chapter 3 - beverly hills high...
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115Algebra 2
Worked-Out Solution Key
Prerequisite Skills (p. 150)
1. The linear inequality that represents the graph shown at
the right is y < 2 3 } 4 x 1 3.
2. The graph of a linear inequality in two variables is the set of all points in a coordinate plane that are solutions of the inequality.
3. x 1 y 5 4 4. y 5 3x 2 3
1
x
y
21
1
x
y
21
5. 22x 1 3y 5 212 1
x
y
21
6. 2x 2 12 5 16 7. 23x 2 7 5 12
2x 5 28 23x 5 19
x 5 14 x 5 2 19
} 3
8. 22x 1 5 5 2x 2 5 9. y ≥ 2x 1 2
5 5 4x 2 5
1
x
y
21
10 5 4x
5 }
2 5 x
10. x 1 4y < 216 11. 3x 1 5y > 25
y < 2 1 } 4 x 2 4 y > 2
3 } 5 x 2 1
1
x
y
22
1
x
y
22
Lesson 3.1
Investigating Algebra Activity 3.1 (p. 152)
1. y1 5 y2 5 3 when x 5 21. The solution is (21, 3).
2. no solution; There are no values of x where y1 5 y2.
3. y1 5 y2 5 5 when x 5 2. The solution is (2, 5).
4. no solution; There are no values of x where y1 5 y2.
5. y1 5 y2 5 2 when x 5 21. The solution is (21, 2).
6. y1 5 y2 for all values of x. There are infi nitely many
solutions.
7. A system of linear equations can have either none, one, or infi nitely many solutions.
3.1 Guided Practice (pp. 153–155)
1. 3x 1 2y 5 24 x 1 3y 5 1
2y 5 23x 2 4 3y 5 1 2 x
y 5 2 3 } 2 x 2 2 y 5
1 } 3 2
1 } 3 x
1x
y
22
From the graph, the lines appear to intersect at (22, 1).
Check:
3x 1 2y 5 24 x 1 3y 5 1
3(22) 1 2(1) 0 24 22 1 3(1) 0 1
26 1 2 0 24 22 1 3 0 1
24 5 24 ✓ 1 5 1 ✓
The solution is (22, 1).
2. 4x 2 5y 5 210 2x 2 7y 5 4
4x 1 10 5 5y 2x 2 4 5 7y
4 } 5 x 1 2 5 y
2 } 7 x 2
4 } 7 5 y
1
x
y
21
From the graph, the lines appear to intersect at (25, 22).
Check:
4x 2 5y 5 210 2x 2 7y 5 4
4(25) 2 5(22) 0 210 2(25) 2 7(22) 0 4
220 1 10 0 210 210 1 14 0 4
210 5 210 ✓ 4 5 4 ✓
The solution is (25, 22).
3. 8x 2 y 5 8 3x 1 2y 5 216
y 5 8x 2 8 2y 5 23x 2 16
y 5 2 3 } 2 x 2 8
4
x
y
22
From the graph, the lines appear to intersect at (0, 28).
Check:
8x 2 y 5 8 3x 1 2y 5 216
8(0) 2 (28) 0 8 3(0) 1 2(28) 0 216
0 1 8 0 8 0 2 16 0 216
8 5 8 ✓ 216 5 216 ✓
The solution is (0, 28).
Chapter 3
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116Algebra 2Worked-Out Solution Key
4. 2x 1 5y 5 6
x
y
1
1
2x 1 5y 5 6
4x 1 10y 5 12
4x 1 10y 5 12
Each point on the line is a solution, and the system has infinitely many solutions. The system is consistent and dependent.
5. 3x 2 2y 5 10
x
y
1
1
3x 2 2y 5 10
3x 2 2y 5 2
3x 2 2y 5 2
The two lines have no point of intersection, so the system has no solution. The system is inconsistent.
6. 22x 1 y 5 5
x
y
1
1
22x 1 y 5 5
y 5 2x 1 2(21, 3)
y 5 2x 1 2
The lines intersect at (21, 3), so (21, 3) is the solution. The system is consistent and independent.
7. y 5 1 p x 1 36 y 5 2.5 p x
y 5 x 1 36 y 5 2.5x
Number of rides
To
tal c
ost
(d
olla
rs)
x
y
100 20 30 40
20
0
40
60
y 5 2.5x
y 5 x 1 36(24, 60)
The lines appear to intersect at (24, 60).
