (−12)
= −4 + 12
=8
One linear equation that has (−4, −12) as a solution is
x − y = 8.
5x − 2y = 5(−4) − 2(−12)
= −20 + 24
=4
x + y = 15
Another linear equation that has (−4, −12) as a solution is
5x − 2y = 4. So, a system of linear equations that has
x + 4 = 15
(−4, −12) as its solution is
−4 = −4
x − y = 8.
5x − 2y = 4
x = 11
The solution is (11, 4). So, the group rents 11 tubes for
people to use and 4 “cooler” tubes.
21. Sample answer: 4x − 5y = 4(3) − 5(5)
24. Sample answer:
3x + y = 3(15) + (−25)
= 45 − 25
= 12 − 25
= −13
One linear equation that has (3, 5) as a solution is
4x − 5y = −13.
−2x − y = −2(3) − 5
= 20
One linear equation that has (15, −25) as a solution is
3x + y = 20.
−2x − y = −2(15) − (−25)
= −30 + 25
= −6 − 5
= −11
Another linear equation that has (3, 5) as a solution is
−2x − y = −11. So, a system of linear equations that has
(3, 5) as its solution is
4x − 5y = −13 .
−2x − y = −11
= −5
Another linear equation that has (15, −25) as a solution is
−2x − y = −5. So, a system of linear equations that has
(15, −25) as its solution is
3x + y = 20.
−2x − y = −5
25. Words
22. Sample answer:
Number
of 5-point
problems
4x + y = 4(−2) + 8
= −8 + 8
=0
One linear equation that has (−2, 8) as a solution is
4x + y = 0.
−3x + 7y = −3(−2) + 7(8)
= 6 + 56
= 62
5
⋅
Number
of 5-point
problems
System x + y = 38
(−2, 8) as its solution is
Step 1
−3x + 7y = 62
.
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+ 2
Number
of 2-point
problems
⋅
Number
of 2-point
problems
= 38
= 100
Variables Let x be the number of 5-point problems on
the test, and let y be the number of 2-point
problems on the test.
Another linear equation that has (−2, 8) as a solution is
−3x + 7y = 62. So, a system of linear equations that has
4x + y = 62
+
5x + 2y = 100
Equation 1
Equation 2
x + y = 38
x − x + y = 38 − x
y = 38 − x
Algebra 1
Worked-Out Solutions
245
Chapter 5
5x + 2y = 100
27. a. x + y + 90 = 180
5x + 2(38 − x) = 100
b. x + y + 90 = 180
Step 2
5x + 2(38) − 2(x) = 100
x + 2 = 3y
5x + 76 − 2x = 100
Step 1
3x + 76 = 100
−76
−76
Equation 2
x + 2 = 3y
x + 2 − 2 = 3y − 2
x = 3y − 2
3x = 24
3x 24
—=—
3
3
x=8
Step 2
x + y + 90 = 180
3y − 2 + y + 90 = 180
4y + 88 = 180
−88
Step 3 x + y = 38
8 + y = 38
−8
−8
y = 30
The solution is (8, 30). So, 8 of the questions on the math
test are worth 5 points, and 30 questions are worth 2 points.
Step 3 x + 2 = 3y
−88
4y = 92
4y 92
—=—
4
4
y = 23
x + 2 = 3(23)
x + 2 = 69
26. Words
Shares of
Stock A
9.5
⋅
Shares of
Stock A
Shares of
Stock B
+
+ 27
⋅
Shares of
Stock B
x + y = 200
System
9.5x + 27y = 4000
−2
= 200
= 4000
Variables Let x be the number of shares of Stock A the
investor owns, and let y be the number of shares
of Stock B the investor owns.
So, x = 67 and y = 23.
28. a. x + y + (y − 18) = 180
b. x + 2y − 18 = 180
Equation 1
3x − 5y = −22
Equation 2
Step 1
Equation 1
Equation 2
x + 2y = 198
x + 2y − 2y = 198 − 2y
x = 198 − 2y
y = 200 − x
9.5x + 27y = 4000
Step 2
3(198) −3 (2y) − 5y = −22
9.5x + 27(200) − 27(x) = 4000
594 − 6y − 5y = −22
9.5x + 5400 − 27x = 4000
594 − 11y = −22
−17.5x + 5400 = 4000
x + y = 200
80 + y = 200
−80
−594
−5400
−17.5x = −1400
−17.5x −1400
—=—
−17.5
−17.5
x = 80
−80
y = 120
The solution is (80, 120). So, the investor owns 80 shares of
Stock A and 120 shares of Stock B.
3x − 5y = −22
3(198 − 2y) − 5y = −22
9.5x + 27(200 − x) = 4000
−5400
x + 2y − 18 = 180
x + 2y − 18 + 18 = 180 + 18
x − x + y = 200 − x
Step 2
−2
x = 67
x + y = 200
Step 1
Step 3
Equation 1
−594
−11y = −616
−11y −616
—=—
−11
−11
y = 56
Step 3
3x − 5y = −22
3x − 5(56) = −22
3x − 280 = −22
+280
+280
3x = 258
3x 258
—=—
3
3
x = 86
So, x = 86 and y = 56.
246
Algebra 1
Worked-Out Solutions
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Chapter 5
29. ax + by = −31 ⇒ a(−9) + b(1) = −31 ⇒ −9a + b = −31
ax − by = −41 ⇒ a(−9) − b(1) = −41 ⇒ −9a − b = −41
−9a + b = −31
Step 1
−9a + 9a + b = −31 + 9a
b = 9a − 31
−9a − b = −41
Step 2
−9a − (9a − 31) = −41
−9a − 9a + 31 = −41
So, a system of linear equations that has (−1, 7) as its
solution and (3, −5) as a solution of Equation 1 but not
Equation 2 is
y = −3x + 4 .
5x − 2y = −19
32. a. The lines appear to intersect at (4, 5).
b. yes; Write an equation of each line. Then solve the system
of linear equations using substitution.
−18a + 31 = −41
−31
−31
33. Words
−18a = −72
−18a −72
—=—
−18
−18
a=4
Number of
pop songs = 3
–36 + b = –31
+36
+36
b=5
When a = 4 and b = 5, the solution of the linear system is
(−9, 1).
30. yes; The equation of a horizontal line is of the form x = a,
and the equation of a vertical line is of the form y = b.
Each equation has either x or y but not both. So, Step 2 is
impossible because an expression for a variable from one
equation cannot be substituted for that same variable in the
other equation. Therefore, the system cannot be solved by
substitution.
System x + y + z = 272
x=3
y + 5 = −3(x) − 3(−3)
y + 5 = −3x + 9
y + 5 − 5 = −3x + 9 − 5
y = −3x + 4
Equation 1
Equation 2
Equation 3
Substitute 3y for x and 32 + y for z in Equation 1 and solve
for y.
x + y + z = 272
3y + y + (32 + y) = 272
5y + 32 = 272
−32
−32
5y = 240
5y 240
—=—
5
5
y = 48
7+5
12
7 − (−5)
m = — = — = — = −3
−1 − 3
−1−3 −4
y − y1 = m(x − x1)
⋅y
z = 32 + y
two given points.
y − (−5) = −3(x − 3)
Number of
rock songs
Variables Let x be the number of pop songs the station plays.
Let y be the number of rock songs the station
plays. Let z be the number of hip-hop songs the
station plays.
31. Sample answer: First find the slope of the line through the
Use point slope form to write an equation of the line.
⋅
Number of
Number of
hip-hop songs = 32 + rock songs
Step 3 –9a + b = –31
–9(4) + b = –31
Number of
Number of
Number of
pop songs + rock songs + hip-hop songs = 272
Substitute 48 for y in Equation 2 and in Equation 3 to solve
for x and z, respectively.
x = 3y
z = 32 + y
x = 3(48)
z = 32 + 48
x = 144
z = 80
So, the radio station plays 144 pop songs, 48 rock songs, and
80 hip-hop songs.
An equation that has both (3, −5) and (−1, 7) as solutions is
y = –3x + 4.
5x – 2y = 5(–1) – 2(7)
= –5 – 14
= –19
An equation that has (−1, 7) as a solution but not (3, −5) is
5x − 2y = −19.
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Algebra 1
Worked-Out Solutions
247
Chapter 5
34. Sample answer:
Words
0.25
10y + x = 27 + 10x + y
Step 2
Number of
quarters + 0.10
⋅
⋅
Number of
Number
+ of dimes =
quarters
Number
of dimes = 2.65
10(11 − x) + x = 27 + 10x + (11 − x)
10(11) − 10(x) + x = 27 + 10x + 11 − x
110 − 10x + x = 9x + 38
Number
of coins
−9x + 110 = 9x + 38
+9x
Variables Let x be the number of quarters you have, and let y
be the number of dimes you have.
System 0.25x + 0.10y = 2.65
x+y=
Number
of coins
+9x
110 = 18x + 38
−38
Equation 1
72 = 18x
Equation 2
72
18
4=x
0.25x + 0.10y = 2.65
Step 3 x + y = 11
0.25(5) + 0.10y = 2.65
4 + y = 11
1.25 + 0.10y = 2.65
−4
−1.25
The solution is (4, 7). So, the original number is 47.
Maintaining Mathematical Proficiency
36. (x − 4) + (2x − 7) = x − 4 + 2x − 7
Then, x + y = 5 + 14 = 19.
= x + 2x − 4 − 7
= 3x − 11
So, a system that represent this situation is
0.25x + 0.10y = 2.65.
x + y = 19
37. (5y − 12) + (−5y − 1) = 5y − 12 − 5y − 1
= 5y − 5y − 12 − 1
= 0 − 13
= −13
Original
Original
tens digit + ones digit = 11
10
38. (t − 8) − (t + 15) = t − 8 − t − 15
Original
Original
ones digit + tens digit =
⋅
(
Original
Original
27 + 10 tens digit + ones digit
)
Variables Let x be the original tens digit, and let y be the
original ones digit.
System
x + y = 11
10y + x = 27 + (10x + y)
Step 1
x + y = 11
x − x + y = 11 − x
y = 11 − x
−4
y=7
0.10y = 1.40
0.10y 1.40
—=—
0.10
0.10
y = 14
35. Words
18x
18
—=—
Let x = 5.
−1.25
−38
Equation 1
Equation 2
= t − t − 8 − 15
= 0 − 23
= −23
39. (6d + 2) − (3d − 3) = 6d + 2 − 3d + 3
= 6d − 3d + 2 + 3
= 3d + 5
40. 4(m + 2) + 3(6m − 4) = 4(m) + 4(2) + 3(6m) − 3(4)
= 4m + 8 + 18m − 12
= 4m + 18m + 8 − 12
= 22m − 4
41. 2(5v + 6) − 6(−9v + 2) = 2(5v) + 2(6) − 6(−9v) − 6(2)
= 10v + 12 + 54v − 12
= 10v + 54v + 12 − 12
= 64v + 0
= 64v
248
Algebra 1
Worked-Out Solutions
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Chapter 5
Section 5.3
Method 2: 2x + y = 6
2x + y = 6
+ 2x − y = 2
2(2) + y = 6
5.3 Explorations (p. 247)
⋅ x + 1 ⋅ y = 4.50 ⇒ x + y = 4.5
1 ⋅ x + 5 ⋅ y = 16.50 ⇒ x + 5y = 16.5
1. a. 1
b.
x + 5y = 16.5
Equation 2
− (x + y = 4.5)
Equation 1
Equation 1
4x + 0 = 8
Equation 2
4x = 8
4x 8
4
4
x=2
y=2
Using both methods, the solution is (2, 2).
4y = 12
Solve this equation to find the value of y and then
substitute this value for y into one of the original
equations and solve to find the value of x.
x + y = 4.5
Equation 1
Sample answer: Method 2 is preferred for this problem
because I am less likely to make a mistake adding the
equations.
c. Method 1: x − 2y = −7
− (x + 2y = 5)
x + 3 = 4.5
−3 −3
− (3x + y = 0)
3x − y = 6
3x − (−3) = 6
0 − 2y = 6
3x + 3 = 6
−2y = 6
−3 −3
6
−2
—=—
3x = 3
y = −3
—=—
3x 3
3
3
x=1
Method 2: 3x − y = 6
3x + y = 0
+ 3x + y = 0
3(1) + y = 0
6x + 0 = 6
6x = 6
6x 6
6
6
x=1
—=—
3+y=0
−3
−3
y = −3
Using both methods, the solution is (1, −3).
Sample answer: Method 2 is preferred for this problem
because it involved less risk of making a careless error
with negative signs.
b. Method 1: 2x + y = 6
− (2x − y = 2)
x − 6 = −7
−4y = −12
2x + y = 6
2x + 2 = 6
0 + 2y = 4
−2 −2
2y = 4
2x = 4
2y 4
—=—
2
2
y=2
2x 4
—=—
2
2
x=2
+6
−4y −12
−4
−4
y=3
The solution is (1.5, 3). So, one drink costs $1.50 and one
sandwich costs $3.
2. a. Method 1: 3x − y = 6
x − 2y = −7
x − 2(3) = −7
0 − 4y = −12
x = 1.5
−2y
−2
−4
—=—
0 + 4y = 12.0
4y = 12
4y 12
—=—
4
4
y=3
4+y=6
−4
x = −1
—=—
Method 2: x −2y = −7
+ x + 2y = 5
2x + 0 = −2
2x = −2
2x −2
2
2
x = −1
—=—
+6
x − 2y = −7
−1 − 2y = −7
+1
+1
−2y = −6
−2y −6
−2
−2
y=3
—=—
Using both methods, the solution is (−1, 3).
Sample answer: Method 2 is preferred for this problem
because it has fewer negative numbers. So, I am less likely
to make a careless error.
3. a. no; The coefficients of the variables have to be the same
or opposites in order to be eliminated by addition or
subtraction. If you multiply each side of Equation 2 by
−2, then the coefficients of the x-terms will be opposites,
and x will be eliminated when you add the equations.
Another option would be to multiply each side of the first
equation by −5.
b. Sample answer:
2x + y = 7
x + 5y = 17
2x + y = 7
Multiply by −2.
−2x − 10y = −34
0 − 9y = −27
−9y = −27
−9y −27
−9
−9
y=3
—=—
x + 5y = 17
x + 5(3) = 17
x + 15 = 17
−15
−15
x=2
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Algebra 1
Worked-Out Solutions
249
Chapter 5
Check 2x + y = 7
?
2(2) + 3 = 7
?
4+3=7
x + 5y = 17
?
2 + 5(3) = 17
?
2 + 15 = 17
7=7✓
Check 3x + 2y = 7
?
3(1) + 2(2) = 7
?
3+4=7
17 = 17 ✓
The solution is (2, 3).
4. First, you may have to multiply one or both equations by a
constant so that at least one pair of like terms has the same
or opposite coefficients. Then, when you add or subtract
the equations, the resulting equation will not contain that
variable. In other words, the variable will be eliminated.
Solve the resulting equation to find the value of the other
variable. Then, substitute this value into one of the original
equations and solve for the variable that had been eliminated.
−3x + 4y = 5
?
−3(1) + 4(2) = 5
?
−3 + 8 = 5
7=7✓
5=5✓
The solution is (1, 2).
2. Step 1
Step 2
x − 3y = 24
x − 3y = 24
3x + y = 12
9x + 3y = 36
Multiply by 3.
10x + 0 = 60
Step 3
10x 60
10
10
x=6
—=—
5. If at least one pair of like terms has the same coefficients,
then you can subtract equations in a system to eliminate one
of the variables. For example, in Exploration 2 part (a), the
x-terms both have a coefficient of 3. So, when you subtract
the equations, the x-terms are eliminated.
If at least one pair of like terms has opposite coefficients,
then you can add equations in a system to eliminate one of
the variables. For example, in Exploration 2 part (a), the
y-terms have coefficients 1 and −1. So, when you add the
equations, the y-terms are eliminated.
If neither pair of like terms has the same or opposite
coefficients, then you have to multiply one or both equations
by a constant. For example, in Exploration 3, neither pair
of like terms has the same coefficients. So, you multiply
Equation 2 by −2. Then, when you add the equations,
the x-terms are eliminated.
6. By the Multiplication Property of Equality, you can multiply
each side of an equation by the same amount and get an
equivalent equation, which means that it has the same
solution(s).
Step 4
3x + y = 12
3(6) + y = 12
18 + y = 12
−18
−18
y = −6
Check
x − 3y = 24
?
6 − 3(−6) = 24
?
6 + 18 = 24
24 = 24 ✓
3. Step 1
x + 4y = 22
x + 4y = 22
4x + y = 13
Multiply by −4.
6y = 12
6y 12
—=—
6
6
y=2
Step 4
3x + 2y = 7
3x + 2(2) = 7
3x + 4 = 7
−4
−4
3x = 3
3x 3
3
3
x=1
—=—
250
Algebra 1
Worked-Out Solutions
−16x − 4y = −52
−15x + 0 = −30
Step 3
−15x = −30
−15x −30
−15
−15
x=2
3x + 2y = 7
0 + 6y = 12
12 = 12 ✓
Step 2
—=—
− 3x + 4y = 5
Step 3
3x + y = 12
?
3(6) + (−6) = 12
?
18 − 6 = 12
The solution is (6, −6).
5.3 Monitoring Progress (pp. 249 –250)
1. Step 2
10x = 60
Step 4
4x + y = 13
4(2) + y = 13
8 + y = 13
−8
−8
y=5
Check x + 4y = 22
?
2 + 4(5) = 22
?
2 + 20 = 22
4x + y = 13
?
4(2) + 5 = 13
?
8 + 5 = 13
22 = 22 ✓
13 = 13 ✓
The solution is (2, 5).
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Chapter 5
4. Step 1
Step 2
5x + 2y = 235,000
Multiply by −2.
−10x − 4y = −470,000
2x + 3y = 160,000
Multiply by 5.
10x + 15y = 800,000
0 + 11y = 330,000
Step 3
11y = 330,000
11y 330,000
11
11
y = 30,000
—=—
Check x + 2y = 13
?
1 + 2(6) = 13
?
1 + 12 = 13
13 = 13 ✓
4. Step 2
9x + y = 2
−4x − y = −17
2x + 3y = 160,000
2x + 90,000 = 160,000
−90,000
5x + 0 = −15
5x = −15
5x −15
—=—
5
5
x = −3
Step 3
−90,000
2x = 70,000
2x 70,000
2
2
x = 35,000
—=—
Step 4
9x + y = 12
9(−3) + y = 12
The solution (35,000, 30,000) is the same. So, a large van
costs $35,000, and a small van costs $30,000.
5.3 Exercises (pp. 251– 252)
Vocabulary and Core Concept Check
1. Sample answer: Write a system in which at least one pair of
like terms has opposite coefficients.
2x − 3y = 2
−5x + 3y = −14
−27 + y = 2
+27
so that the coefficients of the y-terms are 9 and −9. Then
add the equations to eliminate y. Solve the resulting equation
for x. Then substitute the value for x into one of the original
equations, and solve for y.
Monitoring Progress and Modeling with Mathematics
y = 29
Check 9x + y = 2
?
9(−3) + 29 = 2
?
−27 + 29 = 2
−4x − y = −17
?
−4(−3) − 29 = −17
?
12 − 29 = −17
5. Step 2
5x + 6y = 50
x − 6y = −26
6x + 0 = 24
6x = 24
6x 24
—=—
6
6
x=4
Step 3
x + 2y = 13
−x + y = 5
0 + 3y = 18
3y = 18
3y 18
3
3
y=6
—=—
Step 4
x + 2y = 13
x + 2(6) = 13
x + 12 = 13
−12
x=1
−17 = −17 ✓
The solution is (−3, 29).
3. Step 2
−12
+27
2=2✓
2. Sample answer: First, multiply each side of Equation 1 by 3
Step 3
5=5✓
The solution is (1, 6).
Step 4
2x + 3(30,000) = 160,000
−x + y = 5
?
−1 + 6 = 5
Step 4
x − 6y = −26
4 − 6y = −26
−4
−4
−6y = −30
−6y −30
—=—
−6
−6
y=5
Check 5x + 6y = 50
?
5(4) + 6(5) = 50
?
20 + 30 = 50
50 = 50 ✓
x − 6y = −26
?
4 − 6(5) = −26
?
4 −30 = −26
−26 = −26 ✓
The solution is (4, 5).
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Algebra 1
Worked-Out Solutions
251
Chapter 5
6. Step 2
8. Step 2
−x + y = 4
4x − 9y = −21
x + 3y = 4
Step 3
−4x − 3y = 9
0 + 4y = 8
4y 8
—=—
4
4
y=2
Step 3
0 − 12y = −12
x + 3y = 4
Step 4
Step 4
−12y = −12
−12
−12y
— =—
−12
−12
y=1
x + 3(2) = 4
4x − 9y = −21
x+6=4
4x −9(1) = −21
−6
−6
4x − 9 = −21
x = −2
Check −x + y = 4
?
−(−2) + 2 = 4
?
2+2=4
+9
x + 3y = 4
?
−2 + 3(2) = 4
?
−2 + 6 = 4
4=4✓
4x = −12
4x −12
4
4
x = −3
—=—
4=4✓
The solution is (−2, 2).
7. Step 2
−3x − 5y = −7
9=9✓
9. Rewrite Equation 1 in the form ax + by = c.
−7x = 7
7
−7x
—=—
−7
−7
x = −1
−y − 10 = 6x
−y − 10 + 10 = 6x + 10
−y = 6x + 10
Step 4
−y − 6x = 6x − 6x + 10
−4x + 5y = 14
−6x − y = 10
−4(−1) + 5y = 14
Step 2
4 + 5y = 14
−6x − y = 10
−4
5x + y = −10
5y = 10
5y 10
—=—
5
5
y=2
Check −3x − 5y = −7
?
−3(−1) − 5(2) = −7
?
3 − 10 = −7
−4x − 3y = 9
?
−4(−3) − 3(1) = 9
?
12 − 3 = 9
The solution is (−3, 1).
−7x + 0 = 7
−4
Check 4x − 9y = −21
?
4(−3) − 9(1) = −21
?
−12 − 9 = −21
−21 = −21 ✓
−4x + 5y = 14
Step 3
+9
−x + 0 = 0
Step 3
−4x + 5y = 14
?