Check:
y 5 x 1 36 y 5 2.5x
60 0 24 1 36 60 0 2.5(24)
60 5 60 ✓ 60 5 60 ✓
The total costs are equal after 24 rides. If the monthly pass is increased to $36, it will take longer for both options to cost the same.
3.1 Exercises (pp. 156–158)
Skill Practice
1. A consistent system that has exactly one solution is called independent.
2. If the two lines intersect at one point, then the solution is the point of intersection. If the two lines are parallel, there is no solution. If the two lines coincide then there are infinitely many solutions.
3.
1
x
y
21
From the graph, the lines appear to intersect at (1, 21).
y 5 23x 1 2 y 5 2x 2 3
21 0 23(1) 1 2 21 0 2(1)23
21 0 23 1 2 21 0 2 2 3
21 5 21 ✓ 21 5 21 ✓
The solution is (1, 21).
4. 1
x
y
21
From the graph, the lines appear to intersect at (21, 23).
y 5 5x 1 2 y 5 3x
23 0 5(21) 1 2 23 0 3(21)
23 0 25 1 2 23 5 23 ✓
23 5 23 ✓
The solution is (21, 23).
5.
1
x
y
21
From the graph, the lines appear to intersect at (4, 21).
y 5 2x 1 3 2x 2 3y 5 21
21 0 2(4) 1 3 2(4) 2 3(21) 0 21
21 0 24 1 3 24 1 3 0 21
21 5 21 ✓ 21 5 21 ✓
The solution is (4, 21).
Chapter 3, continued
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117Algebra 2
Worked-Out Solution Key
6.
2
x
y
21
From the graph, the lines appear to intersect at (6, 22).
x 1 2y 5 2 x 2 4y 5 14
6 1 2(22) 0 2 6 2 4(22) 0 14
6 2 4 0 2 6 1 8 0 14
2 5 2 ✓ 14 5 14 ✓
The solution is (6, 22).
7.
1
x
y
21
From the graph, the lines appear to intersect at (5, 0).
y 5 2x 2 10 x 2 4y 5 5
0 0 2(5) 2 10 5 2 4(0) 0 5
0 0 10 2 10 5 2 0 0 5
0 5 0 ✓ 5 5 5 ✓
The solution is (5, 0).
8.
1
x
y
22
From the graph, the lines appear to intersect at (12, 0).
2x 1 6y 5 212 x 1 6y 5 12
2(12) 1 6(0) 0 212 12 1 6(0) 0 12
212 1 0 0 212 12 1 0 0 12
212 5 212 ✓ 12 5 12 ✓
The solution is (12, 0).
9.
1
y
x21
From the graph, the lines appear to intersect at (22, 4).
y 5 23x 2 2 5x 1 2y 5 22
4 0 23(22) 2 2 5(22) 1 2(4) 0 22
4 0 6 2 2 210 1 8 0 22
4 5 4 ✓ 22 5 22 ✓
The solution is (22, 4).
10.
1
x
y
21
From the graph, the lines appear to intersect at (26, 5).
y 5 23x 2 13 2x 2 2y 5 24
5 0 23(26) 2 13 2(26)22(5) 0 24
5 0 18 2 13 6 2 10 0 24
5 5 5 ✓ 24 5 24 ✓
The solution is (26, 5).
11.
1
x
y
21
x 2 7y 5 6 23x 1 21y 5 218
7y 5 6 2 x 21y 5 3x 2 18
y 5 x } 7 2
6 } 7 y 5
x } 7 2
6 } 7
The graphs of the equations are the same line. Each point on the line is a solution. There are infi nitely many solutions.
12.
3
x
y
22
y 5 4x 1 3 20x 2 5y 5 215
25y 5 215 2 20x
y 5 4x 1 3
The graphs of the equations are the same line. Each point on the line is a solution, so there are infi nitely many solutions.
13.
1
x
y
21
From the graph, the lines appear to intersect at (3.5, 2.2).
4x 2 5y 5 3 3x 1 2y 5 15
4(3.5) 2 5(2.2) 0 3 3(3.5) 1 2(2.2) 0 15
14 2 11 0 3 10.5 1 4.4 0 15
3 5 3 ✓ 14.9 ø 15 ✓
The solution is approximately (3.5, 2.2).