−4(−1) + 5(2) = 14
?
4 + 10 = 14
−7 = −7 ✓
The solution is (−1, 2).
14 = 14 ✓
−x = 0
0
−x
—=—
−1 −1
x=0
Step 4
5x + y = −10
5(0) + y = −10
0 + y = −10
y = −10
Check −y − 10 = 6x
?
−(−10) − 10 = 6(0)
?
10 − 10 = 0
0=0✓
5x + y = −10
?
5(0) + (−10) = −10
?
0 − 10 = −10
−10 = −10 ✓
The solution is (0, −10).
252
Algebra 1
Worked-Out Solutions
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Chapter 5
12. Step 1
10. Rewrite Equation 2 in the form ax − c = by.
Step 2
7y − 6 = 3x
8x − 5y = 11
7y − 7y − 6 = 3x − 7y
4x − 3y = 5
8x − 5y = 11
−8x + 6y = −10
Multiply by −2.
0+y=1
−6 = 3x − 7y
−6 − 3x = 3x − 3x − 7y
Step 3
−3x − 6 = −7y
y=1
Step 4
4x − 3y = 5
Step 2
4x −3(1) = 5
3x − 30 = y
4x − 3 = 5
−3x − 6 = −7y
+3
0 − 36 = −6y
4x = 8
4x 8
—=—
4
4
x=2
−36 = −6y
Step 3
−36 −6y
−6
−6
6=y
—=—
Check 8x − 5y = 11
?
8(2) − 5(1) = 11
?
16 − 5 = 11
Step 4
3x − 30 = y
3x − 30 = 6
+30
+30
3x 36
3
3
x = 12
13. Step 1
Step 2
11x − 20y = 28
7y − 6 = 3x
?
7(6) − 6 = 3(12)
?
42 − 6 = 36
6=6✓
11x − 20y = 28
3x + 4y = 36
Multiply by 5.
26x + 0 = 208
36 = 36 ✓
11. Step 1
26x 208
26
26
x=8
−2x − 2y = −4
2x + 7y = 9
2x + 7y = 9
0 + 5y = 5
Step 3
Step 4
x+y=2
−1
x=1
Check x + y = 2
?
1+1=2
2=2✓
5y = 5
5y 5
—=—
5
5
y=1
Step 4
3x + 4y = 36
3(8) + 4y = 36
24 + 4y = 36
−24
−24
4y = 12
4y 12
—=—
4
4
y=3
Check 11x − 20y = 28
?
11(8) − 20(3) = 28
?
88 − 60 = 28
x+1=2
2x + 7y = 9
?
2(1) + 7(1) = 9
?
2+7=9
26x = 208
—=—
Step 2
Multiply by −2.
15x + 20y = 180
Step 3
The solution is (12, 6).
−1
5=5✓
The solution is (2, 1).
—=—
x+y=2
4x − 3y = 5
?
4(2) − 3(1) = 5
?
8−3=5
11 = 11 ✓
3x = 36
Check 3x − 30 = y
?
3(12) − 30 = 6
?
36 − 30 = 6
+3
28 = 28 ✓
3x + 4y = 36
?
3(8) + 4(3) = 36
?
24 + 12 = 36
36 = 36 ✓
The solution is (8, 3).
9=9✓
The solution is (1, 1).
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Algebra 1
Worked-Out Solutions
253
Chapter 5
14. Step 1
16. Step 1
Step 2
10x − 9y = 46
−2x + 3y = 10
Multiply by 5.
Step 2
10x − 9y = 46
−2x − 5y = 9
Multiply by 3.
−10x + 15y = 50
3x + 11y = 4
Multiply by 2.
−6x − 15y = 27
6x + 22y = 8
0 + 6y = 96
Step 3
6y = 96
Step 3
6y 96
—=—
6
6
y = 16
0 + 7y = 35
7y 35
—=—
7
7
y=5
Step 4
−2x − 5y = 9
Step 4
−2x + 3y = 10
−2x − 5(5) = 9
−2x + 3(16) = 10
−2x − 25 = 9
+25
−2x + 48 = 10
−48
−2x = 34
34
−2x —
—=
−2
−2
x = −17
−48
−2x = −38
−2x −38
—=—
−2
−2
x = 19
Check
+25
10x − 9y = 46
?
10(19) − 9(16) = 46
?
190 − 144 = 46
−2x + 3y = 10
?
−2(19) + 3(16) = 10
?
−38 + 48 = 10
46 = 46 ✓
10 = 10 ✓
3x + 11y = 4
?
3(−17) + 11(5) = 4
?
−51 + 55 = 4
9=9✓
Step 2
4x − 3y = 8
Multiply by 2.
5x − 2y = −11
Multiply by −3.
8x −6y = 16
Step 2
9x + 2y = 39
Multiply by 4.
6x + 13y = −9
Multiply by −6.
Step 4
210
−70y
−70
−70
y = −3
—=—
Step 4
9x + 2y = 39
9x − 6 = 39
4(−7) − 3y = 8
+6
−28 − 3y = 8
+6
9x = 45
+28
9x
9
−3y = 36
36
−3y
— =—
−3
−3
y = −12
Check
−70y = 210
9x + 2(−3) = 39
4x − 3y = 8
+28
−36x − 78y = 54
Step 3
−7x + 0 = 49
−7x = 49
49
−7x
—=—
−7
−7
x = −7
36x + 8y = 156
0 − 70y = 210
−15x + 6y = 33
Step 3
4=4✓
The solution is (−17, 5).
17. Step 1
The solution is (19, 16).
15. Step 1
Check: −2x −5y = 9
?
−2(−17) − 5(5) = 9
?
34 − 25 = 9
45
9
—=—
x=5
Check
4x − 3y = 8
?
4(−7) − 3(−12) = 8
?
−28 + 36 = 8
5x − 2y = −11
?
5(−7) − 2(−12) = −11
?
−35 + 24 = −11
8=8✓
−11 = −11 ✓
9x + 2y = 39
?
9(5) + 2(−3) = 39
?
45 − 6 = 39
39 = 39 ✓
6x + 13y = −9
?
6(5) + 13(−3) = −9
?
30 − 39 = −9
−9 = −9 ✓
The solution is (5, −3).
The solution is (−7, −12).
254
Algebra 1
Worked-Out Solutions
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Chapter 5
18. Step 1
Step 2
21. Words
12x − 7y = −2
Multiply by 2.
24x − 14y = −4
8x + 11y = 30
Multiply by−3.
−24x − 33y = −90
0 − 47y = −94
−47y −94
Step 3 — = —
−47
−47
y=2
Cost per
quart of =
oil
⋅
Total
cost
Variables Let x be the fee (in dollars) for an oil change, and
let y be the cost per quart of oil used.
+
+
x
x
System
⋅
⋅
5
7
Step 2
=
=
y
y
22.45
25.45
x + 7y = 25.45
Step 4
8x + 11y = 30
−(x + 5y = 22.45)
8x + 11(2) = 30
0 + 2y = 3.00
8x + 22 = 30
−22
−22
Step 4
x + 7y = 25.45
12x − 7y = −2
?
12(1) − 7(2) = −2
?
12 − 14 = −2
8x + 11y = 30
?
8(1) + 11(2) = 30
?
8 + 22 = 30
−2 = −2 ✓
30 = 30 ✓
The solution is (1, 2).
19. The x-terms should have been added, not subtracted.
Step 2
5x − 7y = 16
x + 7y = 8
6x = 24
6x 24
—=—
6
6
x=4
x − 2y = −13
4x + 3y = 8
−4x + 8y = 52
0 + 11y = 60
11y 60
—=—
11
11
60
y=—
11
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x = 14.95
The solution is (14.95, 1.5). So, the fee for an oil change is
$14.95, and each quart of oil costs $1.50.
22. Words Number
of
individual
songs
Cost
Number
per
+
of
individual
albums
song
⋅
⋅
Cost
Total
per = cost
album
6
4
⋅
⋅
x
x
+
+
2
3
Step 2
6x + 2y = 25.92
Multiply by 3.
4x + 3y = 33.93
Multiply by −2.
⋅
⋅
y
y
= 25.92
= 33.93
18x + 6y = 77.76
−8x − 6y = −67.86
10x + 0 = 9.90
Step 2
Multiply by −4.
−10.5 −10.5
Step 1
including −13 on the right.
4x + 3y = 8
x + 10.5 = 25.45
System
20. Each side of Equation 2 should be multiplied by −4,
Step 1
x + 7(1.5) = 25.45
Variables Let x be the cost per individual song, and let y be
the cost per album.
6x + 0 = 24
Step 3
2y = 3
2y 3
—=—
2
2
y =1.5
Step 3
8x = 8
8x 8
—=—
8
8
x=1
Check
Quarts
of oil
used
Fee
for oil +
change
Step 3 10x = 9.9
10x 9.9
10
10
x = 0.99
—=—
Algebra 1
Worked-Out Solutions
255
Chapter 5
24. Step 1
Step 4
6x + 2y = 25.92
−6y + 2 = −4x
6(0.99) + 2y = 25.92
y−2=x
5.94 + 2y = 25.92
−5.94
Step 2
−6y + 2 = −4x
4y − 8 = 4x
Multiply by 4.
−2y − 6 = 0
−5.94
Step 3 −2y − 6 = 0
2y = 19.98
+6
2y 19.98
—=—
2
2
y = 9.99
−2y
6
−2
−2
y = −3
—=—
The solution is (0.99, 9.99). So, the website charges $0.99 to
download a song and $9.99 to download an entire album.
Step 4
23. Step 1 2y = 8 – 5x
y–2=x
–3 – 2 = x
2y 8 – 5x
—= —
2
2
5
y = 4 – —x
2
Step 2
3x + 2y = 4
(
+6
−2y = 6
)
( )
5
3x + 2 4 – —x = 4
2
5
3x + 2(4) – 2 —x = 4
2
3x + 8 – 5x = 4
–2x + 8 = 4
–8
–8
–2x = –4
–2x –4
—=—
–2
–2
x=2
Step 3 2y = 8 – 5x
2y = 8 – 5(2)
2y = 8 – 10
2y = –2
2y –2
—=—
2
2
y = –1
–5 = x
The solution is (–5, –3). Sample explanation: I chose
elimination because the original equations had like terms in
the same respective positions.
25.
1
y–x=2
y = – —4x + 7
1
y–x+x=2+x
m = – —4, b = 7
y=x+2
m = 1, b = 2
8
y
(4, 6)
6
4
1
y = −4 x + 7
y=x+2
2
4
6
x
The solution is (4, 6). Sample explanation: I chose graphing
because Equation 2 was in the slope-intercept form, and it
only took one step to rewrite Equation 1 in slope-intercept
form.
The solution is (2, –1). Sample explanation: I chose
substitution because it took only one step to isolate y in
Equation 2.
256
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Chapter 5
26. Step 1
c. Sample answer: Because the y-value is already isolated
Step 2
1
3x + y = —
3
8
2x – 3y = —
3
in Equation 2, substitution is the most efficient method
for solving this system. First, substitute the expression
5 + x for y in Equation 1 to get x + (5 + x) + 25 = 50.
Solve this equation for x, and check to see if the solution
matches your guess from part (a) for the number of
students who chose breakfast. Then substitute this value
for x in Equation 2. Solve for y, and check to see if the
solution matches your guess from part (a) for the number
of students who chose lunch.
27x + 9y = 3
Multiply by 9.
6x – 9y = 8
Multiply by 3.
33x + 0 = 11
33x = 11
Step 3
33x 11
—=—
33
33
1
x=—
3
Step 4
29. no; Sample answer: If like terms are in the same respective
positions and at least one pair of like terms has the same or
opposite coefficients, then elimination is more efficient than
substitution and will take fewer steps. On the other hand, if
one of the variables in one of the equations is either isolated
already or has a coefficient of 1 or –1, then substitution is
more efficient and will take fewer steps.
1
3x + y = —
3
1
1
3 — +y=—
3
3
1
1+y=—
3
()
–1
30. Sample answer: A system of linear equations that can be
–1
2
y=–—
3
( )
1 2
The solution is —, – — . Sample explanation: I chose
3 3
elimination because both equations had like terms in the
same respective positions, and I was able to rewrite both
equations without fractions so that I could minimize how
much I had to work with fractions.
27. If a = 4 or a = −4, you can solve the linear system by
added or subtracted to eliminate a variable is
2x + y = 10 .
2x − y = 2
31. a. P = 2ℓ + 2w
⇒
P = 2(3ℓ) + 2(2w)
⇒
Step 1
18 = 2ℓ + 2w
about 15 students chose lunch. (Answer does not have to
be exact, but the total should be 25.)
b. Words
Students
who chose
breakfast
Students
who chose
lunch
+
Students
who chose
lunch
=
Multiply by −2.
46 = 6ℓ + 4w
–36 = –4ℓ – 4w
46 = 6ℓ + 4w
10 = 2ℓ + 0
Step 3
10 = 2ℓ
10 2ℓ
—=—
2
2
5=ℓ
Step 4
18 = 2ℓ + 2w
18 = 2(5) + 2w
18 = 10 + 2w
+ 25 = 50
–10
–10
8 = 2w
+
5
46 = 6ℓ + 4w
Step 2
elimination without multiplying first. If the coefficients of
the x-terms are both 4, then you can simply subtract the
equations in order to eliminate x. If the coefficients of the
x-terms are 4 and −4, then you can simply add the equations
to eliminate x.
28. a. Sample answer: About 10 students chose breakfast, and
18 = 2ℓ + 2w
Students
who chose
breakfast
Variables Let x be the number of students who chose
breakfast, and let y be the number of students
who chose lunch.
8
2
2w
2
—=—
4=w
The original rectangle is 4 inches wide and 5 inches long.
b. The new rectangle is 2w = 2(4) = 8 inches wide and
3ℓ = 3(5) = 15 inches long.
System: x + y + 25 = 50
y=5+x
A system of linear equations that represents the numbers
of students who chose breakfast and lunch is
x + y + 25 = 50
y=5+x
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All rights reserved.
.
Algebra 1
Worked-Out Solutions
257
Chapter 5
32. Sample answer: Begin by multiplying each side of
Step 1
Equation 2 by 2. By the Multiplication Property of Equality,
2x + 2y = 12. You can rewrite Equation 1 as Equation 3 by
adding 2x + 2y on the left and adding 12 on the right.
You can rewrite Equation 3 as Equation 1 by subtracting
2x + 2y on the left and subtracting 12 on the right. Because
you can rewrite Equation 1 as Equation 3, and you can
rewrite Equation 3 as Equation 1, System 1 and System 2
have the same solution.
Step 2
2
3
2
3
—x + —y = 20
Multiply by —32 .
x – y = 20
x + y = 30
x – y = 20
2x + 0 = 50
Step 3
2x
2
of 100% fruit juice +
50
2
—=—
Amount (in quarts)
of 20% fruit juice = 6
33. Words Amount (in quarts)
2x = 50
x = 25
Step 4
⋅
100%
Amount
(in quarts)
of 100% + 20%
fruit juice
⋅
Amount
(in quarts)
= 80%
of 20%
fruit juice
⋅6
Variables Let x be the number of quarts of 100% fruit juice
you should use, and let y be the number of quarts
of 20% fruit juice you should use.
System
x+y=6
1.00x + 0.20y = 0.80(6)
⇒
⇒
x+y=6
x + 0.2y = 4.8
Step 2
–25
–y = –5
–y
–1
–5
–1
—=—
y=5
The solution is (25, 5). So, the speed of the current is
5 miles per hour.
3z + x – 2y = –7
– (x + 0.2y = 4.8)
0 + 0.8y = 1.2
⇒
x – 2y + 3z = –7
Subtract Equation 2 from Equation 1.
x + 7y + 3z = 29
0.8y = 1.2
0.8y
0.8
25 – y = 20
–25
35. Sample answer: Rewrite Equation 2 in standard form.
x+y=6
Step 3
x – y = 20
– (x – 2y + 3z = –7)
1.2
0.8
—=—
0 + 9y + 0 = 36
y = 1.5
9y = 36
Both x and z were eliminated. Solve for y.
Step 4
x+y=6
9y = 36
x +1.5 = 6
–1.5
9y 36
9
9
y=4
—=—
–1.5
x = 4.5
The solution is (4.5, 1.5). So, you should mix 4.5 quarts of
100% fruit juice with 1.5 quarts of 20% fruit juice.
40
2
1h
=— h=—h
⋅—
60 min 60
3
60
1h
60 min ⋅ — = — h = 1 h
60 min 60
2
Words — ⋅ ( speed of boat + speed of current ) = 20
3
1 ⋅ ( speed of boat – speed of current ) = 20
34. 40 min
Substitute 4 for y in Equation 3, and solve for x.
5y = 10 – 2x
5(4) = 10 – 2x
20 = 10 – 2x
–10
–10
10 = –2x
10 –2x
—=—
–2
–2
–5 = x
Variables Let x be the speed (in miles per hour) of the boat,
and let y be the speed (in miles per hour) of the
current.
2
System — (x + y) = 20 ⇒
3
1 (x – y) = 20
258
2
3
2
3
—x + —y = 20
⇒
Algebra 1
Worked-Out Solutions
x – y = 20
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Chapter 5
Substitute –5 for x and 4 for y in Equation 1, and solve for z.
39.
–3(4 – 2v) = 6v – 12
x + 7y + 3z = 29
–3(4) – 3(–2v) = 6v – 12
–5 + 7(4) + 3z = 29
–12 + 6v = 6v – 12
–5 + 28 + 3z = 29
–6v
23 + 3z = 29
–23
–12 = –12
Because the statement –12 = –12 is always true, the
equation has infinitely many solutions. The solution is all
real numbers.
–23
3z = 6
3z 6
—=—
3
3
z=2
Check x + 7y + 3z = 29
?
–5 + 7(4) + 3(2) = 29
?
–5 + 28 + 6 = 29
40. y − y1 = m(x – x1)
y − 1 = –2(x – 4)
3z + x – 2y = –7
?
3(2) + (–5) – 2(4) = –7
?
6 – 5 – 8 = –7
29 = 29 ✓
y − 1 = –2(x) – 2(−4)
y − 1 = –2x + 8
+1
An equation of the parallel line is y = –2x + 9.
41. y = mx + b
6 = 5(0) + b
6=0+b
20 = 20 ✓
6=b
So, x = –5, y = 4, and z = 2.
Using m = 5 and b = 6, an equation of the parallel line is
y = 5x + 6.
Maintaining Mathematical Proficiency
36. 5d – 8 = 1 + 5d
–5d
–8 = 1
Because the statement –8 = 1 is never true, the equation has
no solution.
37. 9 + 4t = 12 – 4t
+4t
+4t
9 + 8t = 12
–9
–9
8t = 3
8t
8
3
8
3
t=—
8
3
The equation has one solution, which is t = —.
8
—=—
+1
y = –2x + 9
–7 = –7 ✓
5y = 10 – 2x
?
5(4) = 10 – 2(–5)
?
20 = 10 + 10
–5d
–6v
42.
y – y1 = m(x – x1)
2
y – (–2) = — (x – (–5))
3
2
y + 2 = — (x + 5)
3
2
2
y + 2 = — (x) + — (5)
3
3
10
2
y+2=—x+—
3
3
–2
–2
4
2
y = —x + —
3
3
4
2
An equation of the parallel line is y = —x + —.
3
3
38. 3n + 2 = 2(n – 3)
3n + 2 = 2(n) – 2(3)
3n + 2 = 2n – 6
–2n
–2n
n + 2 = –6
–2
–2
n = –8
The equation has one solution, which is n = –8.
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Algebra 1
Worked-Out Solutions
259
Chapter 5
Section 5.4
y
2.0
5.4 Explorations (p. 253)
Cost of
Initial
1. a. Words Cost =
+
materials
investment
Revenue =
Price per
skateboard ⋅
1.0
⋅ skateboards
Number of
0
C (dollars)
2
0
3
4
5
6
7
8
9
20 40 60 80 100 120 140 160 180 200
because for each skateboard you sell, you spend as much
as you make. So, you are spending money at the same rate
as you have money coming in, and you will never recover
your investment or make any money.
+
Number of
small beads
⋅
Weight per
small bead
Number of
large beads
⋅
Weight per
= Total cost
large bead
⋅ ⋅
20 ⋅ x + 3 ⋅ y = 5
Equations 40 x + 6 y = 10
So, a system of linear equations that represents this
situation is
40x + 6y = 10
.
20x + 3y = 5
b.
40x + 6y = 10
40x − 40x + 6y = 10 − 40x
6y = −40x + 10
−40x + 10
6
20
5
y = −— x + —
3
3
6y
6
—=—
20x + 3y = 5
20x − 20x + 3y = 5 − 20x
0.3
0.4
3.
x
yes; Sample answer: If the linear equations in a system
have the same rate of change, then the lines they describe
are parallel and will never intersect. So, the system has no
solution. If the linear equations in a system describe the
same line, then the system has infinitely many solutions.
The system y = 3x and y = 3x + 1 has no solution. The
system y = 3x and 2y = 6x has infinitely many solutions.
4. a. This system has one solution because the lines intersect at
one point.
b. This system has no solution because the lines are parallel
and will never intersect.
c. This system has infinitely many solutions because both
equations describe the same line.
5.4 Monitoring Progress (pp. 255–256)
1. Solve by elimination.
Step 1
x+y=3
Multiply by −2.
2x + 2y = 6
0=0
2. Solve by substitution.
2x + 2y = 4
2x + 2(−x + 3) = 4
−20x + 5
3
20
5
y = −—x + —
3
3
Algebra 1
Worked-Out Solutions
6
The equation 0 = 0 is always true. So, the solutions are
all the points on the line x + y = 3. The system of linear
equations has infinitely many solutions.
2x + 2(−x) + 2(3) = 4
3y
3
Step 2
−2x − 2y = −6
2x + 2y =
3y = −20x + 5
—=—
260
0.2
because both equations describe the same line, and there
are infinitely many points that are solutions of both
equations. So, there are infinitely many possibilities for
the weights of the beads.
10
b. Sample answer: Your company will never break even
2. a. Words
0.1
c. no; You cannot find the weight of each type of bead
450 470 490 510 530 550 570 590 610 630 650
R (dollars)
0
The two equations describe the same line.
R = 20x
1
20x + 3y = 5
0.5
Equations C = 450 + 20x
x (skateboards) 0
40x + 6y = 10
1.5
Number of
skateboards
2x − 2x + 6 = 4
6=4✗
The equation 6 = 4 is never true. So, the system of linear
equations has no solution.
Copyright © Big Ideas Learning, LLC
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Chapter 5
3. Solve by elimination.
Step 2 −(x + y = 3)
x + 2y = 4
y = −1
Step 3
x+y=3
x+1=3
−1
−1
x=2
The solution is (2, 1).
4. Solve by substitution.
10x + y = 10
10x + (−10x + 2) = 10
10x − 10x + 2 = 10
2 = 10 ✗
Monitoring Progress and Modeling with Mathematics
3. Equation 1
Let y = 0.
−x + y = 1
−x + y = 1
−x + 0 = 1
−0 + y = 1
−x = 1
y=1
1
−x
— =—
−1 −1
x = −1
Equation 2
Let y = 0.
Let x = 0.
x−y=1
x−y=1
x−0=1
0−y=1
x=1
−y = 1
−y
−1
y-intercepts, so they are parallel. So, the system has no solution.
5.4 Exercises (pp. 257 –258)
Vocabulary and Core Concept Check
1. Two lines cannot intersect in exactly two points. So, a system
of linear equations cannot have exactly two solutions.
2. The graph of a linear system that has infinitely many
solutions is a single line because both equations describe that
same line. The graph of a linear system that has no solution
is two parallel lines that never intersect.
1
−1
—=—
The equation 2 = 10 is never true. So, the system of linear
equations has no solution.
5. The lines still have the same slope, but they now have different
Let x = 0.
y = −1
F; Equation 1 has an x-intercept of −1 and a y-intercept
of 1. Equation 2 has an x-intercept of 1 and a y-intercept
of −1. So, graph F matches this system. Because the lines
are parallel, they do not intersect. So, the system of linear
equations has no solution.
4. Equation 1
Equation 2
2x − 2y = 4
−x + y = −2
2x − 2x − 2y = 4 − 2x
−x + x + y = −2 + x
−2y = −2x + 4
y=x−2
−2y −2x + 4
−2
−2
y=x−2
—=—
E; The slope-intercept forms of Equations 1 and 2 are the
same, with a slope of 1 and a y-intercept of −2. So, both
equations in this system describe the same line in graph E.
All points on the line are solutions of both equations. So, the
system of linear equations has infinitely many solutions.
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Algebra 1
Worked-Out Solutions
261
Chapter 5
5. Equation 1
8. Equation 1
Let x = 0.
Let y = 0.
2x + y = 4
2x + y = 4
2(0) + y = 4
2x + 0 = 4
3y = −5x + 17
−3y = −x − 2
0+y=4
2x = 4
y=4
—=—
3y −5x + 17
—=—
3
3
5
17
y = −—x + —
3
3
−3y −x − 2
—=—
−3
−3
1
2
y = —x + —
3
3
5x + 3y = 17
x − 3y = −2
5x − 5x + 3y = 17 − 5x
x − x − 3y = −2 − x
2x 4
2
2
x=2
5
17
A; For Equation 1, the slope is −—3 and the y-intercept is —
,
3
Equation 2
Let x = 0.
Let y = 0.
−4x −2y = −8
−4x − 2y = −8
−4(0) − 2y = −8
−4x −2(0) = −8
0 − 2y = −8
−4x − 0 = −8
−2y = −8
−4x = −8
−2y −8
—=—
−2
−2
y=4
−4x −8
—=—
−4
−4
x=2
and for Equation 2, the slope is —13 and the y-intercept is —23. So,
graph A matches this system. Because the lines intersect in
one point, the system has one solution.
9. Solve by graphing.
y
−4
B; Equations 1 and 2 both have an x-intercept of 2 and a
y-intercept of 4. So, both equations describe the same line
in graph B. All points on the line are solutions of both
equations. So, the system of linear equations has infinitely
many solutions.
Equation 2
x−y=0
5x − 2y = 6
x−x−y=0−x
5x − 5x − 2y = 6 − 5x
−y = −x
−2y = −5x + 6
−y −x
—=—
−1
−1
−2y −5x + 6
—=—
−2
−2
5
y=x
y = — x −3
2
C; For Equation 1, the slope is 1 and the y-intercept is 0,
and for Equation 2, the slope —52 and the y-intercept is −3. So,
graph C matches this system. Because the lines intersect in
one point, the system has one solution.
7. Equation 1
Equation 2
3x − 6y = 9
−2x + 2x + 4y = 1 + 2x
3x − 3x − 6y = 9 − 3x
2x + 1
4
1
1
y = —x + —
2
4
−6y
−6
−3x + 9
−6
1
3
y = —x − —
2
2
4y
4
—=—
1
—2
D; For Equation 1, the slope is and the y-intercept is
1
—2
y = −2x − 4
?
−4 = −2(0) − 4
?
−4 = 0 − 4
y = 2x − 4
?
−4 = 2(0) − 4
?
−4 = 0 − 4
−4 = −4 ✓
−4 = −4 ✓
The solution of the system of linear equations is (0,−4).
10. Solve by graphing.
8
y
y = −6x + 8
4
−8
−4
4
8 x
The lines both have a slope of −6, but they have different
y-intercepts. So, they are parallel. Because parallel lines
do not intersect, there is no point that is a solution of both
equations. So, the system of linear equations has no solution.
11. Solve by elimination.
3x − y = 6
1
—4,
and for Equation 2, the slope is and the y-intercept is
3
−—2.
So, graph D matches this system. Because the lines have
the same slope and different y-intercepts, they are parallel.
Because parallel lines do not intersect, there is no point
that is a solution of both equations. So, the system of linear
equations has no solution.
Algebra 1
Worked-Out Solutions
−4 (0, −4)
−8
−6y = −3x + 9
—=—
y = 2x − 4
y = −6x − 8
−2x + 4y = 1
4y = 2x + 1
4 x
y = −2x − 4
Check
6. Equation 1
262
Equation 2
−3x + y = −6
0=0
The equation 0 = 0 is always true. So, the solutions are
all the points on the line 3x − y = 6. The system of linear
equations has infinitely many solutions.
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Chapter 5
12. Solve by elimination.
17. Equation 1
−x + 2y = 7
Equation 2
y = 7x + 13
−21x + 3y = 39
x − 2y = 7
−21x + 21x + 3y = 39 + 21x
0 = 14 ✗
3y = 21x + 39
The equation 0 = 14 is never true. So, the system of linear
equations has no solution.
13. Solve by elimination.
Step 1
4x + 4y = −8
−2x − 2y = 4
Step 2
4x + 4y = −8
−4x − 4y = 8
Multiply by 2.
0=0
The equation 0 = 0 is always true. So, the solutions are all
the points on the line 4x + 4y = −8. The system of linear
equations has infinitely many solutions.
3y 21x + 39
3
3
y = 7x + 13
—=—
Both equations have a slope of 7 and a y-intercept of 13.
So, the lines are the same. Because the lines are the same,
all points on the line are solutions of both equations. So, the
system of linear equations has infinitely many solutions.
18. Equation 1
Equation 2
y = −6x − 2
12x + 2y = −6
12x − 12x + 2y = −6 − 12x
14. Solve by elimination.
2y = −12x − 6
Step 1 15x − 5y = −20
−3x + y = 4
Step 2 15x − 5y = −20
0=0
The equation 0 = 0 is always true. So, the solutions are all
the points on the line −3x + y = 4. The system of linear
equations has infinitely many solutions.
15. Solve by elimination.
Step 1
Multiply by 2.
6x − 10y = −16
Multiply by −3.
18x − 30y = 48
−18x + 30y = 48
0 = 96
The equation 0 = 96 is never true. So, the system of linear
equations has no solution.
16. Solve by elimination.
Step 1
3x − 2y = −5
Multiply by 5.
15x − 10y = −25
4x + 5y = 47
Multiply by 2.
8x + 10y = 94
23x + 0 = 69
Step 4 3x − 2y = −5
Step 3 23x = 69
3(3) − 2y = −5
23x 69
—=—
23
23
x=3
−9
−9
−2y = −14
−2y −14
−2
−2
y=7
—=—
The solution of the system of linear equations is (3, 7).
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Equation 2
4x + 3y = 27
4x − 3y = −27
4x − 4x + 3y = 27 − 4x
4x − 4x − 3y = −27 − 4x
3y = −4x + 27
−3y = −4x − 27
3y −4x + 27
—=—
3
3
4
y = −— x + 9
3
−3y −4x − 27
—=—
−3
−3
4
y = —x + 9
3
4
Step 2
9 − 2y = −5
The lines both have a slope of −6, but they have different
y-intercepts. So, the lines are parallel. Because parallel lines
do not intersect, there is no point that is a solution of both
equations. So, the system of linear equations has no solution.
19. Equation 1
Step 2
9x − 15y = 24
2y −12x − 6
2
2
y = −6x − 3
—=—
−15x + 5y = 20
Multiply by 5.
The slope of Equation 1 is −—3, and the slope of Equation 2
is —34. So, the lines do not have the same slope, but they both have
a y-intercept of 9. So, they intersect at (0, 9) on the y-axis.
Therefore, the system of linear equations has one solution,
which is (0, 9).
20. Equation 1
Equation 2
−7x + 7y = 1
2x − 2y = −18
−7x + 7x + 7y = 1 + 7x
2x − 2x −2y = −18 − 2x
7y = 7x + 1
−2y = −2x − 18
7x + 1
7
1
y=x+—
7
7y
7
—=—
−2y
−2
−2x − 18
−2
—=—
y=x+9
Both lines have a slope of 1, but they have different
y–intercepts. So, the lines are parallel. Because parallel lines
do not intersect, there is no point that is a solution of both
equations. So, the system of linear equations has no solution.
Algebra 1
Worked-Out Solutions
263
Chapter 5
21. Equation 1
Equation 2
−18x + 6y = 24
3x − y = −2
−18x + 18x + 6y = 24 + 18x
3x − 3x − y = −2 − 3x
6y = 18x + 24
−y = −3x − 2
6y 18x + 24
—=—
6
6
y = 3x + 4
−y −3x − 2
—=—
−1
−1
y = 3x + 2
Both lines have a slope of 3, but they have different
y-intercepts. So, the lines are parallel. Because parallel lines
do not intersect, there is no point that is a solution of both
equations. So, the system of linear equations has no solution.
22. Equation 1
Equation 2
3x − 6y = 30
2x − 2x − 2y = 16 − 2x
3x − 3x − 6y = 30 − 3x
−2y = −2x + 16
−6y = −3x + 30
−2y
−2
—=—
−2x + 16
−2
−6y
−6
−3x + 30
−6
1
y = — x −5
2
y=x−8
Equation 1 has a slope of 1 and a y-intercept of −8. Equation 2
has a slope of —21 and a y-intercept of −5. Because the equations
do not have the same slope or y-intercept, they must intersect
in one point that is not on the y-axis. So, the system of linear
equations has one solution.
23. Only lines that have the same slope are parallel and will
never intersect. These lines will intersect at one point if they
are extended. So, the system has one solution.
Equation 1
−4x + y = 4
4x + y = 12
4x − 4x + y = 12 − 4x
y = 4x + 4
y = −4x + 12
y
12
−2
3
4 —12
⋅
⋅
x
+
4
x
+
6
Cost per
cup of = Total
⋅ almonds
Cost
⋅
⋅
Step 1
(1, 8)
2
4 x
−4x + y = 4
?
−4(1) + 8 = 4
?
−4 + 8 = 4
4=4✓
y
=
6
y
=
9
Step 2
3x + 4y = 6
Multiply by 3.
9
—2 x
Multiply by −2.
+ 6y = 9
9x + 12y =
18
−9x − 12y = −18
0=0
The equation 0 = 0 is always true. In this context, x and y
must be positive. So, the solutions are all points on the line
3x + 4y = 6 in Quadrant I. The system of linear equations
has infinitely many solutions.
26. Words Team A: Distance
Team A’s
(in miles) =
speed
traveled
Team B: Distance
Team B’s
(in miles) =
speed
traveled
⋅
Time
(in hours)
−2
it takes to
finish
⋅
Time
(in hours)
it takes to
finish
Variables Let x be how long (in hours) it takes to finish
the race, and let y be how far (in miles) the team
travels for the remainder of the race.
System Team A: y = 6x − 2
Because the two teams are traveling at the same speed,
Team B will not catch up to Team A. The graphs of these
lines are parallel because they have the same slope. So,
there is no point that is a solution of both equations, and the
system of linear equations has no solution.
4x + y = 12
Check
System
⋅
Cost
Amount
per
(in cups)
cup of +
of
dried
almonds
fruit
Team B: y = 6x
−4x + y = 4
4
−4
Amount
(in cups)
25. Words
of dried
fruit
Equation 2
−4x + 4x + y = 4 + 4x
8
y-intercepts. So, the lines are parallel. Because parallel lines
do not intersect, there is no point that is a solution of both
equations. So, the system of linear equations has no solution.
Solve by elimination.
2x − 2y = 16
—=—
24. The lines do have the same slope of 3, but they have different
4x + y = 12
?
4(1) + 8 = 12
?
4 + 8 = 12
12 = 12 ✓
The lines intersect at (1, 8), which is the solution of the
system.
27. Words Number
of
coach
tickets
⋅
Number
Cost
of
per
+ business
coach
class
ticket
tickets
⋅
Cost per
Money
business
=
collected
class
ticket
Variables Let x be the cost (in dollars) of one coach ticket,
and let y be the cost (in dollars) of one business
class ticket.
System 150
170
⋅
⋅
x
+
80
x
+
100
⋅
⋅
y
= 22,860
y
= 27,280
A system of equations is 150x + 80y = 22,860 and
170x + 100y = 27,280.
264
Algebra 1
Worked-Out Solutions
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All rights reserved.
Chapter 5
Equation 1
31. a. never; The y-intercept of y = ax + 4 is 4, and the
150x + 80y = 22,860
y-intercept of y = bx − 2 is −2. Because the lines do not
have the same y-intercept, they are not the same line. So,
they cannot have infinitely many solutions.
150x −150x + 80y = 22,860 − 150x
80y = 22,860 −150x
b. sometimes; If a = b, then the lines will have the same
15
1143
y = −—x + —
8
4
slope and therefore be parallel. So, the system of linear
equations would have no solution.
c. always; When a < b, the lines do not have the same slope.
So, the graphs are neither parallel lines nor the same line.
Therefore, they must intersect in exactly one point, which
means the system always has one solution.
Equation 2
170x + 100y = 27,280
170x −170x + 100y = 27,280 − 170x
100y = −170x + 27,280
1364
17
y = − —x + —
10
5
32. no; You cannot determine the exact costs with the
information given.
15
1143
Equation 1 has a slope of −— and a y-intercept of — .
8
4
17
1364
Equation 2 has a slope of −— and a y-intercept of — .
10
5
Because the equations do not have the same slope or
y-intercept, they must intersect in one point that is not
on the y-axis. So, the system of linear equations has
one solution.
28. Sample answer: y = x + 1
y = −x + 3
y=
1
—2 x
−1
y = −x + 3
−4
(−4, −3)
30. a. Sample answer: Team C’s runner passed Team B’s runner
at about 40 meters.
b. yes; Team C’s runner would have passed Team A’s runner
eventually. The lines that represent Team A’s and Team C’s
runners have different slopes, so they will intersect.
c. no; Team B’s runner could not have passed Team A’s
runner. Because Team B’s runner and Team A’s runner are
running at the same speed, the lines that represent them
have the same slope and are therefore parallel. So, the
lines will never intersect.
Copyright © Big Ideas Learning, LLC
All rights reserved.
2
x
+
10
⋅
⋅
y
=
y
= 190
38
–15x – 10y = –190
Multiply by −5.
15x + 10y = 190
0=0
Maintaining Mathematical Proficiency
33.
∣ 2x + 6 ∣ = ∣ x ∣
2x + 6 = x
same slope, then the lines are neither parallel nor the same.
So, the lines must intersect in one point. This point of
intersection is the one and only solution of the system.
+
The equation 0 = 0 is always true. So, the system of linear
equations has infinitely many solutions. In this context, x and
y must be positive. So, the solutions are all points on the line
3x + 2y = 38 in Quadrant I. Therefore, there are infinitely
many possibilities for the exact cost of one admission and
one skate rental.
( 83 , 13 )
29. The system has one solution. If two lines do not have the
x
15x + 10y = 190
Equation 3
1
y = 2x − 1
⋅
⋅
⋅
Cost
Total
per
=
cost
skate
rental
Step 2
3x + 2y = 38
Equation 2
x
3
⋅
Number
Cost per
+ of skate
admission
rentals
Step 1
Equation 1
y
2
−2
System
15
(1, 2)
2
y=x+1
Number
of
admissions
Solve by elimination.
As shown in the graph, each pair of lines intersects in one
point, but there is no point that is a solution of all three
equations.
4
Words
–2x
2x + 6 = –x
or
–2x
–2x
–2x
6 = –x
6 = –3x
—=—
–x
–1
—=—
–6 = x
–2 = x
6
–1
Check
6
–3
∣ 2x + 6 ∣ = ∣ x ∣
–3x
–3
∣ 2x + 6 ∣ =∣ x ∣
?
∣ 2( –6 ) + 6 ∣ = ∣ –6 ∣
?
∣ –12 + 6 ∣ = 6
?
∣ –6 ∣ = 6
∣ 2( –2 ) + 6 ∣ = ∣ –2 ∣
?
6=6✓
2=2✓
?
∣ –4 + 6 ∣ = 2
?
∣2∣ = 2
The solutions are x = –6 and x = –2.
Algebra 1
Worked-Out Solutions
265
Chapter 5
34. ∣ 3x – 45 ∣ = ∣ 12x ∣
3x – 45 = 12x
–3x
or
–3x
Check ∣ 2x + 1 ∣ = ∣ 3x – 11 ∣
?
∣ 2( 12 ) + 1 ∣ = ∣ 3( 12 ) – 11 ∣
?
∣ 24 + 1 ∣ = ∣ 36 – 11 ∣
?
∣ 25 ∣ = ∣ 25 ∣
3x – 45 = –12x
–3x
–3x
–45 = 9x
–45 = –15x
–45 9x
—=—
9
9
–5 = x
–45 –15x
—=—
–15
–15
3=x
Check ∣ 3x – 45 ∣ = ∣ 12x ∣
?
∣ 3( 3 ) – 45 ∣ = ∣ 12( 3 ) ∣
?
∣ –15 – 45 ∣ = ∣ –60 ∣
?
∣ –60 ∣ = 60
?
∣ 9 – 45 ∣ = ∣ 36 ∣
?
∣ –36 ∣ = 36
36 = 36 ✓
The solutions are x = –5 and x = 3.
35. ∣ x – 7 ∣ = ∣ 2x – 8 ∣
x – 7 = 2x – 8
–x
x – 7 = −( 2x – 8 )
or
–x
x – 7 = –2x + 8
–7 = x – 8
+8
+2x
+8
+2x
3x – 7 = 8
1=x
+7
+7
3x = 15
3x 15
3
3
x=5
—=—
Check ∣ x – 7 ∣ = ∣ 2x – 8 ∣
?
∣ 1 – 7 ∣ = ∣ 2( 1 ) – 8 ∣
?
∣ –6 ∣ = ∣ 2 – 8 ∣
?
6 = ∣ –6 ∣
∣ x – 7 ∣ = ∣ 2x – 8 ∣
?
∣ 5 – 7 ∣ = ∣ 2( 5 ) – 8 ∣
?
∣ –2 ∣ = ∣ 10 – 8 ∣
2 = ∣2∣
6=6✓
2=2✓
The solutions are x = 1 and x = 5.
36. ∣ 2x + 1 ∣ = ∣ 3x –11 ∣
2x + 1 = 3x – 11
–2x
2x + 1 = – ( 3x – 11 )
or
–2x
1 = x – 11
+11
2x + 1 = –3x + 11
+3x
+11
12 = x
+3x
5x + 1 = 11
–1
–1
5x = 10
5x
5
10
5
—=—
x=2
266
Algebra 1
Worked-Out Solutions
?
∣ 4 + 1 ∣ = ∣ 6 – 11 ∣
?
∣ 5 ∣ = ∣ –5 ∣
5=5✓
The solutions are x = 12 and x = 2.
∣ 3( –5 ) – 45 ∣ = ∣ 12( –5 ) ∣
60 = 60 ✓
?
∣ 2( 2 ) + 1 ∣ = ∣ 3( 2 ) – 11 ∣
25 = 25 ✓
∣ 3x – 45 ∣ = ∣ 12x ∣
?
∣ 2x + 1 ∣ = ∣ 3x – 11 ∣
5.1–5.4 What Did You Learn? (p. 259)
1. You know the total number of songs played, the relationship
between the number of pop songs played and the number
of rock songs played, and the relationship between the
number of hip-hop songs played and the number of rock
songs played. The solution can be found by writing a system
of three linear equations in three variables that represents
the problem. Then, substitute an expression for x and an
expression for z into the first equation that contains all three
variables. Solve this equation for y. Substitute the value of
y into each of the other two equations and solve for x and z,
respectively.
2. Sample answer: An Internet site offers commercial-free
viewing of individual episodes of a TV show for one price
or access to an entire season of a TV show for another price.
If you knew how many individual shows and seasons you
purchased in Month 1 and Month 2 and the total charge for
each of those months, you could write a system of linear
equations, similar to the one in Exercise 22, that could be
solved to find the cost of viewing one episode and the cost
for access to an entire season.
3. Sample answer: What are the slope and y-intercept of the
line that describes the first receipt? the second receipt? How
are these two equations related? What does that tell you
about the system of linear equations?
5.1–5.4 Quiz (p. 260)
1. The lines appear to intersect at (3, 1).
1
Check y = −—3 x + 2
? 1
1 = −—3 (3) + 2
?
1 = −1 + 2
y=x−2
?