Chapter 3, continued
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118Algebra 2Worked-Out Solution Key
14.1
x
y
21
From the graph, the lines appear to intersect at (22, 23).
7x 1 y 5 217 3x 2 10y 5 24
7(22) 1 (23) 0 217 3(22) 2 10(23) 0 24
214 2 3 0 217 26 1 30 0 24
217 5 217 ✓ 24 5 24 ✓
The solution is (22, 23).
15. C;
x
y1
1
7x 1 2y 5 25
24x 2 y 5 2
(1, 26)
From the graph, the lines appear to intersect at (1, 26).
24x 2 y 5 2 7x 1 2y 5 25
24(1) 2 (26) 0 2 7(1) 1 2(26) 0 25
24 1 6 0 2 7 2 12 0 25
2 5 2 ✓ 25 5 25 ✓
The solution is (1, 26).
16. The student did not check the solution in the second equation.
3x 2 2y 5 2 x 1 2y 5 6
3(0) 2 2(21) 0 2 0 1 2(21) 0 6
0 1 2 0 2 22 Þ 6
2 5 2 ✓
The solution is not (0, 21).
17.
x
y
1
1
y 5 21 (2, 21)
3x 1 y 5 5
From the graph, the lines appear to intersect at (2, 21).
y 5 21 3x 1 y 5 5
21 5 21 ✓ 3(2) 1 (21) 0 5
6 2 1 0 5
5 5 5 ✓
The solution is (2, 21). The system is consistent and independent.
18.
x
y
1
1
(3, 2)
2x 2 y 5 4
x 2 2y 5 21
From the graph, the lines appear to intersect at (3, 2).
2x 2 y 5 4 x 2 2y 5 21
2(3) 2 2 0 4 3 2 2(2) 0 21
6 2 2 0 4 3 2 4 0 21
4 5 4 ✓ 21 5 21 ✓
The solution is (3, 2). The system is consistent and independent.
19.
x
y
2
1
y 5 3x 2 2
y 5 3x 1 2
y 5 3x 1 2 y 5 3x 2 2
The graphs of the equations are two parallel lines. The system has no solution. The system is inconsistent.
20.
x
y
1
1
y 5 2x 2 1
26x 1 3y 5 23
y 5 2x 2 1 26x 1 3y 5 23
3y 5 6x 2 3
y 5 2x 2 1
The graphs of the equations are the same line. The system has infi nitely many solutions. The system is consistent and dependent.
Chapter 3, continued
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119Algebra 2
Worked-Out Solution Key
21.
x
y
1
2
220x 1 12y 5 224
5x 2 3y 5 6
220x 1 12y 5 224 5x 2 3y 5 6
12y 5 20x 2 24 23y 5 25x 1 6
y 5 5 } 3 x 2 2 y 5
5 } 3 x 2 2
The graphs of the equations are the same line. The system has infi nitely many solutions. The system is consistent and dependent.
22.
x
y
2
1
4x 2 5y 5 0
3x 2 5y 5 25(5, 4)
From the graph, the lines appear to intersect at (5, 4).
4x 2 5y 5 0 3x 2 5y 5 25
4(5) 2 5(4) 0 0 3(5) 2 5(4) 0 25
20 2 20 0 0 15 2 20 0 25
0 5 0 ✓ 25 5 25 ✓
The solution is (5, 4).
The system is consistent and independent.
23.
x
y
2
1
3x 1 7y 5 62x 1 9y 5 4
(2, 0)
From the graph, the lines appear to intersect at (2, 0).
3x 1 7y 5 6 2x 1 9y 5 4
3(2) 1 7(0) 0 6 2(2) 1 9(0) 0 4
6 1 0 0 6 4 1 0 0 4
6 5 6 ✓ 4 5 4 ✓
The solution is (2, 0).
The system is consistent and independent.
24.
x
y
2
21
4x 1 5y 5 3
6x 1 9y 5 9
(23, 3)
From the graph, the lines appear to intersect at (23, 3).
4x 1 5y 5 3 6x 1 9y 5 9
4(23) 1 5(3) 0 3 6(23) 1 9(3) 0 9
212 1 15 0 3 218 1 27 0 9
3 5 3 ✓ 9 5 9 ✓
The solution is (23, 3).