1=3−2
1=1✓
1=1✓
The solution is (3, 1).
2. The lines appear to intersect at (−2, −2).
Check
y = —12 x − 1
y = 4x + 6
? 1
−2 = —2 (−2) − 1
?
−2 = −1 − 1
?
−2 = 4(−2) + 6
?
−2 = −8 + 6
−2 = −2 ✓
−2 = −2 ✓
The solution is (−2, −2).
Copyright © Big Ideas Learning, LLC
All rights reserved.
Chapter 5
3. The lines appear to intersect at (0, 1).
Check y = 1
1=1✓
x + 4y = 10
6. Step 1
y = 2x + 1
?
1 = 2(0) + 1
?
1=0+1
x + 4y − 4y= 10 − 4y
x = 10 − 4y
3x − 5y = 13
Step 3 x + 4y = 10
3(10 − 4y) − 5y = 13
x + 4(1) = 10
3(10) − 3(4y) − 5y = 13
x + 4 = 10
Step 2
1=1✓
The solution is (0, 1).
30 − 12y − 5y = 13
4. Substitute x − 4 for y in Equation 2 and solve for x.
−2x + y = 18
30 − 17y = 13
Step 3 y = x − 4
−2x + (x − 4) = 18
−30
y = −22 − 4
−2x + x − 4 = 18
−17y −17
−17
−17
y=1
+4
Check
y=x−4
?
−26 = −22 − 4
−2x + y = 18
?
−2(−22) + (−26) = 18
?
44 − 26 = 18
−26 = −26 ✓
13 = 13 ✓
7. Step 2
Step 3
y − x + x = −5 + x
y=x−5
2y + x = −4
Step 3 y − x = −5
2(x − 5) + x = −4
y − 2 = −5
+2
2x − 10 + x = −4
+10
3x = 6
3x 6
3
3
x=2
2y + x = −4
?
2(−3) + 2 = −4
?
−6 + 2 = −4
x+y=4
Step 4
−2x = −4
−2x −4
—=—
−2
−2
x=2
x+y=4
2+y=4
−2
−2
y=2
Check x + y = 4
−3x − y = −8
?
2+2=4
?
−3(2) − 2 = −8
4=4✓
—=—
Check
+2
y = −3
3x − 10 = −4
+10
10 = 10 ✓
−2x + 0 = −4
y − x = −5
2(x) − 2(5) + x = −4
x + 4y = 10
?
6 + 4(1) = 10
?
6 + 4 = 10
−3x − y = −8
The solution is (−22, −26).
Step 2
3x − 5y = 13
?
3(6) − 5(1) = 13
?
18−5 = 13
The solution is (6, 1).
18 = 18 ✓
5. Step 1
x=6
−30
—=—
−x = 22
22
−x
—=—
−1 −1
x = −22
Check
−4
−17y = −17
y = −26
−x − 4 = 18
+4
−4
?
−6 − 2 = −8
−8 = −8 ✓
y − x = −5
?
−3 −2 = −5
The solution is (2, 2).
−5 = −5 ✓
−4 = −4 ✓
The solution is (2, −3).
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Algebra 1
Worked-Out Solutions
267
Chapter 5
8. Step 1
11. Solve by elimination.
Step 2
x + 3y = 1
−2x − 6y = −2
Multiply by −2.
5x + 6y = 14
Step 1 6x + 2y = 16
5x + 6y = 14
2x − y = 2
Step 2 6x + 2y = 16
4x − 2y = 4
Multiply by 2.
3x + 0 = 12
Step 4 x + 3y = 1
Step 3
4 + 3y = 1
−4
−4
3x = 12
3x 12
—=—
3
3
x=4
3y = −3
3y −3
—=—
3
3
y = −1
Check
10x 20
10
10
x=2
—=—
12 + 2y = 16
−12
2y = 4
x + 3y = 1
?
4 + 3(−1) = 1
?
4−3=1
5x + 6y = 14
?
5(4) + 6(−1) = 14
?
20 − 6 = 14
14 = 14 ✓
2y 4
—=—
2
2
y=2
The solution is (2,2).
12. Solve by elimination.
Step 1
Step 2
3x − 3y = −2
The solution is (4, −1).
Multiply by 2.
−6x + 6y = 4
9. Step 1
6x − 6y = −4
−6x + 6y = 4
Step 2
2x − 3y = −5
Multiply by 2.
5x + 2y = 16
Multiply by 3.
0=0
4x − 6y = −10
15x + 6y = 48
19x + 0 = 38
5x + 2y = 16
Step 3
5(2) + 2y = 16
10 + 2y = 16
−10
19x = 38
19x 38
—=—
19
19
x=2
−10
The equation 0 = 0 is always true. So, the solutions are
all points on the line 3x − 3y = −2. The system of linear
equations has infinitely many solutions.
13. a. Words
14
+ 4
8
+
2y = 6
2y
2
6
2
y=3
2x − 3y = −5
?
2(2) − 3(3) = −5
?
4 − 9 = −5
6
⋅
Growing
time
(in years)
=
Height
(in inches)
⋅
Growing
time
(in years)
=
Height
(in inches)
Variables Let x be how long (in years) the trees are
growing, and let y be the height (in inches) of
the trees.
—=—
Check
10x = 20
Step 3
6(2) + 2y = 16
−12
1=1✓
Step 4
10x + 0 = 20
Step 4 6x + 2y = 16
5x + 2y = 16
?
5(2) + 2(3) = 16
?
10 + 6 = 16
−5 = −5 ✓
System 14 + 4x = y
8 + 6x = y
A system of linear equations that represents this situation
is y = 4x + 14 and y = 6x + 8.
16 = 16 ✓
The solution is (2, 3).
10. Solve by elimination.
Step 2
x−y=1
−(x − y = 6)
0 = −5
The equation 0 = −5 is never true. So, the system of linear
equations has no solution.
268
Algebra 1
Worked-Out Solutions
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Chapter 5
Step 3 x + y = 3
y
Height (inches)
b.
1+y=3
40
y = 4x + 14
−1
30
(3, 26)
20
10
0
−1
y=2
The solution is (1,2). So, you spend 1 hour driving at
55 miles per hour on highways, and you spend 2 hours
driving at 40 miles per hour on the rest of the roads.
y = 6x + 8
0
1
2
3
b. You drive 55x = 55(1) = 55 miles on highways and
x
4
40y = 40(2) = 80 miles on the rest of the roads.
Time (years)
Check
y = 4x + 14
?
26 = 4(3) + 14
?
26 = 12 + 14
y = 6x + 8
?
26 = 6(3) + 8
?
26 = 18 + 8
26 = 26 ✓
26 = 26 ✓
15. Words
7
The solution is (3, 26). So, in 3 years, both trees will be
26 inches tall.
14. a. Words
Time
(in hours)
on highway
+
Time
(in hours)
on other roads
=3
Number of
touchdowns
⋅
Number of
field goals
+
Number of
+ 3
touchdowns
⋅
=6
Number of
= 26
field goals
Variables Let x be the number of touchdowns the home
team scores, and let y be the number of field goals
the home team scores.
System x + y = 6
7x + 3y = 26
Solve by elimination.
55
⋅
Time
(in hours) + 40
on highway
⋅
Time
(in hours) = 135
on other roads
Variables Let x be how much time (in hours) you spend
driving at 55 miles per hour on highways, and
let y be how much time (in hours) you spend
driving at 40 miles per hour on the rest of
the roads.
System x + y = 3
Step 2
x+y=6
Multiply by −3.
7x + 3y = 26
−3x − 3y = −18
7x + 3y = 26
4x + 0 = 8
Step 4 x + y = 6
2+y=6
−2
−2
Step 3
4x = 8
4x 8
—=—
4
4
x=2
y=4
55x + 40y = 135
The solution is (2, 4). So, the home team scores
2 touchdowns and 4 field goals.
Solve by substitution.
Step 1
Step 1
x+y=3
x−x+y=3−x
y=3−x
Step 2
55x + 40y = 135
55x + 40(3−x) = 135
55x + 40(3) − 40(x) = 135
55x + 120 − 40x = 135
15x + 120 = 135
−120
−120
15x = 15
15x
15
15
15
—=—
x=1
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Algebra 1
Worked-Out Solutions
269
Chapter 5
Section 5.5
c. Method 1
2
−—3 x
5.5 Explorations (p. 261)
1. a. A linear equation that uses the left side is y = 2x − 1.
A linear equation that uses the right side is y =
b.
1
−—2 x
+ 4.
2
2
+—3 x
+4
+4
4
x
6
Method 2
4
5
c. The two sides of the equation are equal to each other.
If you set one side of the equation equal to y, the transitive
property allows you to set the other side of the equation
equal to y.
1
1
y = −4 x + 1
1
+—4 x
6
y
−4
1
—3 x
= −1
⋅
6
Method 2
y
+x
2
+0.5
y = −x + 2.5
1
−1
3 3x
3
3
1=x
—=—
2
1
x
y = 2x − 0.5
2
−4
Method 2
+3x
y
3
+3x
−1.5
y = 3x + 4
(−3, 2)
−4
11
5
1.5 = 4x + 1.5
1
−4
5
11
−3x + 1.5 = x + 1.5
Method 2
+4=3
22
5
f. Method 1
—3 x + 4 = —3 x + 3
1
—3 x
⋅— = — ⋅—x
Using either method, the solution is x = 1.
Using either method, the solution is x = −4.
1
y = 3x − 3
(1, 1.5)
x
x = −4
−—3 x
4 x
3.0 = 3x
4
3
1
11
x
—
5
+0.5
⋅ — x = — ⋅ (−3)
−—3 x
22
—
5
+x
−2
3
b. Method 1
+3
2.5 = 3x − 0.5
−2
—4 x = −3
2
−x + 2.5 = 2x − 0.5
2
−6
−4
Using either method, the solution is x = 2.
(−4, 2)
−4
(2, 3)
−2
+3
=
y
2=x
1
3
2
—
11
y = 2x + 4
—4 x + 4 = 1
11
e. Method 1
Method 2
1
−—5 x
7
3=3✓
—2 x + 4 = −—4 x + 1
4
4
7
y = 5x + 5
x−3
—5 = —
5
1
2x − 1 = −—2 x + 4
? 1
2(2) − 1 = −—2 (2) + 4
?
4 − 1 = −1 + 4
3
4
2
y = −3 x − 1
+ = 3x − 3
The x-value of the point of intersection is 2.
4
y = 3x − 4
7
—5
4
—5 x
4
—3
1
−6
d. Method 1
−—5 x
−4
(3, −3)
Using either method, the solution is x = 3.
2
2. a. Method 1
6 x
4
2
(2, 3)
−2
1
−2
3=x
2
+—4 x
y
−1 = x − 4
1
Check
−4
−2
y = 2x − 1
y =− 2 x + 4
−1=
+—3 x
y
6
Method 2
1
—3 x
y = x + 1.5
−1.5
1
y
(0, 1.5)
y = −3x + 1.5
0 = 4x
2
1
y = 3x + 3
−2
2
−2
x
0 4x
4
4
0=x
—=—
−1
1
2 x
−1
Using either method, the solution is x = 0.
⋅
3 —13 x = 3 (−1)
x = −3
Using either method, the solution is x = −3.
270
Algebra 1
Worked-Out Solutions
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Chapter 5
3. Sample answer: First, use the left side of the original
3.
equation to write a linear equation. Then use the right side
of the original equation to write a second linear equation.
Graph these two linear equations, and find the x-value of
the point of intersection. This value is the solution of the
original equation.
5.5 Monitoring Progress (pp. 262–264)
1
1. —2 x − 3 = 2x
y
2
y = 2x
y = 2x
The graphs intersect at (−2, −4).
1
—2 x − 3 = 2x
Check
1
—2
−4
−2
x
2
System 1
System 2
y = 2x + 2
y = 2x + 2
y=x−2
y = −x + 2
y
y = 2x + 2
−4
−2
y
2
4
x
y=x−2
−4
y = −x + 2
−4
−2
−4
1
y = 2x − 3
y = 2x + 2
(0, 2)
2
4 x
−2
(−4, −6)
The graphs intersect
at (−4, −6).
The graphs intersect
at (0, 2).
Check
Check
∣ 2x + 2 ∣ = ∣ x − 2 ∣
?
∣ 2(−4) + 2 ∣ = ∣ −4 − 2 ∣
?
∣ −8 + 2 ∣ = ∣ −6 ∣
?
∣ −6 ∣ = 6
∣ 2x + 2 ∣ = ∣ x − 2 ∣
?
∣ 2(0) + 2 ∣ = ∣ 0 − 2 ∣
?
∣ 0 + 2 ∣ = ∣ −2 ∣
?
∣2∣ = 2
6=6✓
(−2, −4)
?
(−2) − 3 = 2(−2)
?
−1 − 3 = −4
2x + 2 = −(x − 2)
2x + 2 = −x + 2
an equation with variables on both sides, you will always
be able to get an exact answer, even if the solution is not
a whole number. However, you sometimes have to work
with fractions and decimals, which can be tedious, and it is
possible to make careless errors. When you use the graphical
method to solve an equation with variables on both sides,
you see a visual representation of how the value of each
expression changes as the value of x changes. However, it
can be tedious to find the appropriate scale for the axes, and
the graphical method may only provide an estimate of the
solution, especially if the solution is not a whole number.
y = —12x − 3
Equation 2
2x + 2 = x − 2
4. Sample answer: When you use the algebraic method to solve
Graph the system.
∣ 2x + 2 ∣ = ∣ x − 2 ∣
Equation 1
2=2✓
So, the solutions are x = −4 and x = 0.
−4 = −4 ✓
So, the solution of the equation is x = −2.
2. −4 + 9x = −3x + 2
Graph the system.
4
y = −4 + 9x
2
y
y = −3x + 2
y = −4 + 9x
(0.5, 0.5)
The graphs intersect
at (0.5, 0.5).
Check −4 + 9x = −3x +2
?
−4 + 9(0.5) = −3(0.5) + 2
?
−4 + 4.5 = −1.5 + 2
−1
2
−2
3 x
y = −3x + 2
−4
0.5 = 0.5 ✓
So, the solution of the equation is x = 0.5.
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Algebra 1
Worked-Out Solutions
271
Chapter 5
4. ∣ x − 6 ∣ = ∣ −x + 4 ∣
2. For the first system, set y equal to both expressions inside the
Equation 1
absolute value symbols, 2x − 4 and −5x + 1. For the second
system, set y equal to the expression inside the first absolute
value symbol, 2x − 4, and the opposite of the expression
inside the second absolute value symbol, 5x − 1.
Equation 2
x − 6 = −x + 4
x − 6 = −(−x + 4)
x−6=x−4
System 1
System 2
y=x−6
y=x−6
y = −x + 4
y=x−4
y
Monitoring Progress and Modeling with Mathematics
3. The graphs of y = −2x + 3 and y = x intersect at (1, 1).
Check
y
y = −x + 4
−4
−2
2
−2x + 3 = x
?
−2(1) + 3 = 1
?
−2 + 3 = 1
4 x
−2
2
x
4
2
(5, −1)
−2
y=x−4
y=x−6
y=x−6
−4
The graphs intersect
at (5, −1).
−8
The lines are parallel and
do not intersect.
1=1✓
So, the solution of the original equation is x = 1.
4. The graphs of y = −3 and y = 4x + 1 intersect at (−1, −3).
Check
−3 = 4x + 1
?
−3 = 4(−1) + 1
?
−3 = −4 + 1
Check
∣ x − 6 ∣ = ∣ −x + 4 ∣
?
∣ 5 − 6 ∣ = ∣ −5 + 4 ∣
−3 = −3 ✓
?
∣ −1 ∣ = ∣ −1 ∣
So, the solution of the original equation is x = −1.
1=1✓
1
5. The graphs of y = −x − 1 and y = —3 x + 3 intersect at
So, the solution is x = 5.
5. Words
Cost
per
mile
Company A
⋅
(−3, 2).
Company C
Cost
Flat
Miles +
per
=
fee
mile
⋅
Check
−x − 1 = —13 x + 3
Flat
Miles +
fee
?
−(−3) − 1 = —13 (−3) + 3
Variable Let x be the number of miles traveled.
?
3 − 1 = −1 + 3
Equation 3.25x + 125 = 3.3x + 115
Use a graphing calculator to graph the system.
y = 3.25x + 125
y = 3.3x + 115
2=2✓
So, the solution of the original equation is x = −3.
3
6. The graphs of y = −—2 x – 2 and y = −4x + 3 intersect at
(2, −5).
900
Check
3
−—2 x − 2 = −4x + 3
Intersection
Y=775
0 X=200
0
250
Because the graphs intersect at (200, 775), the solution of
the equation is x = 200. So, the total costs are the same after
200 miles.
?
3
−—2 (2) − 2 = −4(2) + 3
?
−3 − 2 = −8 + 3
−5 = −5 ✓
So, the solution of the original equation is x = 2.
5.5 Exercises (pp. 265–266)
Vocabulary and Core Concept Check
1. The solution of the equation is x = 6.
272
Algebra 1
Worked-Out Solutions
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Chapter 5
7. x + 4 = −x
10. −2x + 6= 5x − 1
Graph the system.
y=x+4
6
y = −x
Graph the system.
y
y = −x
y = 5x − 1
y=x+4
(−2, 2)
y
y = 5x − 1
y = −2x + 6
(1, 4)
4
2
y = −2x + 6
2
−2
4 x
2
−2
2
4
6 x
−2
The graphs intersect
at (−2, 2).
Check
x + 4 = −x
?
−2 + 4 = −(−2)
The graphs intersect
at (1, 4).
Check
−2x + 6 = 5x − 1
?
−2(1) + 6 = 5(1) − 1
?
−2 + 6 = 5 − 1
2=2✓
So, the solution of the original equation is x = −2.
4=4✓
So, the solution of the equation is x = 1.
8. 4x = x + 3
Graph the system.
y = 4x
y=x+3
1
11. —2 x – 2 = 9 − 5x
y
4
y=x+3
Graph the system.
(1, 4)
4
y=
y = 4x
−2
4 x
2
1
—2 x
y
1
−2
y = 2x − 2
2
y = 9 − 5x
−2
4
(2, −1)
6 x
y = 9 − 5x
The graphs intersect
at (1, 4).
−4
Check
4x = x + 3
?
4(1) = 1 + 3
The graphs intersect
4=4✓
1
—2 x − 2 = 9 − 5x
?
1
—2 (2) − 2 = 9 − 5(2)
So, the solution of the equation is x = 1.
?
1 − 2 = 9 − 10
9. x + 5 = −2x − 4
−1 = −1 ✓
Graph the system.
y
y=x+5
4
y = −2x − 4
Check
at (2, −1).
y=x+5
So, the solution of the equation is x = 2.
(−3, 2)
−4
x
y = −2x − 4
The graphs intersect
at (−3, 2).
Check
x + 5 = −2x − 4
?
−3 + 5 = −2(−3) − 4
?
2=6−4
2=2✓
So, the solution of the equation is x = −3.
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Algebra 1
Worked-Out Solutions
273
Chapter 5
1
12. −5 + —4 x = 3x + 6
14.
Graph the system.
y = −5 +
1
—4 x
y = 3x + 6
−6(x + 4) = −3x − 6
−6(x) − 6(4) = −3x − 6
y
y = 3x + 6
−6x − 24 = −3x − 6
Graph the system.
−8
−4
(−4, −6)
4
8 x
−4
−8
y
24
y = −6x − 24
y =−3x − 6
(−6, 12)
1
y = −5 + 4 x
−24
−12
24 x
12
y = −3x − 6
y = −6x − 24
The graphs intersect
Check
at (−4, −6).
−5 + —4 x = 3x + 6
?
1
−5 + —4 (−4) = 3(−4) + 6
?
−5 − 1 = −12 + 6
−24
1
The graphs intersect
at (−6, 12).
−6 = −6 ✓
So, the solution of the equation is x = −4.
12 = 12 ✓
13. 5x − 7 = 2(x + 1)
So, the solution of the equation is x = −6.
5x − 7 = 2(x) + 2(1)
15. 3x − 1 = −x + 7
5x − 7 = 2x + 2
y = 5x − 7
8
−4
(3, 8)
4
−4
6
y = −x + 7
2
y = 3x − 1
2
y = 5x − 7
5x − 7 = 2(x + 1)
?
5(3) − 7 = 2(3 + 1)
?
15 − 7 = 2(4)
8=8✓
(2, 5)
y = −x + 7
4
8 x
Check
So, the solution of the equation is x = 3.
y
y = 3x − 1
y = 2x + 2
−8
The graphs intersect
at (3, 8).
Graph the system.
y
Graph the system.
y = 2x + 2
Check
−6(x + 4) = −3x − 6
?
−6(−6 + 4) = −3(−6) − 6
?
−6(−2) = 18 − 6
The graphs intersect
at (2, 5).
4
x
6
Check
3x − 1 = −x + 7
?
3(2) − 1 = −2 + 7
?
6−1=5
5=5✓
So, the equation has one solution, which is x = 2.
16. 5x − 4 = 5x + 1
Graph the system.
y
y = 5x − 4
274
Algebra 1
Worked-Out Solutions
2
y = 5x + 1
y = 5x + 1
The lines have the same slope
but different y-intercepts. So,
they are parallel. Because
parallel lines never intersect,
there is no point that is
a solution of both linear
equations. So, the original
equation has no solution.
−4
−2
2
4 x
y = 5x − 4
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Chapter 5
17.
−4(2 − x) = 4x − 8
−4(2) − 4(−x) = 4x − 8
1
1
3
—2 (8x) + —2 (3) = 4x + —2
−8 + 4x = 4x − 8
Graph the system.
y
y = −8 + 4x
−2
6 x
4
y = 4x − 8
−2
Because the lines have the
same slope and the same
y-intercept, the lines are the
same. So, all points on the
line are solutions of both
equations, which means
the original equation has
infinitely many solutions.
−4
y = −4(2 − x)
y = 4x − 8
−2x − 3 = 2x − 4
2
Check
4 x
−4
−4
1
y = 2 (8x + 3)
−2
2
4 x
second graph, the lines intersect at (1, −3).
−2
y = −2x − 3
3
y = 4x + 2
21. The lines intersect at (−2, −6) in the first graph, and in the
y
y = −2x − 3
−4
4
y = 4x + —32
The lines have the same slope
and the same y-intercept. So,
the lines are the same, which
means all points on the line
are solutions of both linear
equations. So, the original
equation has infinitely many
solutions.
−2x − 3 = 2(x) − 2(2)
Graph the system.
y
y = 4x + —32
−6
−8
4x + —32 = 4x + —32
Graph the system.
18. −2x − 3 = 2(x − 2)
y = 2x − 4
3
1
—2 (8x + 3) = 4x + —2
20.
∣ x − 4 ∣ = ∣ 3x ∣
?
∣ −2 − 4 ∣ = ∣ 3(−2) ∣
?
∣ −6 ∣ = ∣ −6 ∣
y = 2x − 4
(0.25, −3.5)
∣ x − 4 ∣ = ∣ 3x ∣
?
∣ 1 − 4 ∣ = ∣ 3(1) ∣
?
∣ −3 ∣ = ∣ 3 ∣
6=6✓
3=3✓
So, the solutions are x = −2 and x = 1.
Check
The graphs intersect
at (0.25, −3.5).
−2x − 3 = 2(x − 2)
?
−2(0.25) − 3 = 2(0.25 − 2)
?
−0.5 − 3 = 2(−1.75)
−3.5 = −3.5 ✓
So, the original equation has one solution, which is x = 0.25.
19. −x − 5 =
1
−—3 (3x
+ 5)
1
y = −x − 5
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∣ 2x + 4 ∣ = ∣ x − 1 ∣
?
∣ 2(−5) + 4 ∣ =
?
∣ −10 + 4 ∣ =
?
∣ −6 ∣ =
∣ −5 − 1 ∣
∣ −6 ∣
∣ 2x + 4 ∣ = ∣ x − 1 ∣
?
∣ 2(−1) + 4 ∣ = ∣ −1 − 1 ∣
?
∣ −2 + 4 ∣ = ∣ −2 ∣
?
∣2∣ = 2
6
2=2✓
5
y = −x − 3
Graph the system.
The lines have the same slope
and different y-intercepts.
So, the lines are parallel.
Because parallel lines do not
intersect, there is no point that
is a solution of both linear
equations. So, the original
equation has no solution.
Check
So, the solutions are x = −5 and x = −1.
−x − 5 = −x − —53
y = −x −
second graph, the lines intersect at (−1, 2).
6=6✓
1
−x − 5 = −—3 (3x) − —3 (5)
5
—3
22. In the first graph, the lines intersect at (−5, −6), and in the
y
2
−6
−2
2 x
−2
y = −x − 5
Algebra 1
Worked-Out Solutions
275
Chapter 5
23.
∣ 2x ∣ = ∣ x + 3 ∣
25. ∣ −x + 4 ∣ = ∣ 2x − 2 ∣
Equation 1
2x = x + 3
Equation 2
Equation 1
Equation 2
2x = −(x + 3)
−x + 4 = 2x − 2
−x + 4 = −(2x − 2)
2x = −x − 3
−x + 4 = −2x + 2
System 1
System 2
System 1
System 2
y = 2x
y = 2x
y = −x + 4
y = −x + 4
y=x+3
y = −x − 3
y = 2x − 2
y = −2x + 2
y
4
(3, 6)
6
y
y
y = −x − 3
−4
y = 2x
−2
2
4 x
2
?
The graphs intersect
at (2, 2).
?
∣ 2(3) ∣ = ∣ 3 + 3 ∣
∣ 2(−1) ∣ = ∣ −1 + 3 ∣
?
?
∣6∣ = ∣6∣
∣ −2 ∣ = ∣ 2 ∣
6=6✓
2=2✓
So, the solutions are x = 3 and x = −1.
Equation 2
2x − 6 = x
2x − 6 = −x
System 1
System 2
y = 2x − 6
y = 2x − 6
y=x
y = −x
y
x
2
4
The graphs intersect
at (−2, 6).
Check
∣ −x + 4 ∣ = ∣ 2x − 2 ∣
?
∣ −2 + 4 ∣ =
?
∣2∣ =
?
2=
∣ 2(2) − 2 ∣
∣4 − 2∣
∣ −x + 4 ∣ = ∣ 2x − 2 ∣
?
∣ −(−2) + 4 ∣ = ∣ 2(−2) − 2 ∣
?
∣ 2 + 4 ∣ = ∣ −4 − 2 ∣
?
∣2∣
∣ 6 ∣ = ∣ −6 ∣
6=6✓
y
−2
2
4
8 x
6 x
4
(2, −2)
y = −x
−2
−4
−4
y = 2x − 6
The graphs intersect
at (6, 6).
Check
−2
(6, 6)
4
−4
−4
So, the solutions are x = 2 and x = −2.
Equation 1
y=x
Check
2
y = −2x + 2
2=2✓
24. ∣ 2x − 6 ∣ = ∣ x ∣
−8
6 x
y = −x + 4
∣ 2x ∣ = ∣ x + 3 ∣
Check
4
2
y = 2x − 2
−4
The graphs intersect
at (−1, −2).
∣ 2x ∣ = ∣ x + 3 ∣
−2
−2
(−1, −2)
4 x
The graphs intersect
at (3, 6).
8
y = −x + 4
y = 2x
4
2
Check
6
(−2, 6)
(2, 2)
2
2
y=x+3
y
y = 2x − 6
The graphs intersect
at (2, −2).
∣ 2x − 6 ∣ = ∣ x ∣
?
∣ 2(6) − 6 ∣ =
?
∣ 12 − 6 ∣ =
?
∣6∣ =
−6
∣ 2x − 6 ∣ = ∣ x ∣
Check
∣6∣
?
∣ 2(2) − 6 ∣ = ∣ 2 ∣
?
6
∣4 − 6∣ = 2
6
∣ −2 ∣ = 2
6=6✓
?
2=2✓
So, the solutions are x = 6 and x = 2.
276
Algebra 1
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Chapter 5
26. ∣ x + 2 ∣ = ∣ −3x + 6 ∣
27. ∣ x + 1 ∣ = ∣ x − 5 ∣
Equation 1
Equation 2
Equation 1
Equation 2
x + 2 = −3x + 6
x + 2 = −(−3x + 6)
x+1=x−5
x + 1 = −(x − 5)
x + 2 = 3x − 6
x + 1 = −x + 5
System 1
System 2
System 1
System 2
y=x+2
y=x+2
y=x+1
y=x+1
y = −3x + 6
y = 3x − 6
y=x−5
y = −x + 5
y
8
y=x+2
y = −3x + 6
y
y=x+1 2
−8
4
8 x
−4
y = 3x − 6
4 x
The graphs intersect
at (1, 3).
The graphs intersect
at (4, 6).
Check
Check
∣ x + 2 ∣ = ∣ −3x + 6 ∣
∣ x + 2 ∣ = ∣ −3x + 6 ∣
?
∣1 + 2∣ =
?
∣3∣ =
?
3=
?
∣4 + 2∣ =
?
∣6∣ =
?
6=
∣ −3(1) + 6 ∣
∣ −3 + 6 ∣
∣3∣
3=3✓
2
4
y = −x + 5
−2
2
−4
y=x−5
The graphs do not
intersect. So, this
system has no solution.
∣ −12 + 6 ∣
6=6✓
(2, 3)
2
4
x
The graphs intersect
at (2, 3).
Check ∣ x + 1 ∣ = ∣ x − 5 ∣
?
∣2 + 1∣ = ∣2 − 5∣
?
∣ 3 ∣ = ∣ −3 ∣
∣ −3(4) + 6 ∣
∣ −6 ∣
y=x+1
x
−4
4
−4
y
(4, 6)
y=x+2
(1, 3)
y
3=3✓
So, the solution is x = 2.
So, the solutions are x = 1 and x = 4.
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Algebra 1
Worked-Out Solutions
277
Chapter 5
30. 4∣ x + 2 ∣ = ∣ 2x + 7 ∣
28. ∣ 2x + 5 ∣ = ∣ −2x + 1 ∣
Equation 1
Equation 1
Equation 2
2x + 5 = −2x + 1
2x + 5 = −(−2x + 1)
2x + 5 = 2x − 1
System 1
System 2
y = 2x + 5
y = 2x + 5
y = −2x + 1
y = 2x − 1
y
y
6
(−1, 3)
−2
System 1
System 2
y = 4x + 8
y = 4x + 8
y = 2x + 7
y = −2x − 7
The graphs intersect
at (−1, 3).
−4
y
−2
2
y = 2x + 7
y = 4x + 8
4 x
The graphs do not intersect.
So, this system has no
solution.
∣ 2x + 5 ∣ = ∣ −2x + 1 ∣
4
−6
Check
?
∣ −2 + 5 ∣ = ∣ 2 + 1 ∣
Check
So, the solution is x = −1.
29. ∣ x − 3 ∣ = 2 ∣ x ∣
Equation 1
Equation 2
x − 3 = 2(x)
x − 3 = 2(−x)
x − 3 = 2x
x − 3 = −2x
System 1
System 2
y=x−3
y = 2x
y = −2x
y
2=2✓
2
−2
(1, −2)
y = 2.1x + 0.6
The solution is x = 1.8.
?
∣ −3 − 3 ∣ =
2 ∣ −3 ∣
?
∣ −6 ∣ = 2(3)
The graphs intersect at
(1, −2).
6
Graph the system.
y = −1.4x + 6.9
(−3, −6)
6
Intersection
X=-2
Y=-.9
−4
The solution is x = −2.
32. 2.1x + 0.6 = −1.4x + 6.9
4 x
−2
−6
y = −0.2x − 1.3
y
y = −2x
−4
4
y = 0.7x + 0.5
y=x−3
4 x
4∣ x + 2 ∣ = ∣ 2x + 7 ∣
?
4∣ −2.5 + 2 ∣ = ∣ 2(−2.5) + 7 ∣
?
4∣ −0.5 ∣ = ∣ −5 + 7 ∣
?
4(0.5) = ∣ 2 ∣
Graph the system.
4
∣x − 3∣ = 2 ∣x∣
The graphs intersect
at (−2.5, −2).
The solutions are x = −0.5 and x = −2.5.
−6
Check
−8
31. 0.7x + 0.5 = −0.2x − 1.3
y=x−3
The graphs intersect at
(−3, −6).
x
−4
6=6✓
3=3✓
y=x−3
4
4∣ x + 2 ∣ = ∣ 2x + 7 ∣
?
4∣ −0.5 + 2 ∣ = ∣ 2(−0.5) + 7 ∣
?
4∣ 1.5 ∣ = ∣ −1 + 7 ∣
?
4(1.5) = ∣ 6 ∣
?
∣3∣ = ∣3∣
−2
y
−4
2 x
The graphs intersect
at (−0.5, 6).
?
y = 2x
−6
(−2.5, −2)
2
∣ 2(−1) + 5 ∣ = ∣ −2(−1) + 1 ∣
−4
y = −2x − 7
(−0.5, 6)
2
4 x
2
Check
4x + 8 = −2x − 7
y = 4x + 8
y = 2x − 1
−4
4(x) + 4(2) = −2x − 7
4x + 8 = 2x + 7
y = 2x + 5
y = −2x + 1
4(x + 2) = −(2x + 7)
4(x) + 4(2) = 2x + 7
6
y = 2x + 5
Equation 2
4(x + 2) = 2x + 7
−6
Intersection
X=1.8
Y=4.38
−2
6
Check ∣ x − 3 ∣ = 2 ∣ x ∣
?
∣1 − 3∣ =
2 ∣1∣
?
∣ −2 ∣ = 2(1)
6=6✓
2=2✓
So, the solutions are x = −3 and x = 1.
278
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Chapter 5
33. Words
36. no; The x-value of the coordinate pair is the solution of the
Company A
Cost
per
guest
⋅
equation. So, the solution of the equation −x + 4 = 2x −8
is x = 4. When x = 4, the equation is true because each side
of the equation is equal to 0.
Company B
Cost
Flat
Guests +
per
=
fee
guest
⋅
Flat
Guests +
fee
mx + b = −2x − 1
37. Sample answer:
m(−3) + b = −2(−3) − 1
Variable Let x be the number of guests at the wedding
reception.
−3m + b = 6 − 1
−3m + b = 5
Equation 20x + 500 = 16x + 800
Use a graphing calculator
to graph the system.
Let m = 1.
3000
−3(1) + b = 5
−3 + b = 5
y = 20x + 500
+3
y = 16x + 800
Intersection
Y=2000
0 X=75
0
100
Because the graphs intersect at (75, 2000), the solution of
the equation is x = 75. So, the total costs are the same for
75 guests.
+3
b=8
So, if m = 1 and b = 8, then the solution of
x + 8 = −2x − 1 is x = −3.
38. a. Sample answer: The graphs intersect at approximately
(7, 5.7).
34. Words
b. Sample answer: The company will break even after being
Current
age in
+7
dog
years
⋅
Current
age
Human
= in cat + 4
years
years
⋅
Human
years
39. P = x + (x − 2) + 6 = 2x + 4
Variable Let x be how many human years have passed.
A = —12 (x − 2)(6) = 3(x − 2) = 3x − 6
Equation 16 + 7x = 28 + 4x
Use a graphing calculator
to graph the system.
Graph the system.
100
y
24
y = 2x + 4
y = 3x − 6
y = 16 + 7x
y = 28 + 4x
Intersection
Y=44
0 X=4
0
10
Because the graphs intersect at (4, 44), the solution of the
equation is x = 4. So, after 4 human years, your dog will be
44 in dog years, and your cat will be the same age of 44 in
cat years.
35. Use a graphing calculator to graph the system.
d = ∣ −5t + 100 ∣
y = 2x + 4
(10, 24)
16
The graphs intersect
at (10, 24). So, x = 10.
y = 3x − 6
8
4
8
12
x
40. Sample answer: Let b = 15,000.
b + mx = 20,000 − 1500x
15,000 + m(5) = 20,000 − 1500(5)
15,000 + 5m = 20,000 − 7500
∣
d = ∣ −—
3 t + 50
10
15,000 + 5m = 12,500
−15,000
100
Intersection
Y=50
0 X=30
0
open for about 7 years. In other words, the company
will be open for business for about 7 years before it
has generated enough revenue to recover the start-up
expenses.
50
The graphs intersect between 15 and 20 seconds, but that is
when you and your friend are running in opposite directions.
The intersection at (30, 50) represents when you catch up to
your friend. So, the solution is x = 30, meaning you catch up
to your friend after 30 seconds.
Copyright © Big Ideas Learning, LLC
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−15,000
5m = −2500
5m −2500
—=—
5
5
m = −500
So, a car with an initial value of $15,000 that decreases in
value at a rate of $500 per year will also have a value of
$12,500 in exactly 5 years.
Algebra 1
Worked-Out Solutions
279
Chapter 5
41. a. The solution will be negative. The lines will intersect to
47.
the left of the y-axis as in the example shown, where
x = −2 is the solution of the equation 2x + 4 = —12 x + 1.
−f (x)
1
y = x + 1,
2
y = ax + b
4
−1
0
1
6
0
−6
−6
0
6
−6x
y
y = 2x + 4,
y = cx + d
x
y
2
(−2, 0)
f(x) = 6x
2
−4
4 x
2
−8
−4
8 x
4
g(x) = −f(x)
−2
b. The solution will be positive. The lines will intersect to
the right of the y-axis as in the example shown, where
x = 2 is the solution of the equation 2x − 6 = —12 x − 3.
The function g is of the form y = −f (x). So, the graph of g is
a reflection in the x-axis of the graph of f.
y
x
−4
−2
1
y = x − 3, −2
2
48.
2
(2, −2)
y = ax + b
0
2
4
−6
−4
−2
4x
−4
0
4
9
1
−7
y
f(x) = −2x + 1
−8
6
−4
8 x
4
−4
8
−8
0
2
g(x) = f(4x)
9
44. n ≥ 9
0
4
8
12
The function g is of the form y = f (ax), where a = 4. So, the
graph of g is a horizontal shrink of the graph of f by a factor
of 1 ÷ 4 = —14.
16
−6
−16 −12
46.
−8
−4
0
49.
y
x
−4
1
5
42. y > 5
45. c < −6
0
y = 2x − 6,
y = cx + d
Maintaining Mathematical Proficiency
43. x ≤ −2
−1
f (4x)
−4
−6
x
−2
−2
g(x) = f(x + 2)
x
−1
1
3
x−1
−2
0
2
f (x − 1)
−3
−2
−1
2
f(x) = x − 5
−4
The function g is of the form y = f (x − h), where h = −2.
So, the graph of g is a horizontal translation 2 units left of
the graph of f.
−2
y
1
f(x) = 2 x − 2
x
−4 g(x) = f(x − 1)
−6
The function g is of the form y = f (x − h), where h = 1.
So, the graph of g is a horizontal translation 1 unit right of
the graph of f.
280
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Chapter 5
Section 5.6
b.
4
5.6 Explorations (p. 267)
2
1. a. The dashed line crosses the y-axis at (0, −3), and the
slope of the line is 1. So, an equation represented by the
dashed line is y = x − 3.
−4
−2
−4
y = 1x + (−3)
y=x−3
Test (0, 0).
b. The solutions are all the points below the line y = x − 3.
1
y ≤ −—2 x + 1
c. An inequality represented by the graph is y < x − 3.
? 1
0 ≤ −—2 (0) + 1
Sample answer: The point (4, 0) is in the shaded region,
and to make the inequality true for that point, the < symbol
is needed.
?
0≤0+1
0≤1✓
2. a. Check students’ work.
b. Check y ≥
1
Sample answer: Graph y = −—2 x + 1 with a solid line
−3
because the inequality symbol ≤ indicates that the points
on the line are solutions. Test the point (0, 0) to determine
whether it is a solution of the inequality. Because the point
(0, 0) is a solution, shade the half-plane that contains (0, 0).
?
0 ≥ —14 (0) − 3
?
0≥ 0−3
c.
2
0≥−3✓
The point (0, 0) is a solution of the inequality y ≥ —14 x − 3.
3. a.
6
4 x
2
−2
y = mx + b
1
—4 x
y
−6
−2
y
2 x
−2
y
4
−6
2
Test (0, 0).
−4
−2
2 x
−2
Test (0, 0).
y>x+5
?
0>0+5
0>5✗
Sample answer: Graph y = x + 5 with a dashed line
because the inequality symbol > indicates that the
points on the line are not solutions. Test the point (0, 0)
to determine whether it is a solution of the inequality.
Because the point (0, 0) is not a solution, shade the halfplane that does not contain (0, 0).
Copyright © Big Ideas Learning, LLC
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y ≥ −x − 5
?
0 ≥ −0 − 5
0 ≥ −5 ✓
Sample answer: Graph y = −x − 5 with a solid line
because the inequality symbol ≥ indicates that the points
on the line are solutions. Test the point (0, 0) to determine
whether it is a solution of the inequality. Because the point
(0, 0) is a solution, shade the half-plane that contains (0, 0).
4. To graph a linear inequality in two variables, first graph the
boundary line for the inequality. Use a dashed line for < or >.
Use a solid line for ≤ or ≥. Next, test a point that is not on
the boundary line to determine whether it is a solution of
the inequality. When the test point is a solution, shade the
half-plane that contains the test point. When the test point is
not a solution, shade the half-plane that does not contain the
test point.
Algebra 1
Worked-Out Solutions
281
Chapter 5
5. Sample answer: You are selling tickets for your band’s first
show. Adult tickets cost $10 each, and child tickets cost $5
each. You and the other band members set a goal of selling at
least $500 worth of tickets. You can write and graph a linear
inequality to represent how many of each type of ticket you
must sell in order to reach your goal. Let x be how many
adult tickets you sell, and let y be how many child tickets you
sell. Then, a linear inequality that represents this situation is
10x + 5y ≥ 500.
5.6 Monitoring Progress (pp. 268–270)
1. x + y > 0
?
−2 + 2 > 0
0 > 0✗
So, (−2, 2) is not a solution of the inequality.
2.
4x − y ≥ 5
?
4(0) − 0 ≥ 5
?
0−0≥5
0≥5✗
x + y ≤ −4
7.
x − x + y ≤ −4 − x
y ≤ −x − 4
5x − 2y ≤ −1
?
5(−4) − 2(−1) ≤ −1
?
−20 + 2 ≤ −1
−2
x + y ≤ −4
?
0 + 0 ≤ −4
−6
0 ≤ −4 ✗
x − 2y < 0
8.
⋅
2
4 x
−2
−4
⋅
4x + 3y ≤ 12
4x − 4x + 3y ≤ 12 − 4x
3y ≤ −4x + 12
3y −4x + 12
—≤—
3
3
4
y ≤ −— x + 4
3
Test (0, 0).
y > −1
0 > −1 ✓
4 x
−4
6.
4
y
2
−2
Test (0, 0).
x ≤ −4
0 ≤ −4 ✗
Tomatoes (pounds)
y
2
⋅
Pounds
Amount
of
≤ you can
tomatoes
spend
Variables Let x be the weight (in pounds) of red peppers,
and let y be the weight (in pounds) of tomatoes.
⋅
−2
−6
−2
Inequality 4 x + 3 y ≤ 12
2
−8
−4
Pounds
Cost per
of red + pound of
peppers
tomatoes
So, (5, −7) is a solution of the inequality.
−2
2
−4 < 0 ✓
11 ≤ 15 ✓
−4
y
4
x − x − 2y < 0 − x
−2y < − x
−2y > −x
— —
−2 −2
1
y > —x
2
Test (0, 2).
Cost per
pound
of red
peppers
−2x − 3y < 15
?
−2(5) − 3(−7) < 15
?
−10 + 21 ≤ 15
4
2 x
9. Words
−18 ≤ −1 ✓
5.
−2
x − 2y < 0
?
0 − 2(2) < 0
?
0−4<0
So, (−4, −1) is a solution of the inequality.
4.
−6
Test (0, 0).
So, (0, 0) is not a solution of the inequality.
3.
y
2
y
4
Test (0, 0).
4x + 3y ≤ 12
?