The system is consistent and independent.
25.
x
y
1
1
8x 1 9y 5 15
5x 2 2y 5 17
(3, 21)
From the graph, the lines appear to intersect at (3, 21).
8x 1 9y 5 15 5x 2 2y 5 17
8(3) 1 9(21) 0 15 5(3) 2 2(21) 0 17
24 2 9 0 15 15 1 2 0 17
15 5 15 ✓ 17 5 17 ✓
The solution is (3, 21).
The system is consistent and independent.
26.
x
y
2
8
(8, 22)
x 2 3y 5 1012
x 1 2y 5 2214
From the graph, the lines appear to intersect at (8, 22).
1 }
2 x 2 3y 5 10
1 }
4 x 1 2y 5 22
1 }
2 (8) 2 3(22) 0 10
1 }
4 (8) 1 2(22) 0 22
4 1 6 0 10 2 2 4 0 22
10 5 10 ✓ 22 5 22 ✓
The solution is (8, 22).
The system is consistent and independent.
Chapter 3, continued
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120Algebra 2Worked-Out Solution Key
27.
x
y
2
2
3x 2 2y 5 215
x 2 y 5 2523
3x 2 2y 5 215 x 2 2 } 3 y 5 25
22y 5 23x 2 15 2 2 } 3 y 5 2x 2 5
y 5 3 } 2 x 1
15 } 2 y 5
3 } 2 x 1
15 } 2
The graphs of the equations are the same line. The system has infi nitely many solutions. The system is consistent and dependent.
28.
x
y
1
1
52 x 2 y 5 24
145x 2 2y 5
5 }
2 x 2 y 5 24 5x 2 2y 5
1 } 4
2y 5 2 5 } 2 x 2 4 22y 5 25x 1
1 } 4
y 5 5 } 2 x 1 4 y 5
5 } 2 x 2
1 } 8
The graphs of the equations are parallel lines. The system has no solution, so it is inconsistent.
29. A;
x
y
1
1
3x 1 4y 5 26
212x 1 16y 5 10
212x 1 16y 5 10 3x 1 4y 5 26
There is exactly one solution of the system. The system is consistent and independent.
30. a. Sample answer: 3x 1 2y 5 9
2x 1 y 5 5
b. Sample answer: 3x 2 y 5 2
26x 1 2y 5 10
c. Sample answer: 2x 2 3y 5 4
24x 1 6y 5 28
31.
1
x
y
22
y 5 x 1 2 y 5 x
There is no solution.
32.
1
x
y
21
From the graph, the lines appear to intersect at (2.5, 1.5).
y 5 x 2 1 y 5 2x 1 4
1.5 0 2.5 2 1 1.5 0 22.5 1 4
1.5 0 1.5 1.5 5 1.5 ✓
1.5 5 1.5 ✓
The solution is (2.5, 1.5).
33.
1
x
y
22
From the graph, the lines appear to intersect at (24, 2) and (4, 2).
y 5 x 22 y 5 2
2 0 24 2 2 2 5 2 ✓
2 0 4 2 2
2 5 2 ✓
2 0 4 2 2
2 0 4 2 2 2 5 2 ✓
The solutions are (24, 2) and (4, 2).
34. a. The system is consistent and independent when a Þ c.
b. The system is consistent and dependent when a 5 c and b 5 d.
c. The system is inconsistent when a 5 c and b Þ d.
Problem Solving
35. Let x 5 hrs as lifeguard.
20 4 6 8 10 12 14
20
468
10121416
(6, 8)
Hours as a lifeguard
Ho
urs
as
a ca
shie
r
x
y
Let y 5 hrs as cashier.
x 1 y 5 14
8x 1 6y 5 96
You worked 6 hours as a lifeguard and 8 hours as acashier last week.
Chapter 3, continued
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121Algebra 2
Worked-Out Solution Key
36. Let x 5 warnings.
x
y
x 1 y 5 375
y 1 37 5 x
00
80
120
240
320
80 160 240 320
(206, 169)
Number of warnings
Nu
mb
er o
f ti
cket
s
Let y 5 speeding tickets.
x 1 y 5 375
y 1 37 5 x
The state trooper issued 206 warnings and 169 speeding tickets.
37. Let y 5 total cost (in dollars).
Let x 5 days.
0 4 62 8 10 12 14 160
5025
75100
150125
175
(11, 132)
Number of days
Tota
l co
st (
do
llars
)
x
y y 5 12x
y 5 121 1 x
Option A: y 5 121 1 x
Option B: y 5 12x
The plans are equal when used for 11 days. If the daily cost of Option B increases, the plans will be equal in fewer days. The graph of the new option B equation will be the rotation of y 5 12x counterclockwise and the x-intersection with the graph of the option A equation will be less than 11.
38. a. Let y 5 total cost (in dollars).
Let x 5 years since purchase.
Refrigerator A: y 5 600 1 50x
Refrigerator B: y 5 1200 1 40x
b.
B
A
(60, 3600)
Co
st (
do
llars
)
0200 40 60 80
1000
2000
3000
4000
Number of years
x
y
In 60 years the total costs of owning the refrigerators will be equal.
c. No. It is not likely that the refrigerators would be in use for 60 years. You can conclude that, for the life of the refrigerators, the cost of owning refrigerator A will always be less than the cost of owning refrigerator B.
39. a. m 5 20.09583x 1 50.84 (Men)
b. w 5 20.1241x 1 57.08 (Women)
c.
0 80 160 240 32008
16243240485664
Years since 1972
Win
nin
g t
imes
(se
con
ds)
x
y
men
women
The lines appearto intersect at(220.7, 29.7).
You can predict that the women’s performance will catch up to the men’s performance about 221 years after 1972, or in 2193.
d. No. The lines may be good models for the years contained in the data set, but eventually the swimming times will start leveling off as swimmers approach the maximum swimming speed for a human. So, it is not reasonable to assume that the winning times will continue to decrease indefi nitely at the given rates.
40. a. Distance from park(ft)
5 Total distance(ft)
2 Speed(ft/sec)
p Time(sec)
d 5 5000 2 25 p t
An equation is d 5 5000 2 25t.
b. Let d 5 0 to fi nd when you reach the park.
0 5 5000 2 25t
d 5 5000 2 25t
d 5 3000 2 15tDis
tan
ce f
rom
par
k (f
eet)
0500 100 150 200
1000
2000
4000
5000
3000
Time (seconds)
t
d
(200, 0)
25t 5 5000
t 5 200
Use the verbal model from part (a) to write the equation d 5 3000 2 rt for your friend’s distance d from the park after t seconds, where r represents your friend’s speed. Let d 5 0 and t 5 200 and solve for r.
0 5 3000 2 r (200)
r 5 15
Your friend travels at a speed of 15 feet per second.
c. An equation for your friend’s distance is
d 5 3000 2 15t.
Mixed Review
41. 8x 1 1 5 3x 2 14
8x 5 3x 2 15
5x 5 215
x 5 23
42. 24(x 1 3) 5 5x 1 9
24x 2 12 5 5x 1 9
24x 2 21 5 5x
221 5 9x
2 7 } 3 5 x
43. x 1 2 5 3
} 2 x 2 5 } 4
x 1 13
} 4 5 3 } 2 x
13
} 4 5
1 } 2 x
13
} 2 5 x
Chapter 3, continued
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122Algebra 2Worked-Out Solution Key
44. x 2 18 5 9
x 2 18 5 9 or x 2 18 5 29
x 5 27 or x 5 9
45. 2x 1 5 5 12
2x 1 5 5 12 or 2x 1 5 5 212
2x 5 7 or 2x 5 217
x 5 7 } 2 or x 5 2
17 } 2
46. 5x 2 18 5 17
5x 2 18 5 17 or 5x 2 18 5 217
5x 5 35 or 5x 5 1
x 5 7 or x 5 1 } 5
47. 3x 2 2y 5 8 48. 25x 1 y 5 212
22y 5 23x 1 8 y 5 5x 2 12
y 5 3 } 2 x 2 4 When x 5 9:
When x 5 22: y 5 5(9) 2 12
y 5 3 } 2 (22) 2 4 y 5 45 2 12
y 5 23 2 4 y 5 33
y 5 27
49. 8x 2 3y 5 10 50. 8x 2 2y 5 7
23y 5 28x 1 10 22y 5 28x 1 7
y 5 8 } 3 x 2
10 } 3 y 5 4x 2
7 } 2
When x 5 8: When x 5 21:
y 5 8 } 3 (8) 2
10 } 3 y 5 4(21) 2
7 } 2
y 5 64
} 3 2 10
} 3 y 5 24 2 7 } 2
y 5 18 y 5 2 15
} 2
51. 16x 1 9y 5 224 52. 212x 1 9y 5 260
9y 5 216x 2 24 9y 5 12x 2 60
y 5 2 16
} 9 x 2 8 } 3 y 5
4 } 3 x 2
20 } 3
When x 5 26: When x 5 27:
y 5 2 16
} 9 (26) 2 8 } 3 y 5
4 } 3 (27) 2
20 } 3
y 5 32
} 3 2 8 } 3 y 5 2
28 } 3 2
20 } 3
y 5 8 y 5 216
53. C 5 5 } 9 (F 2 32)
C 5 5 } 9 (101 2 32)
C 5 5 } 9 (69)
C 5 115
} 3 ø 38.3
Your dog’s temperature is about 38.38C. Your dog does not have a fever.