4(0) + 3(0) ≤ 12
?
0 + 0 ≤ 12
3
2
0 ≤ 12 ✓
1
0
0
1
2
3
4 x
Red peppers (pounds)
x
−2
−4
282
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Chapter 5
Check
4x + 3y ≤ 12
?
4(1) + 3(1) ≤ 12
?
4 + 3 ≤ 12
4x + 3y ≤ 12
?
4(1.5) + 3(2) ≤ 12
?
6 + 6 ≤ 12
7 ≤ 12 ✓
7.
6≤6✓
12 ≤ 12 ✓
Sample answer: One possible solution is (1, 1) because it lies
in the shaded half-plane. Another possible solution is (1.5, 2)
because it lies on the solid line. So, you can buy 1 pound of
red peppers and 1 pound of tomatoes, or 1.5 pounds of red
peppers and 2 pounds of tomatoes.
−6x + 4y ≤ 6
?
−6(−3) + 4(−3) ≤ 6
?
18 − 12 ≤ 6
So, (−3, −3) is a solution of the inequality.
8.
3x − 5y ≥ 2
?
3(−1) − 5(−1) ≥ 2
?
−3 + 5 ≥ 2
2≥2✓
5.6 Exercises (pp. 271–272)
So, (−1, −1) is a solution of the inequality.
Vocabulary and Core Concept Check
1. To tell whether an ordered pair is a solution of a linear
9.
inequality, substitute the values into the linear inequality.
If they make the inequality true, then the ordered pair is a
solution of the linear inequality. If they make the inequality
false, then the ordered pair is not a solution of the linear
inequality.
2. The graph of a linear inequality in two variables and a linear
equation in two variables both have lines. The graph of a
linear equation in two variables is always a solid line, and
only the points on the line are solutions of the equation. The
graph of a linear inequality in two variables has a boundary
line that is either solid or dashed, and you shade on one side
or the other of the boundary line to indicate all of the points
that are solutions of the inequality.
Monitoring Progress and Modeling with Mathematics
3. x + y < 7
?
2+3< 7
5<7✓
So, (2, 3) is a solution of the inequality.
4. x − y ≤ 0
?
5−2≤ 0
3≤0✗
So, (5, 2) is not a solution of the inequality.
5.
x + 3y ≥ −2
?
−9 + 3(2) ≥ −2
?
−9 + 6 ≥ −2
−3 ≥ −2 ✗
So, (−9, 2) is not a solution of the inequality.
6.
8x + y > −6
?
8(−1) + 2 > −6
?
−8 + 2 > −6
−x − 6y > 12
?
−(−8) − 6(2) > 12
?
8 − 12 > 12
−4 > 12 ✗
So, (−8, 2) is not a solution of the inequality.
10.
−4x − 8y < 15
?
−4(−6) − 8(3) < 15
?
24 − 24 < 15
0 < 15 ✓
So, (−6, 3) is a solution of the inequality.
11. no; (0, −1) is not a solution because it lies in the half-plane
that is not shaded.
12. yes; (−1, 3) is a solution because it lies in the shaded
half-plane.
13. yes; (1, 4) is a solution because it lies in the shaded half-plane.
14. no; (0, 0) is not a solution because it lies on the dashed line.
15. no; (3, 3) is not a solution because it lies on the dashed line.
16. no; (2, 1) is not a solution because it lies in the half-plane that
is not shaded.
17. Test (12, 14). 8x + 12y ≤ 250
?
8(12) + 12(14) ≤ 250
?
96 + 168 ≤ 250
264 ≤ 250 ✗
no; (12, 14) is not a solution of the inequality 8x + 12y ≤ 250.
So, the carpenter cannot buy twelve 2-by-8 boards and fourteen
4-by-4 boards.
−6 > −6 ✗
So, (−1, 2) is not a solution of the inequality.
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Worked-Out Solutions
283
Chapter 5
3x + 2y ≥ 93
?
3(20) + 2(18) ≥ 93
?
60 + 36 ≥ 93
18. Test (20, 18).
24.
y
Test (0, 0).
y≤5
0≤5✓
6
Test (0, 0).
x<9
0<9✓
8
4
12
16 x
−2
yes; (20, 18) is a solution of the inequality 3x + 2y ≥ 93.
So, you earn an A on the test.
8
y
2
96 ≥ 93 ✓
19.
4
−4
25.
−4
4
Test (0, 0).
y > −2x − 4
y
2
2
4 x
2
−4
−4
−2
20.
4 x
2
8
y
?
0 > −2(0) − 4
?
0>0−4
0 > −4 ✓
−6
Test (0, 0).
y>6
0>6✗
26.
y
4
Test (0, 0).
2
4
−4
2
−2
2
4 x
y ≤ 3x − 1
?
0 ≤ 3(0) − 1
?
0≤0−1
0 ≤ −1 ✗
−4
−2
21.
4 x
2
4
Test (0, 0).
x<2
0<2✓
y
2
−4x + y < −7
27.
−4x + 4x + y < −7 + 4x
y < 4x −7
−4
−2
4 x
y
−4
−2
Test (0, 0).
−2
4 x
−2
−4
−4
22.
Test (0, 0).
x ≥ −3
0 ≥ −3 ✓
y
4
2
−6
−4
3x − y ≥ 5
3x − 3x − y ≥ 5 − 3x
−2
−y ≥ −3x + 5
−y −3x + 5
—≤—
−1
−1
y ≤ 3x − 5
−4
y
23.
−4
−2
2
−4
0 < −7 ✗
−6
28.
2 x
−2
−4x + y < −7
?
−4(0) + 0 < −7
4 x
Test (0, 0).
y > −7
0 > −7 ✓
y
−2
Test (0, 0).
2
−8
−2
−12
−4
4
6 x
3x − y ≥ 5
?
3(0) − 0 ≥ 5
0≥5
−16
284
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Chapter 5
33. Words
5x − 2y ≤ 6
29.
5x − 5x − 2y ≤ 6 − 5x
Cost per
arcade
game
−2y ≤ −5x + 6
−2y −5x + 6
—≥—
−2
−2
5
y ≥ —x − 3
2
y
4
Inequality
Test (0, 0).
⋅
2.25y ≤ −0.75x + 20
−0.75x + 20
2.25
1
80
y ≤ −— x + —
3
9
2.25y
2.25
— ≤ ——
Test (0, 0).
−x + 4y > −12
0.75x + 2.25y ≤ 20
?
0.75(0) + 2.25(0) ≤ 20
?
0 + 0 ≤ 20
−x + x + 4y > −12 + x
4y > x −12
x −12
4
1
y > —x − 3
4
4y
4
—>—
−4
0 ≤ 20 ✓
Test (0, 0).
y
−2
2
−x + 4y > −12
?
−0 + 4(0) > −12
4 x
0 > −12 ✓
−2
−4
Number of snacks
2
y
16
12
8
4
0
0
31. The line should be dashed for <.
y
2
−2
8
16
24
32 x
Number of games
−6
4
⋅
0.75x − 0.75x + 2.25y ≤ 20 − 0.75x
−2
30.
Number
Amount
of
≤ you can
snacks
spend
0.75 x + 2.25 y ≤ 20
0≤6✓
4 x
2
⋅
0.75x + 2.25y ≤ 20
5x − 2y ≤ 6
?
5(0) − 2(0) ≤ 6
−2
Number
Cost
of arcade + per
games
snack
Variables Let x be the number of arcade games you can play,
and let y be the number of snacks you can buy.
2
−4
⋅
2
Check
0.75x + 2.25y ≤ 20
?
0.75(8) + 2.25(4) ≤ 20
?
6 + 9 ≤ 20
15 ≤ 20 ✓
4 x
−2
32. The half-plane on the other side of the boundary line should
0.75x + 2.25y ≤ 20
?
0.75(16) + 2.25(2) ≤ 20
?
12 + 4.5 ≤ 20
16.5 ≤ 20 ✓
Sample answer: Two possible solutions are (8, 4) and (16, 2)
because they lie in the shaded half-plane. So, you can play
8 games and buy 4 snacks for a total of $15.00, or you can
play 16 games and buy 2 snacks for a total of $16.50.
be shaded.
4
y
Test (0, 0).
2
−2
2
−2
4 x
y ≤ 3x − 2
?
0 ≤ 3(0) − 2
?
0≤0−2
0 ≤ −2 ✗
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Algebra 1
Worked-Out Solutions
285
Chapter 5
34. Words
36. The slope of the boundary line through (0, 2) and (2, 3)
Cost per
adult
ticket
⋅
Number
Cost per
of adult + student
tickets
ticket
⋅
Number
of student ≥ 1500
tickets
Variables Let x be the number of adult tickets the drama
club must sell, and let y be the number of student
tickets the club must sell.
⋅
⋅
10x + 6y ≥ 1500
10x − 10x + 6y ≥ 1500 − 10x
6y ≥ − 10x + 1500
6y −10x + 1500
— ≥ ——
6
6
5
y ≥ −— x + 250
3
Student tickets
100
0
37. The boundary line passes through (0, −2) and (2, −3).
10x + 6y ≥ 1500
?
10(0) + 6(0) ≥ 1500
?
0 + 0 ≥ 1500
200
0 ≥ 1500 ✗
0
50
100
150
200 x
Adult tickets
Check
10x + 6y ≥ 1500
?
10(75) + 6(200) ≥ 1500
?
750 + 1200 ≥ 1500
10x + 6y ≥ 1500
?
10(150) + 6(300) ≥ 1500
?
1500 + 1800 ≥ 1500
1950 ≥ 1500 ✓
1
y = —x + 2
2
4>2✓
1
So, an inequality that represents the graph is y > — x + 2.
2
Test (0, 0).
300
y = mx + b
1
Test (0, 4). y > — x + 2
2
?1
4 > — (0) + 2
2
?
4>0+2
Inequality 10 x + 6 y ≥ 1500
y
400
3−2 1
is m = — = —.
2−0 2
The boundary line crosses the y-axis at (0, 2). So, b = 2.
3300 ≥ 1500 ✓
Sample answer: Two possible solutions are (75, 200) and
(150, 300) because they lie in the shaded half-plane. So, you
can sell 75 adult tickets and 200 student tickets for a total
of $1950, or you can sell 150 adult tickets and 300 student
tickets for a total of $3300.
35. The slope of the boundary line through (0, 1) and (1, 3) is
3−1 2
m = — = —, or 2.
1−0 1
The boundary line crosses the y-axis at (0, 1). So, b = 1.
y = mx + b
y = 2x + 1
−3 − (−2) −3 + 2 −1
The slope is m = — = — = —.
2−0
2−0
2
The boundary line crosses the y-axis at (0, −2). So, b = −2.
y = mx + b
1
y = − —x − 2
2
1
Test (0, −4). y ≤ −— x − 2
2
? 1
−4 ≤ −— (0) − 2
2
?
−4 ≤ 0 − 2
−4 ≤ −2 ✓
1
So, an inequality that represents this graph is y = −— x − 2.
2
38. The boundary line passes through (0, −3) and (1, 0).
0 − (−3) 0 + 3 3
The slope is m = — = — = —, or 3.
1−0
1−0 1
The boundary line crosses the y-axis at (0, −3). So, b = −3.
y = mx + b
y = 3x − 3
Test (2, 0), y ≤ 3x − 3
?
0 ≤ 3(2) − 3
?
0≤6−3
0≤3✓
So, an inequality that represents this graph is y ≤ 3x − 3.
Test (0, 4).
y > 2x + 1
?
4 > 2(0) + 1
?
4>0+1
4>1✓
So, an inequality that represents the graph is y > 2x + 1.
286
Algebra 1
Worked-Out Solutions
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Chapter 5
39. a. Words
Weight per
Weight of
+
small box
delivery person
+
Weight per
large box
⋅
⋅ small boxes
Number of
Number of
Weight
≤
large boxes
limit
Variables Let x be the number of small boxes the
delivery person can take on the elevator, and
let y be the number of large boxes.
200 + 75
Inequality
c. D; Test (0, 0).
3x − 2y > 6
?
3(0) − 2(0) > 6
0>6✗
The shaded half-plane of Graph D does not contain (0, 0).
Also, the boundary line is dashed for >.
d. B; Test (0,0).
⋅ x + 40 ⋅ y ≤ 2000
3x − 2y ≥ 6
?
3(0) − 2(0) ≥ 6
0≥6✗
200 + 75x + 40y ≤ 2000
200 − 200 + 75x + 40y ≤ 2000 − 200
75x + 40y ≤ 1800
75x − 75x + 40y ≤ 1800 − 75x
40y ≤ −75x + 1800
−75x + 1800
40
15
y ≤ −—x + 45
8
40y
40
— ≤ ——
Test (0, 0).
200 + 75x + 40y ≤ 2000
?
200 + 75(0) + 40(0) ≤ 2000
?
200 + 0 ≤ 2000
200 ≤ 2000 ✓
41. Sample answer: You have to know which side of the
boundary line to shade. So, you need to choose a point on
one side or the other of the boundary line to know which
region contains the solutions of the inequality.
−1 − 1 −2
m = — = —, or −1
3−1
2
The line passes through (0, 2). So, b = 2.
42. Sample answer:
y = mx + b
y = −x + 2
Test (0, 0). y ≥ −x + 2
?
0≥ −0+2
?
0≥ 0+2
0≥2✗
y
80
Small boxes
The shaded half-plane of Graph B does not contain (0, 0).
Also, the boundary line is solid for ≥.
So, y ≥ −x + 2 is an inequality in which (1, 1), (3, −1), and
(−1, 3) are solutions that lie on the line. Also, (0, 0), (0, −1),
and (0, 1) are not solutions because they lie in the unshaded
half-plane.
60
40
20
4
0
0
10
20
30
40 x
Large boxes
b. Sample answer: The shaded region contains points whose
coordinates are not whole numbers, but it is not possible
to load only part of a box on the elevator. Also, it is
possible that even though a certain number of boxes are
allowed on the elevator based on their weight, they may
be too big in size to fit inside the elevator.
40. a. C; Test (0, 0).
3x − 2y ≤ 6
?
3(0) − 2(0) ≤ 6
(−1, 3)
(0, 1)
(0, 0)
−2
y
(1, 1)
2
x
4
(0, −1) (3, −1)
−4
y ≥ −x + 2
43. no; If the point (0, 0) is on the boundary line, then you have
to choose a different test point that is not on the boundary
line.
0≤6✓
Graph C has (0, 0) in the shaded half-plane and has a solid
line for ≤.
b. A; Test (0, 0).
3x − 2y < 6
?
3(0) − 2(0) < 6
0<6✓
Graph A has (0, 0) in the shaded half-plane and has a
dashed line for <.
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Algebra 1
Worked-Out Solutions
287
Chapter 5
44. The slope of the boundary line is
47. d = −8 − (−5) = −8 + 5 = −3
5 − (−5) 5 + 5 10
m = — = — = —, or 2.
2 − (−3) 2 + 3
5
y − y1 = m (x − x1)
Position
8
y − 5 = 2(x − 2)
3
4
5
−8
+5
−4
8 x
4
8
1
3
2
48. d = —2 −—2 = —2 , or 1
Position
−8
1
2
3
1
−—2
Term
So, an equation of the boundary line is y = 2x + 1.
An inequality is y ≤ 2x + 1.
−—2
3
4
5
6
7
8
1
—2
3
—2
5
—2
7
—2
9
—
2
—2
2
+—2
Check
Test (−2, −3).
7
+(−3) +(−3) +(−3)
(−3, −5) −4
y = 2x + 1
y ≤ 2x + 1
Test (6, 5).
?
−3 ≤ 2(−2) + 1
?
−3 ≤ −4 + 1
y ≤ 2x +1
?
5 ≤ 2(6) + 1
?
5 ≤ 12 + 1
−3 ≤ −3 ✓
5 ≤ 13 ✓
8 − (−16) 8 + 16 24
m = — = — = —, or 3.
1 − (−7)
1+7
8
y − y1 = m(x − x1)
y − 8 = 3(x − 1)
8
y − 8 = 3(x) − 3(1)
y − 8 = 3x − 3
−8
+8
(1, 8)
−4
(−7, −16)
8 x
4
2x + y ≤ 4
2x − y ≤ 0
2x − 2x − y ≤ 0 − 2x
y ≤ −2x + 4
−y ≤ −2x
y
—≥—
Test (3, 14). y < 3x + 5
?
14 < 3(3) + 5
?
14 < 9 + 5
14 < 14 ✗
Inequality 2
2x − 2x + y ≤ 4 − 2x
Check
0 < −16 ✗
B; The point (1,0) is not
in the shaded region of
Graph B, and it is not a
solution of Inequality 2.
A; The point (1,0) is in
the shaded half-plane
of Graph A, and it is a
solution of Inequality 1.
2. a. Inequality 1
−16
So, an equation of the boundary line is y = 3x + 5.
An inequality is y < 3x + 5.
Test (−7, 0). y < 3x + 5
?
0 < 3(−7) + 5
?
0 < −21 + 5
Inequality 2
Test (1,0). 2x − y ≤ 0
?
2(1) − 0 ≤ 0
?
2−0≤0
2≤0✗
2≤4✓
−8
y = 3x + 5
5.7 Explorations (p.273)
y
16
2
+—2
Section 5.7
Test (1,0). 2x + y ≤ 4
?
2(1) + 0 ≤ 4
?
2+0≤4
45. The slope of the boundary line is
2
+—2
11
11
The next three terms are —72, —92, and —
.
2
1. Inequality 1
So, (6, 5) and (−2, −3) are solutions of y ≤ 2x + 1.
+8
6
The next three terms are −20, −23, and −26.
y − 5 = 2x − 4
+5
2
(2, 5)
4
y − 5 = 2(x) − 2(2)
Term
y
1
−5 −8 −11 −14 −17 −20 −23 −26
−y
−1
4
−2x
−1
y ≥ 2x
2
−4
−2
2
4
x
So, (−7, 0) and (3, 14) are not solutions of y < 3x + 5.
−4
Maintaining Mathematical Proficiency
46. d = 8 − 0 = 8
Position
1
2
3
4
5
6
7
8
Term
0
8
16
24
32
40
48
56
+8
The next three terms are 40, 48, and 56.
288
Algebra 1
Worked-Out Solutions
+8
+8
The shaded half-planes overlap as shown.
b. Sample answer: The two regions that are shaded with
only one color contain points whose coordinates are
the solutions of one inequality but not the other. The
region that is shaded with both colors contains points
whose coordinates are solutions of both inequalities. The
unshaded region contains points whose coordinates are
not solutions of either inequality.
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Chapter 5
3. Sample answer: In order to graph a system of linear
inequalities, graph each inequality in the same coordinate
plane. Look for the intersection, or overlapping portion of
the shaded half-planes that are solutions of the inequalities.
The point in this intersection have coordinates that are
solutions of the system.
4.
y > 2x − 3
y
4
y ≥ —12 x + 1
2
−4
−2
2
4 x
−2
4. The solution of the system is represented by the region
where the shaded half-planes of the inequalities overlap.
5. no; If the boundary lines are parallel and their half-planes do
not overlap, then the system has no solution.
5.
−2x + y < 4
2x + y > 4
−2x + 2x + y < 4 + 2x
2x − 2x + y > 4 − 2x
y < 2x + 4
6. The blue line is vertical and passes through the point (2, 0).
So, an equation for this line is x = 2. Because the shaded
region is to the left of this solid boundary line, an inequality
is x ≤ 2. The red line is horizontal and passes through the
point (0, 3). So, an equation for this line is y = 3. Because
the shaded region is below this solid boundary line, an
inequality is y ≤ 3. So, a system of linear inequalities
represented by the graph is x ≤ 2 and y ≤ 3.
6
y > −2x + 4
y
4
2
2
4
6 x
−2
5.7 Monitoring Progress (pp. 274–277)
6. Inequality 1: The vertical boundary line passes through (3, 0).
y>x−4
?
5 > −1 − 4
1. y < 5
5<5✗
So, an equation of the line is x = 3. Because the shaded region
is to the left of the boundary line, the inequality is x < 3.
5 > −5 ✓
Because (−1, 5) is not a solution of each inequality, it is not
a solution of the system.
2. y ≥ 3x + 1
y > x −1
?
4> 1−1
?
4 ≥ 3(1) + 1
?
4≥ 3+1
4>0✓
Inequality 2: The slope of the other boundary line is −1,
and the y-intercept is 2. So, an equation of the line is
y = −x + 2. Because the shaded region is below the dashed
boundary line, the inequality is y < −x + 2.
So, the system of linear inequalities represented by the graph
is x < 3 and y < −x + 2.
1
7. Inequality 1: One of the lines has a slope of −—2 and a
1
y-intercept of 1. So, an equation of the line is y = −—2x + 1.
4≥4✓
Because the shaded region is above this solid boundary
1
line, the inequality is y ≥ −—2 x + 1.
Because the ordered pair (1, 4) is a solution of each
inequality, it is a solution of the system.
Inequality 2: The slope of the other boundary line is 2,
and the y-intercept is −3. So, an equation of this line is
y = 2x − 3. Because the shaded region is below this solid
boundary line, the inequality is y ≤ 2x − 3.
3. y ≥ − x + 4
x+y≤0
x−x+y≤0−x
y ≤ −x
8
y
8. Sample answer: Another ordered pair in the solution region
is (3, 4.5). So, you can spend 3 hours at the mall and
4.5 hours at the beach.
4
−8
−4
So, the system of linear inequalities represented by the
1
graph is y ≥ −—2 x + 1 and y ≤ 2x − 3.
4
8 x
−4
−8
Check
x+y≤8
?
3 + 4.5 ≤ 8
x≥2
3≥2 ✓
y>4
4.5 > 4 ✓
7.5 ≤ 8 ✓
9. The boundary line at x = 2 will now be at x = 3. So, the
shaded region will be one small triangular region of the
graph between this line and the other two existing lines.
Because the solution (2.5, 5) will no longer be inside the
shaded region, it will no longer be a solution.
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Algebra 1
Worked-Out Solutions
289
Chapter 5
5.7 Exercises (pp. 278–280)
Vocabulary and Core Concept Check
1. You can substitute the values into each inequality of the
11. Graph the system.
y
8
y ≥ −3
4
y ≥ 5x
system and verify that the values make the inequalities true.
−4
−6
−2
2 x
2. The ordered pair that does not belong is (1, −2). This ordered
pair is on both of the boundary lines, and because one of the
boundary lines is dashed, this ordered pair is a solution of only
one of the inequalities. So, it is not a solution of the system.
The other three ordered pairs are solutions of the system
because each one is in the shaded region.
12. Graph the system.
y < −1
4
y
2
x>4
Monitoring Progress and Modeling with Mathematics
2
6
8 x
2
4 x
2
4 x
2
4 x
6
8 x
3. no; The ordered pair (−4, 3) is a solution of one inequality
but not the other because it is on the solid boundary line, but
is not in the shaded region below the dashed boundary line.
So, it is not a solution of the system.
4. yes; The ordered pair (−3, −1) is a solution of each
inequality because it is in the shaded region below both
boundary lines. So, it is a solution of the system.
13. Graph the system.
y
4
y < −2
y>2
−4
5. no; The ordered pair (−2, 5) is not in the shaded region. So,
−2
it is not a solution of the system.
−4
6. no; The ordered pair (1, 1) is not in the shaded region. So, it
is not a solution of the system.
7. y < 4
2<4 ✓
y>x+3
?
2 > −5 + 3
14. Graph the system.
y≥x+1
2 > −2 ✓
2
−4
Because the ordered pair (−5, 2) is a solution of each
inequality, it is a solution of the system.
8.
y > −2
−1 > −2 ✓
y>x−5
?
−1 > 1 − 5
−1 > −4 ✓
Because the ordered pair (1, −1) is a solution of each
inequality, it is a solution of the system.
9. y ≤ x + 7
?
0≤0+7
0≤7 ✓
y ≥ 2x + 3
?
0 ≥ 2(0) + 3
?
0≥0+3
0≥3 ✗
Because (0, 0) is not a solution of each inequality, it is not a
solution of the system.
10.
y ≤ −x + 1
?
−3 ≤ −4 + 1
−3 ≤ −3 ✓
y ≤ 5x − 2
?
−3 ≤ 5(4) − 2
?
−3 ≤ 20 − 2
y
4
y<x−1
−4
15. Graph the system.
y ≥ −5
y − 1 < 3x
y
2
−4
−2
y − 1 + 1 < 3x + 1
y < 3x + 1
−4
−6
16. Graph the system.
x+y>4
x−x+y>4−x
4
2
y>−x+4
y ≥ —32 x − 9
y
2
4
−2
−4
−3 ≤ 18 ✓
Because the ordered pair (4, −3) is a solution of each
inequality, it is a solution of the system.
290
Algebra 1
Worked-Out Solutions
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Chapter 5
17. Graph the system.
21. Inequality 1: The vertical boundary line passes through
x+y>1
−x − y < − 3
x−x+y>1−x
−x + x − y < −3 + x
y > −x + 1
4
(−1, 0). So, an equation of the line is x = −1. Because the
shaded region is to the right of the solid boundary line, the
inequality is x ≥ −1.
−y < x − 3
Inequality 2: The horizontal boundary line passes through
(0, 3). So, an equation of the line is y = 3. Because the
shaded region is below this dashed boundary line, the
inequality is y < 3.
−y x − 3
—>—
−1
−1
y > −x + 3
y
2
−2
2
4
So, the system of linear inequalities represented by the graph
is x ≥ −1 and y < 3.
6 x
−2
22. One vertical line passes through (2,0), and the other passes
through (4,0). So, equations for the lines are x = 2 and
x = 4, respectively. Because the shaded region is to the
right of the dashed boundary line x = 2 and to the left of the
dashed boundary line x = 4, the system of linear inequalities
represented by the graph is x > 2 and x < 4.
−4
18. Graph the system.
2x + y ≤ 5
y + 2 ≥ −2x
2x − 2x + y ≤ 5 − 2x
y + 2 − 2 ≥ −2x − 2
y ≤ −2x + 5
23. Inequality 1: One of the lines has a slope of −3 and a
y-intercept of 2. So, an equation of the line is y = −3x + 2.
Because the shaded region is above this solid boundary line,
the inequality is y ≥ −3x + 2.
y ≥ −2x − 2
y
Inequality 2: The slope of the other boundary line is —23,
and the y-intercept −2. So, an equation of this line is
y = —23x − 2. Because the shaded region is above this
solid boundary line, the inequality is y ≥ —23 x − 2.
4
2
−4
−2
2
4 x
So, the system of linear inequalities represented by the graph
is y ≥ −3x + 2 and y ≥ —23 x − 2.
−2
19. Graph the system.
y
4
x<4
24. Inequality 1: One of the lines has a slope of 5 and a
y-intercept of 1. So, an equation of the line is y = 5x + 1.
Because the shaded region is below this solid boundary line,
the inequality is y ≤ 5x + 1.
2
y>1
y ≥ −x + 1
−2
6 x
2
−2
−4
20. Graph the system.
x + y ≤ 10
x − x + y ≤ 10 − x
y ≤ −x + 10
8
So, the system of linear inequalities represented by the graph
is y ≤ 5x + 1 and y > x − 2.
y
6
25. This system has no solution. Both lines have a slope of
4
x−y≥2
x−x−y≥2−x
− y ≥ −x + 2
−y −x + 2
−1
−1
y≤x−2
—≤—
Inequality 2: The slope of the other boundary line is 1,
and the y-intercept is −2. So, an equation of this line is
y = x − 2. Because the shaded region is above this dashed
boundary line, the inequality is y > x − 2.
2
4
6
8 x
−2. One line has a y-intercept of −1, and the other has a
y-intercept of −3. So, the equations of the boundary lines
are y = −2x − 1 and y = −2x − 3. Because the graph has
no shaded region, the half-planes of the inequalities must
not intersect. So, the solutions of the inequality bounded
by y = −2x − 1 must be above the dashed line, and the
inequality must be y > −2x − 1. Also, the solutions of the
inequality bounded by y = −2x − 3 must be below the
dashed line, and the inequality must be y < −2x − 3. So,
the system of linear inequalities represented by the graph is
y > −2x − 1 and y < −2x − 3.
y>2
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Algebra 1
Worked-Out Solutions
291
1
26. Both of the boundary lines in this graph have a slope of —3 .
One line has a y-intercept of 0, and the other has a y-intercept
of −2. So, the equations of the boundary lines are y = —13 x and
y = —13x − 2. Because the shaded region lies above both solid
lines, the inequalities must be y ≥ —13 x and y ≥ —13x − 2.
So, the system of linear inequalities represented by the graph
is y ≥ —13x and y ≥ —13 x − 2.
Strawberries (pounds)
Chapter 5
y
8
6
4
2
0
0
should be below the line on the graph, and the solutions of
y ≥ x + 3 should be above the line on the graph. Because
the lines have the same slope of 1, they are parallel. So, the
half-planes will not intersect.
2
4
8 x
b. Sample Answer: One ordered pair in the solution region
is (3, 2). So, you can buy 3 pounds of blueberries and
2 pounds of strawberries.
c. yes; The ordered pair (4, 1) is in the solution region.
y
4
30. a. Words Grocery store
−2
4 x
2
⋅
earnings
(in dollars)
per hour
2
−4
6
Blueberries (pounds)
27. The system should have no solution. The solutions of y ≤ x − 1
⋅
−4
28. The shaded region is correctly below the line y ≤ 3x + 4, but
it should be above the line y > —12 x + 2.
Music
lesson
hours
worked
≥ 120
4
Grocery
store hours +
worked
−2
2
x
−2
29. a. Words Blueberries’
cost per
pound
⋅
⋅
Pounds of +
blueberries
Pounds of
≤
strawberries
x≥8
x + y ≤ 20
10x + 15y ≥ 120
x + y ≤ 20
10x −10x + 15y ≥ 120 − 10x
x − x + y ≤ 20 − x
15y ≥ −10x + 120
15y −10x + 120
15
15
2
y ≥ −—3x + 8
x+y≥3
4x + 3y ≤ 21
x+y≥3
4x −4x + 3y ≤ 21 − 4x
x−x+y≥3−x
−4x + 21
3
4
y ≤ −—x + 7
3
3y
3
292
Algebra 1
Worked-Out Solutions
y≥−x+3
y
20
Hours teaching music
⋅ x + 3 ⋅ y ≤ 21
y ≤ − x + 20
—≥—
Variables Let x be the weight (in pounds) of blueberries
you can buy, and let y be the weight
(in pounds) of strawberries you can buy.
—≤—
≤ 20
System 10x + 15y ≥ 120
Amount you
can spend
3y ≤ − 4x + 21
Music
lesson
hours
worked
Variables Let x be the number of hours you work at the
grocery store, and let y be the number of hours
you work teaching music lessons.
Strawberries’
cost per
pound
Pounds of
Pounds of
Total
blueberries + strawberries ≥ amount
System 4
Music lesson
earnings
(in dollars)
per hour
Grocery
store hours ≥ 8
worked
y
−4
Grocery
store hours +
worked
16
12
8
4
0
0
4
8
12
16
20 x
Hours at grocery store
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Chapter 5
b. Sample answer: One ordered pair in the solution region
33.
is (12, 4). So, you can work 12 hours at the grocery store
and 4 hours teaching music lessons.
(−1, 1)
c. no; The ordered pair (8, 1) is not in the shaded solution
(6, 1)
−2
region.
31. a. Words
y
2
2
x
4
(−1, −3)
Number of
surfperch
≤ 15
Number of
rockfish
≤ 10
Number of
surfperch
+
(6, −3)
a. A system of linear inequalities represented by the shaded
rectangle is y ≥ −3, y ≤ 1, x ≥ −1, and x ≤ 6.
b. The area of the rectangle is
Number of
rockfish
A =ℓw = 7(4) = 28 square units.
≤ 20
34.
y
Variables Let x be the number of surfperch you can
catch, and let y be the number of rockfish you
can catch.
(2, 5)
4
2
System x ≤ 15
−2
y ≤ 10
6 x
4
(−2, −3)
x + y ≤ 20
(6, −3)
a. The base of the triangle is defined by the line y = −3. The
x − x + y ≤ 20 − x
shaded solution region is above this line. So, an inequality
is y ≥ −3.
y ≤ − x + 20
y
20
−3 − 5 −8
5 − (−3) 5 + 3 8
m = — = — = — m = — = —, or −2
4
2 − (−2) 2 + 2 3
6−2
16
12
y − y1 = m(x − x1)
y − y1 = m(x − x1)
y − 5 = 2(x − 2)
y − 5 = −2(x − 2)
8
y − 5 = 2(x) − 2(2)
y − 5 = −2(x)−2(−2)
4
y − 5 = 2x − 4
y − 5 = −2x + 4
0
+5
0
8
4
12
16
b. yes; The ordered pair (11, 9) is on one of the solid
boundary lines of the shaded solution region. So, it is a
solution of the system.
x−y≤4
x−y≥4
x−x−y≤4−x
x−x−y≥4−x
−y ≤ −x + 4
−y −x + 4
−1
−1
y≥x−4
—≥—
+5
+5
y = 2x + 1
20 x
Surfperch caught
32.
2
−2
−y ≥ − x + 4
−y −x + 4
−1
−1
y≤x−4
—≤—
+5
y = −2x + 9
So, the equations of the lines that define the other two
sides of the triangle are y = 2x + 1 and y = −2x + 9.
Because the shaded area is below each of these lines,
the inequalities are y ≤ 2x + 1 and y ≤ −2x + 9.
The system of linear inequalities represented by the
shaded triangle is y ≤ 2x + 1, y ≤ −2x + 9, and y ≥ −3.
b. The area of the triangle is
1
1
A = —bh = — (8)(8) = 32 square units.
2
2
The equations each describe a half-plane with the same solid
boundary line, but the solutions are on opposite sides of the
line. So, the intersection of the regions is the line y = x − 4
only. In other words, only the points on the line y = x − 4
are solutions of both inequalities.
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Algebra 1
Worked-Out Solutions
293
35. a. Words
Amount
spent on
housing
+
Amount
spent on
savings
Hours your friend drives
Chapter 5
≤ —12 (2000)
⋅
Amount spent
on savings
≥ 10% 2000
Amount spent
on housing
≤ 30% 2000
⋅
x + y ≤ 1000
y ≥ 0.1 (2000)
x − x + y ≤ 1000 − x
y ≥ 0.1(2000)
x ≤ 0.3(2000)
y ≥ 200
x ≤ 600
y ≤ − x + 1000
Money spent on
savings (in dollars)
y
1200
800
400
0
0
400
800
1200
x
Money spent on
housing (in dollars)
Sample answer: One ordered pair in the solution region is
(400, 400). So, you can spend $400 each on housing and
savings.
36. Words
Hours you
Hours friend
< 15
+
drive
drives
70
⋅
8
4
0
0
4
8
16 x
12
Sample answer: One ordered pair in the shaded solution
region is (4, 8). So, one possibility is for you to drive 4 hours
and your friend to drive 8 hours in one day.
37. Sample answer: When you solve systems of linear
x ≤ 0.3 (2000)
x + y ≤ 1000
12
Hours you drive
Variables Let x be how much you spend on housing and
let y be how much you spend on savings.
System
y
16
Hours you
+ 60
drive
⋅
inequalities, you are finding coordinate pairs of at least two
variables that make multiple inequalities true. The same is
true for systems of linear equations in that you are finding
coordinate pairs of at least two variables for at least two
equations. Both types of systems can have infinitely many
solutions or no solution. Graphing can be used to solve
both kinds of systems, and the graphs of both kinds of
systems involve straight lines. Graphing is the only method
that can be used to solve systems of linear inequalities,
however, and there are two other methods (substitution and
elimination) for solving systems of linear equations. When
you solve a system of linear inequalities, you are looking
for an overlapping region of coordinate pairs that make
the inequalities true. When you solve a system of linear
equations, you are usually looking for a single coordinate
pair that makes the equations true. The only time you get
more than one solution for a system of linear equations is
when the equations in the system describe the same line.
In this case, all of the points on the line are solutions of
the system.
38. The systems that have point C as a solution, but not points
Hours friend
≥ 600
drives
Hours you
Hours friend
>
drive
drives
Variables Let x be how many hours you drive, and
let y be how many hours your friend drives.
System x + y < 15
70x + 6y ≥ 600
A, B, and D are y > −3x + 4 and y ≤ 2x + 1 or y > −3x + 4
and y < 2x + 1. Sample answer: Point C is below the line
y = 2x + 1 and above the line y = −3x + 4. So, one
inequality must be either y < 2x + 1 or y ≤ 2x + 1, and the
other must be either y > −3x + 4 or y ≥ −3x + 4. Because
point B is on the line y = −3x + 4 and point B cannot be a
solution, it must be that y > −3x + 4. The other inequality
could be either y < 2x + 1 or y ≤ 2x + 1.
39. A system of linear inequalities that is equivalent to
y>x
⎧
x + y < 15
70x + 60y ≥ 600
x − x + y < 15 − x
70x − 70x + 60y ≥ 600 − 70x
y < − x + 15
y>x
∣ y ∣ < x is ⎨y < x, if x > 0 .
⎩y > −x, if x > 0
60y ≥ −70x + 600
60y −70x + 600
60
60
7
y ≥ −—x + 10
6
4
y
2
—≥—
−2
2
4
6 x
−2
−4
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Chapter 5
40. no; It is possible for a system of linear inequalities in which
the boundary lines are parallel to have infinitely many
solutions. Sample answer: Examples of systems with parallel
boundary lines and infinitely many solutions are Exercises 22
and 26, where the shaded solution region is either between
the lines or above (or below) both of the lines.
8x + 16y ≤ 100
?
8(9) + 16(1) ≤ 100
?
72 + 16 ≤ 100
Check
intersection can be at most a half-plane.
8x + 16y ≤ 100
?
8(3) + 16(8) ≤ 100
x + y ≤ 10
?
3 + 8 ≤ 10
?
24 + 128 ≤ 100
11 ≤ 10 ✗
42. Sample answer: A system of linear inequalities whose
solutions are all in Quadrant I is y > 0 and x > 0.
solutions have one positive coordinate and one negative
coordinate is x > 0 and y < 0.
44. Sample answer: A system of linear inequalities that has no
solution is y > x and y < x.
10 ≤ 10 ✓
88 ≤ 100 ✓
41. no, The solution of each inequality is a half-plane, and so the
43. Sample answer: A system of linear inequalities whose
x + y ≤ 10
?
9 + 1 ≤ 10
152 ≤ 100 ✗
So, the coordinate pair (9, 1) is in the solution region,
but (3, 8) is not.
47. Sample answer: A system of linear equations that has exactly
one solution is y ≥ ∣ x ∣ and y ≤ 0. The solution is (0, 0).
48. a. A system for this situation is
45 a. Sample answer: In order for the system to have no
solution, another inequality is −4x + 2y < 6.
b. Sample answer: In order for the system to have infinitely
many solutions, another inequality is −2x + y > 3.
0.5x + 0.25y ≤ 20, 2x + 3y ≤ 120, x ≥ 0, and y ≥ 0.
0.5x + 0.25y ≤ 20
2x + 3y ≤ 120
0.5x − 0.5x + 0.25y ≤ 20 − 0.5x
2x − 2x + 3y ≤ 120 − 2x
0.25y ≤ −0.5x + 20
0.25y −0.5x + 20
—≤—
0.25
0.25
46. Sample answer: If the T-shirts cost $8 each, and the
sweatshirts are $16 each, and your gift card is for $100 and
can be used to buy up to 10 clothing items, then you can buy
9 T-shirts and 1 sweatshirt, but you cannot buy 3 T-shirts and
8 sweatshirts.
Number of
Number of
Words 8
T-shirts + 16
sweatshirts ≤ 100
⋅
Number of
Number of
T-shirts + sweatshirts ≤ 10
Variables Let x be how many T-shirts you can buy, and let y
be how many sweatshirts you can buy.
System 8x + 16y ≤ 100
30
20
10
0
10
20
30
40
50 x
Necklaces
8x + 16y ≤ 100
x + y = ≤ 10
8x − 8x + 16y ≤ 100 − 8x
16y ≤ −8x + 100
16y
8x + 100
— ≤ −—
16
16
1
y ≤ −—x + 6.25
2
y
Sweatshirts
40
0
x + y ≤ 10
12
x − x + y ≤ 10 − x
y ≤ − x + 10
b. The boundary lines x = 0 and y = 0 intersect at (0, 0).
2
The boundary lines x = 0 and y = −—3x + 40 intersect at
2
(0, 40) because 40 = −—3(0) + 40.
The boundary lines y = 0 and y = −2x + 80 intersect at
(40, 0) because 0 = −2(40) + 80.
2
The boundary lines y = −2x + 80 and y = −—3x + 40
intersect at (30, 20) because 20 = −2(30) + 80 and
2
20 = −—3 (30) + 40.
So, the vertices of the graph of the system are (0, 0),
(0, 40), (40, 0), and (30, 20).
(3, 8)
8
y ≤ −2x + 80
y
50
Key chains
⋅
3y ≤ −2x + 120
3y −2x + 120
—≤—
3
3
2
y ≤ −—x + 40
3
4
0
(9, 1)
0
4
8
12
x
T-shirts
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Algebra 1
Worked-Out Solutions
295
Chapter 5
c. For (0, 0):
5.5 –5.7 What Did You Learn? (p. 281)
R = 10x + 8y
R = 10(0) + 8(0)
1. Sample answer: You and your friend are running to a fence
and back. So, for every point on your way toward the fence,
you are the same distance from the fence as another point on
your return from the fence. Every output of an absolute value
function is paired with two different input values. So, it can
be a good model for your distance from the fence during
the race.
R = $0
For (0, 40):
R = 10x + 8y
R = 10(0) + 8(40)
R = 0 + 320
R = $320
For (40, 0):
R = 10x + 8y
2. Sample answer: If your answer is inaccurate, the delivery
person might overload the elevator, which could lead to a
dangerous situation and put the delivery person or others
at risk.
R = 10(40) + 8(0)
R = 400 + 0
R = $400
3. Sample answer: Use a verbal model to write inequalities
For (30, 20): R = 10x + 8y
to represent each part of the problem, and designate two
variables to represent the two unknown values. Simplify the
inequalities, and rewrite one so that it is in slope-intercept
form. Graph all three inequalities on the same coordinate
plane and shade the intersection. Finally, choose a point in
the shaded region as a solution of the problem.
R = 10(30) + 8(20)
R = 300 + 160
R = $460
The vertex (30, 20) results in the maximum revenue. So, you
will make the most money if you make and sell 30 necklaces
and 20 key chains.
Maintaining Mathematical Proficiency
Chapter 5 Review (pp. 282–284)
1.
⋅ ⋅ ⋅ ⋅
50. (−13) ⋅ (−13) ⋅ (−13) = (−13)
51. x ⋅ x ⋅ x ⋅ x ⋅ x ⋅ x = x
49. 4
4 4 4 4 = 45
y
−2
2
−2
3
y = −3x + 1
x
4
(2, −5)
−4
6
−6
y=x−7
52. y = mx + b
y = 1x + (−6)
Check Equation 1
y=x−6
So, an equation is y = x − 6.
53. y = mx + b
y = −3x + 5
2.
4x − 2y = 6
y
4x − 4x − 2y = 6 − 4x
1
y = −—4 x − 1
So, an equation is y =
−2y = −4x + 6
1
−—4x
− 1.
55. y = mx + b
−2y −4x + 6
−2
−2
y = 2x − 3
—=—
y = —43x + 0
y = —43x
So, an equation is y =
−5 = −5 ✓
The solution is (2, −5).
54. y = mx + b
1
y=x−7
?
−5 = 2 − 7
−5 = −5 ✓
So, an equation is y = −3x + 5.
y = −—4 x + (−1)
Equation 2
y = −3x + 1
?
−5 = −3(2) + 1
?
−5 = −6 + 1
Check Equation 1
4
—3 x.
−4
−2
2
−2
4 x
(1, −1)
y = 2x − 3
Equation 2
y = −4x + 3
?
−1 = −4(1) + 3
?
−1 = −4 + 3
−1 = −1 ✓
y = −4x + 3
4x − 2y = 6
?
4(1) − 2(−1) = 6
?
4+2=6
6=6✓
The solution is (1, −1).
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Chapter 5
5x + 5y = 15
2x − 2y =10
5x − 5x + 5y = 15 − 5x
2x − 2x − 2y = 10 − 2x
3.
5y = −5x + 15
5y −5x + 15
5
5
y = −x + 3
—=—
y
−2y = −2x + 10
5. Step 1
Step 2
x−y=1
Step 3
x + 4y = 6
x−y=1
x−y+y=1+y
(y + 1) + 4y = 6
x−1=1
x = y +1
5y + 1 = 6
−2y −2x + 10
−2
−2
y=x−5
—=—
−1
+1
−1
+1
x=2
5y = 5
5y 5
—=—
5
5
y = −x + 3
2
y=1
2
(4, −1)
x
Check x + 4y = 6
?
2 + 4(1) = 6
?
2+4=6
−2
−4
y=x−5
Check Equation 1
5x + 5y = 15
?
5(4) + 5(−1) = 15
?
20 − 5 = 15
15 = 15 ✓
Equation 2
2x − 2y = 10
?
2(4) − 2(−1) = 10
?
8 + 2 = 10
The solution is (2, 1)
6. Step 1
y + 3x = 6
y + 3x − 3x = 6 − 3x
10 = 10 ✓
y = 6 − 3x
Step 2
2x + 3y = 4
4. Substitute 5x + 7 for y in Equation 1 and solve for x.
2x + 3(6 − 3x) = 4
3x + y = −9
2x + 3(6) − 3(3x) = 4
3x + (5x + 7) = −9
2x + 18 − 9x = 4
8x + 7 = −9
−7x + 18 = 4
−7
−18
8x = −16
8x
8
1=1✓
6=6✓
The solution is (4, −1).
−7
x−y=1
?
2−1=1
−16
8
—=—
−7x
−7
−14
−7
—=—
x = −2
x=2
Substitute −2 for x in Equation 2 and solve for y.
Step 3
y = 5x + y
y + 3x = 6
y = 5(−2) + 7
y + 3(2) = 6
y = −10 + 7
y+6=6
y = −3
Check
−18
−7x = −14
3x + y = −9
?
3(−2) + (−3) = −9
?
−6 − 3 = −9
−9 = −9 ✓
The solution is (−2, −3).
y = 5x + 7
?
−3 = 5(−2) + 7
?
−3 = −10 + 7
−3 = −3 ✓
−6
−6
y=0
Check
2x + 3y = 4
?
2(2) + 3(0) = 4
?
4+0=4
4=4✓
y + 3x = 6
?
0 + 3(2) = 6
?
0+6=6
6=6✓
The solution is (2, 0).
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Algebra 1
Worked-Out Solutions
297
Chapter 5
7. Words 4
Tubes
+ 0.50
of paint
⋅
⋅
Paint
= 20
brushes
9. Step 1
Step 2
x + 6y = 28
x + 6y = 28
2x − 3y = −19
Paint
Tubes
brushes = 2 ∙ of paint
Multiply by 2.
5x + 0 = −10
Step 3 5x = −10
Variables Let x be the number of tubes of paint you
purchase, and let y be the number of paint brushes
you purchase.
System 4x + 0.5y = 20
Equation 1
y = 2x
Equation 2
−10
5
5x
5
—=—
x = −2
Step 4
x + 6y = 28
Substitute 2x for y in Equation 1.
−2 + 6y = 28
4x + 0.5y = 20
+2
4x + 0.5(2x) = 20
+2
6y = 30
4x + x = 20
6y
6
30
6
—=—
5x = 20
y=5
20
5
5x
5
4x − 6y = −38
—=—
Check x + 6y = 28
?
−2 + 6(5) = 28
?
−2 + 30 = 28
x=4
Substitute 4 for x in Equation 2.
y = 2x
2x − 3y = −19
?
2(−2) − 3(5) = −19
?
−4 − 15 = −19
28 = 28 ✓
y = 2(4)
y=8
−19 = −19 ✓
So, the solution is (−2, 5).
The solution is (4, 8). So, you purchase 4 tubes of paint and
8 paint brushes.
8. Step 2
9x − 2y = 34
10. Step 1
Step 2
8x − 7y = −3
Multiply by 3.
6x − 5y = −1
Multiply by −4.
0 − y = −5
5x + 2y = −6
Step 3
14x = 28
Step 3
−y
−1
Step 4 6x − 5y = −1
6x − 5(5) = −1
5x + 2y = −6
6x − 25 = −1
5(2) + 2y = −6
+25
10 + 2y = −6
+25
6x = 24
−10
6x
6
24
6
—=—
2y = −16
−16
2
x=4
—=—
y = −8
Check
9x − 2y = 34
?
9(2) − 2(−8) = 34
?
18 + 16 = 34
−5
−1
y=5
Step 4
2y
2
−y = −5
—=—
14x 28
—=—
14
14
x=2
−10
24x − 21y = −9
−24x + 20y = 4
5x + 2y = −6
?
5(2) + 2(−8) = −6
?
10 − 16 = −6
34 = 34 ✓
Check 8x − 7y = −3
?
8(4) − 7(5) = −3
?
32 − 35 = −3
6x − 5y = −1
?
6(4) − 5(5) = −1
?
24 − 25 = −1
−3 = −3 ✓
−1 = −1 ✓
So, the solution is (4, 5).
−6 = −6 ✓
The solution is (2, −8).
298
Algebra 1
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Chapter 5
11. Solve by substitution.
15.
Substitute y + 2 for x in Equation 2.
∣ x + 1 ∣ = ∣ −x − 9 ∣
Equation 1
−3x + 3y = 6
x + 1 = −x − 9
−3( y + 2) + 3y = 6
−3y − 6 + 3y = 6
0−6=6
System 1
System 2
y=x+1
y=x+1
y = −x − 9
−6 = 6 ✗
−6
−4
(−5, −4)
12. Solve by elimination.
12
y=x+9
−2
4
15x − 30y = −45
Multiply by 5.
Multiply by 3.
y
x
y=x+1
Step 2
−5x + 10y = 10
y=x+9
y
The equation −6 = 6 is never true. So, the system has
no solution.
3x − 6y = −9
x + 1 = −(−x − 9)
x+1=x+9
−3( y) − 3(2) + 3y = 6
Step 1
Equation 2
− 15x + 30y = 30
0 = −15
The equation 0 = −15 is never true. So, the system has
no solution.
y=x+1
−6
−8
y = −x − 9
The graphs intersect
at (−5, −4).
−4
4
8 x
The lines are parallel. So,
they will never intersect, and
this system has no solution.
Check
13. Solve by substitution.
∣ x + 1 ∣ = ∣ −x − 9 ∣
Step 1
?
∣ −5 + 1 ∣ = ∣ −(−5) − 9 ∣
3y = 3x + 24
?
∣ −4 ∣ = ∣ 5−9 ∣
3y 3x + 24
3
3
y=x+8
—=—
?
4 = ∣ −4 ∣
4=4✓
Step 2
So, the solution is x = −5.
−4x + 4y = 32
−4x + 4(x + 8) = 32
−4x + 4(x) + 4(8) = 32
−4x + 4x + 32 = 32
0 + 32 = 32
32 = 32
Because the equation 32 = 32 is always true, every solution
of −4x + 4y = 32 is also a solution of 3x + 24 = 3y. So,
the system has infinitely many solutions.
14. Graph the system.
y = —13x + 5
y = −2x − 2
6
(−3, 4)
4
The graphs intersect at (−3, 4).
y
1
y = 3x + 5
2
Check
1
—3 x + 5 = −2x − 2
1
—3
−4
−2
2
x
y = −2x − 2
?
(−3) + 5 = −2(−3) − 2
?
−1 + 5 = 6 − 2
4=4✓
So, the solution of the equation is x = −3.
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Algebra 1
Worked-Out Solutions
299
Chapter 5
16. ∣ 2x − 8 ∣ = ∣ x + 5 ∣
19.
Equation 1
2x − 8 = x + 5
5x + 10y < 40
2x − 8 = −(x + 5)
System 2
y = 2x − 8
y = 2x − 8
y=x+5
y = −x − 5
y
16
10y
10
1
−4
2
y = 2x − 8
16
8
4 x
y≥x+1
y = −x − 5
x
The graphs intersect
at (13, 18)
The graphs intersect
at (1, −6).
Check
Check
∣ 2x − 8 ∣ = ∣ x + 5 ∣
?
∣ 2(13) − 8 ∣ =
?
∣ 26 − 8 ∣ =
?
∣ 18 ∣ =
∣ 13 + 5 ∣
∣ 18 ∣
18
18 = 18 ✓
2
2
21. Graph the system.
6
y > −2x + 3
∣1 + 5∣
−2
2
6
4
6 x
2
x
−2
22.
Test (0, 0). y > −4
x + 3y > 6
6
x − x + 3y > 6 − x
0 > −4 ✓
2
y
y ≥ —14 x − 1
∣6∣
6=6✓
y
4 x
−2
So, the solutions of the equation are x = 13 and x = 1.
17.
y
−4
∣ 2x − 8 ∣ = ∣ x + 5 ∣
?
∣ 2(1) − 8 ∣ =
?
∣2 − 8∣ =
?
∣ −6 ∣ =
4
y≤x−3
y = 2x − 8
−6
24
4 x
0 < 40 ✓
(1, −6)
−x + 6
3
1
y > −—x + 2
3
y
4
3y > −x + 6
4 x
3y
3
−2
—>—
−6
18.
2
−2
20. Graph the system.
8
−4
−2
5x + 10y < 40
?
5(0) + 10(0) < 40
−2
y=x+5
2
Test (0,0).
y
(13, 18)
−5x + 40
10
—<—
y < −—2x + 4
−2
24
y
10y < −5x + 40
2x − 8 = −x − 5
System 1
6
5x − 5x + 10y < 40 − 5x
Equation 2
−4
−2
−2
2x + y < 7
−9x + 3y ≥ 3
4
−9x + 9x + 3y ≥ 3 + 9x
3y
3
9x + 3
3
y < −2x + 7
2
3y ≥ 9x + 3
—≥—
2x − 2x + y < 7 − 2x
y
−4
−2
2
4 x
y ≥ 3x + 1
Test (0, 0). −9x + 3y ≥ 3
?
−9(0) + 3(0) ≥ 3
−4
0≥3✗
300
Algebra 1
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Chapter 5
Chapter 5 Test (p. 285)
3. Sample answer: Solve by substitution because y is already
isolated on one side of the equation.
1. Sample answer: Solve by elimination because the
Substitute 4x + 4 for y in Equation 2.
coefficients of the x-terms are opposites.
–8x + 2y = 8
Step 2
8x + 3y = –9
–8x + 2(4x + 4) = 8
–8x + y = 29
–8x + 2(4x) + 2(4) = 8
0 + 4y = 20
–8x + 8x + 8 = 8
4y = 20
0+8=8
4y 20
4
4
y=5
8=8
Step 3
—=—
Because the equation 8 = 8 is always true, the solutions
of the system are all points on the line y = 4x + 4. So, the
system has infinitely many solutions.
Step 4
–8x + y = 29
4. Sample answer: Solve by substitution because x is already
–8x + 5 = 29
–5
isolated on one side of the equation.
Substitute y – 11 for x
in Equation 2.
–5
–8x = 24
x − 3y = 1
–8x 24
–8
–8
x = –3
—=—
Check
8x + 3y = –9
?
8(–3) + 3(5) = –9
?
–24 + 15 = –9
–8x + y = 29
?
–8(–3) + 5 = 29
?
24 + 5 = 29
–9 = –9 ✓
–2y 12
–2
−2
y = –6
—=—
29 = 29 ✓
Check
is in slope-intercept form, and the other will be after only
one step.
y
1
—2 x + y = –6
3
–17 = –17 ✓
The solution is (–17, –6).
1
y = —35x + 5
−4
(−10, −1)
x
−4
1
y = − 2 x − 6 −8
The graphs appear to intersect at (–10, –1).
Check
1
—2 x + y = –6
?
1
—2 (–10) + (–1) = –6
?
–5 – 1 = –6
–6 = –6 ✓
x – 3y = 1
?
–17 – 3(–6) = 1
?
–17 + 18 = 1
1=1✓
—2 x – —2 x + y = –6 – —2 x
1
x = y – 11
?
–17 = –6 – 11
8
y = 5x + 5
y = – —2x – 6
x = –17
+11
–2y = 12
2. Sample answer: Solve by graphing because one equation
1
x = –6 − 11
–2y − 11 = 1
The solution is (–3, 5).
1
x = y − 11
(y − 11) − 3y = 1
+11
Substitute –6 for y
in Equation 1.
y = —35 x + 5
?
–1 = —35 (–10) + 5
?
–1 = –6 + 5
5. Sample answer: Solve by elimination because both equations
are in standard form, and none of the terms have a coefficient
of 1 or –1.
Step 1
Step 2
6x – 4y = 9
Multiply by 3.
9x – 6y = 15
Multiply by −2.
18x – 12y = 27
–18x + 12y = –30
0 + 0 = –3
0 = –3 ✗
The equation 0 = –3 is never true. So, the system has
no solution.
–1 = –1 ✓
The solution is (–10, –1).
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Algebra 1
Worked-Out Solutions
301
Chapter 5
6. Sample answer: Solve by substitution because y is already
isolated on one side of the equation.
10.
Substitute 5x – 7 for y in Equation 2.
−2
5x − 5x + y > 4 − 5x
x–7=–1
x=6
11.
Substitute 6 for x in Equation 1.
−3x + y > −2
6
2
y = 30 – 7
y = 23
Check
y = 5x – 7
?
23 = 5(6) – 7
?
23 = 30 – 7
−4
–4x + y = –1
?
–4(6) + 23 = –1
?
–24 + 23 = –1
23 = 23 ✓
12. a. Words
10
–1 = –1 ✓
5
7. Sample answer:
⋅
Cost per
gallon of + 2
gasoline
⋅
Cost per
gallon of + 1
gasoline
y
4
−4
System 10x + 2y = 45.5
8 x
5x + y = 22.75
(4, −3)
y≤x+4
−8
⋅
Cost per
quart of = 45.50
oil
⋅
Cost per
quart of = 22.75
oil
Equation 1
Equation 2
Solve by elimination.
y ≥ −x
Step 1
So, a system of inequalities that has the points (1, 2) and
(4, −3) in its solution region, but does not have (−2, 8) as a
solution is y ≤ x + 4 and y ≥ −x.
10x + 2y = 45.5
5x + y = 22.75
10x + 2y = 45.5
Multiply by −2.
two equivalent equations, and you solve each of those equations
in the same way that you solve 4x + 3 = −2x + 9. You write
and graph a linear equation for each side of the equation. The
x-value of the point where the two linear equations intersect
is the solution of the original equation. The absolute value
equation has two solutions, one for each system, whereas the
equation 4x + 3 = −2x + 9 only has one solution.
9. 2y ≤ x + 4
6
y
1
y ≤ —x + 2
2
−2
2
−2
Algebra 1
Worked-Out Solutions
0=0
Because the equation 0 = 0 is always true, every solution
of 10x + 2y = 45.5 is also a solution of 5x + y = 22.75.
So, the system has infinitely many solutions, and you do
not have enough information to find the answer.
b. Words
8
⋅
Cost per
gallon of + 2
gasoline
⋅
Cost per
quart of = 38.40
oil
Equation 8x + 2y = 38.40
x+4
—≤—
2
2
2y
−4
−10x − 2y = −45.5
0+0=0
8. Sample answer: When solving ∣ 2x + 1 ∣ = ∣ x − 7 ∣, you write
302
x
Variables Let x be the cost of 1 gallon of gasoline, and
let y be the cost of 1 quart of oil.
(1, 2)
−8
−2
−2
The solution is (6, 23).
(−2, 8)
y
4
y > 3x − 2
y = 5(6) – 7
6 x
−6
−3x + 3x +y > −2 + 3x
y = 5x – 7
4
−4
y > −5x + 4
+7
2
−2
5x + y > 4
–4x + (5x – 7) = –1
y
2
y<−x+1
–4x + y = –1
+7
x+y<1
x−x+y<1−x
x
This equation is not equivalent to either of the equations
from part (a). So, you can write a system of equations
with this equation and one of the equations from part (a),
and the solution of the system will be the cost of 1 gallon
of gasoline and 1 quart of oil.
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Chapter 5
c. Solve by elimination.
b. Words
Step 2
Step 3
10x + 2y = 45.5
10x +2y = 45.5
−(8x + 2y = 38.40)
System 12x + 3y ≤ 60
35.5 + 2y = 45.5
2x 7.1
2
2
x = 3.55
—=—
−35.5
x+y≥6
−35.5
x+y≥6
x−x+y≥6−x
2y = 10
2y 10
—=—
2
2
y=5
y ≥ −x + 6
Graph the system.
y ≥ −x + 6
y ≤ −4x +20
The solution is (3.55, 5). So, 1 gallon of gasoline costs $3.55,
and 1 quart of oil costs $5.
y
20
13. Sample answer: Solving a system of linear equations by
14. a. Words
12
⋅
Number of
+ 3
trophies
⋅
12
8
4
0
0
1
2
3
4
5 x
Trophies
Number of
≤ 60
medals
Variables Let x be the number of trophies you can buy,
and let y be the number of medals you can buy.
Inequality 12x + 3y ≤ 60
Sample answer: One of the points in the shaded region is
(3, 4). So, you can buy 3 trophies and 4 medals.
15.
8x + 4y = 12
8x − 8x + 4y = 12 − 8x
12x + 3y ≤ 60
12x − 12x + 3y ≤ 60 − 12x
3y ≤ −12x + 60
3y −12x + 60
—≤—
3
3
y ≤ −4x + 20
y
20
16
Medals
16
Medals
graphing gives you a visual picture of how the output values
of each equation change as the input values change and can
be quick when an estimate is sufficient. However, it can be
tedious to find the best scale for the axes that will allow you
to see where the solution occurs, and if one or more of the
coordinates of the solution is not a whole number, then the
graphing method may only give you an estimate.
≥ 6
Inequality x + y ≥ 6
10(3.55) + 2y = 45.5
2x + 0 = 7.1
Number of
Number of
+
trophies
medals
4y = −8x +12
4y
−8x + 12
—=—
4
4
3y = −6x − 15
3y −6x − 15
—=—
3
3
y = −2x − 5
y = −2x + 3
The slopes of both graphs are −2, but the graph of
8x + 4y = 12 has a y-intercept of 3, and the graph of
3y = −6x −15 has a y-intercept of −5. So, the graphs
are parallel and will never intersect. So, the system has
no solution.
12
8
4
0
0
1
2
3
4
5 x
Trophies
Sample answer: One of the points in the shaded region is
(2, 8). So, you can buy 2 trophies and 8 medals.
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Algebra 1
Worked-Out Solutions
303
Chapter 5
Chapter 5 Standards Assessment (pp. 286 –287)
ax − 8 = 4 − x
7. a.
1. D; To find the x-intercept, let y = 0.
a(−2) − 8 = 4 − (−2)
−9x + 2y = −18
−2a − 8 = 4 + 2
−9x + 2(0) = −18
−2a − 8 = 6
−9x + 0 = −18
+8
−9x = −18
−9x
−9
−18
−9
−2a
−2
—=—
a = −7
To find the y-intercept, let x = 0.
When a = −7, the solution is x = −2.
b.
−9(0) + 2y = −18
ax − 8 = 4 − x
a(12) − 8 = 4 − 12
0 + 2y = −18
12a − 8 = −8
2y = −18
2y −18
—=—
2
2
y = −9
+8
C(8) = 100 + 5(8) = 100 + 40 = 140
C(12) = 100 +5(12) = 100 + 60 = 160
C(16) = 100 +5(16) = 100 + 80 = 180
0
12
12a
12
—=—
a=0
When a = 0, the solution is x = 12.
c.
ax − 8 = 4 − x
a(3) − 8 = 4 − 3
3a − 8 = 1
+8
So, the numbers in the range of the function are 130, 140,
160, and 180.
+8
3a = 9
3a
3
9
3
—=—
3. Because the shaded region is above both boundary lines,
one of which is solid and one of which is dashed, the system
is y ≥ 3x − 2 and y > −x + 5.
+8
12a = 0
So, the intercepts of −9x + 2y = −18 are (2, 0) and (0, −9).
2. C(6) = 100 +5(6) = 100 + 30 = 130
14
−2
—=—
x=2
−9x + 2y = −18
+8
−2a = 14
a=3
When a = 3, the solution is x = 3.
4. no; If you were to write a system of equations representing
the two sides of this equation, the lines would have the same
slope. So, they are either the same line and the equation has
infinitely many solutions, or they are parallel lines and the
equation has no solution.
5. The phrases you should use are “reflection in the x-axis,”
“horizontal translation,” and “vertical translation.”
6. The two equations that form a system of linear equations
that has no solution are y = 3x + 2 and y = 3x + —12 . These
equations have the same slope (m = 3) and different
y-intercepts (b = 2 and b = —12, respectively). So, they are
parallel lines and will never intersect.
8. C; The only ordered pair that is in the shaded half-plane is
(−1, −1). The point (−1, 1) is on the dashed line. So, it is
not a solution of the linear inequality. The other two points
are in the unshaded half-plane. So, they are not solutions
either.
9. The systems
4x − 5y = 3
4x − 5y = 3
12x − 15y = 9
2x + 15y = −1
−4x − 30y = 2
2x + 15y = −1
are equivalent.
If you multiply Equation 1 of the first system by 3, you get
3(4x − 5y = 3) ⇒ 12x − 15y = 9. If you multiply Equation 2
of the first system by −2, you get
−2(2x + 15y = −1) ⇒ −4x − 30y = 2.
A = —21bh
10. P = x + 16 + 13
P > 9 + 16 + 13
A = —21(x)(13)
P > 38
A > —12 (9)(13)
So, P > 38 and A > 58.5
describe the perimeter and
area of the triangle.
304
Algebra 1
Worked-Out Solutions
A > 58.5
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