3.1 Graphing Calculator Activity (p. 159)
1. The solution is about (2.3,20.3).
2. The solution is about (2.71, 9.57).
3. The solution is about (11.08, 16.25).
4. The solution is (24,243).
5. The solution is about (251.43, 26.14).
6. The solution is about (212.21, 1.97).
7. Let x 5 days in San Antonio.
Let y 5 say in Dallas.
x 1 y 5 7
275x 1 400y 5 2300
Using a graphing calculator, the solution is (4, 3). You should spend 4 days in San Antonio and 3 days in Dallas.
8. Let x 5 number of adult tickets sold.
Let y 5 number of child tickets sold.
x 1 y 5 800
7x 1 5y 5 4600
Using a graphing calculator, the solution is (300, 500). The movie theater admitted 300 adults and 500 children that day.
Lesson 3.2
3.2 Guided Practice (pp. 161–163)
1. 4x 1 3y 5 22
x 1 5y 5 29 → x 5 29 2 5y
When x 5 29 2 5y:
4x 1 3y 5 22
4(29 2 5y) 1 3y 5 22
236 2 20y 1 3y 5 22
217y 5 34
y 5 22
When y 5 22:
x 5 29 2 5y
x 5 29 2 5(22)
x 5 29 1 10
x 5 1
The solution is (1, 22).
Check: 4x 1 3y 5 22 x 1 5y 5 29
4(1) 1 3(22) 0 22 1 1 5(22) 0 29
4 2 6 0 22 1 2 10 0 29
22 5 22 ✓ 29 5 29 ✓
Chapter 3, continued
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123Algebra 2
Worked-Out Solution Key
Chapter 3, continued
2. 3x 1 3y 5 215 3 3 9x 1 9y 5 245
5x 2 9y 5 3 5x 2 9y 5 3
14x 5 242
x 5 23
When x 5 23:
3x 1 3y 5 215
3(23) 1 3y 5 215
29 1 3y 5 215
y 5 22
The solution is (23, 22).
Check: 3x 1 3y 5 215 5x 2 9y 5 3
3(23) 1 3(22) 0 215 5(23) 2 9(22) 0 3
29 2 6 0 215 215 1 18 0 3
215 5 215 ✓ 3 5 3 ✓
3. 3x 2 6y 5 9 → x 5 2y 1 3
24x 1 7y 5 216
When x 5 2y 1 3:
24x 1 7y 5 216
24(2y 1 3) 1 7y 5 216
28y 2 12 1 7y 5 216
2y 5 24
y 5 4
When y 5 4:
3x 2 6y 5 9
3x 2 6(4) 5 9
3x 5 33
x 5 11
The solution is (11, 4).
Check: 3x 2 6y 5 9 24x 1 7y 5 216
3(11) 2 6(4) 0 9 24(11) 1 7(4) 0 216
33 2 24 0 9 244 1 28 0 216
9 5 9 ✓ 216 5 216 ✓
4.ShortSleeve Cost($/shirt)
p Short Sleeve Shirts(shirts)
1
LongSleeveCost($/shirt)
p LongSleeveShirts(shirts)
5
TotalCost($)
5 p x 1 7 p y 5 3715
ShortSleeve SellingPrice($/shirt)
p
Short Sleeve Shirts(shirts)
1
LongSleeveSellingPrice($/shirt)
p
LongSleeveShirts(shirts)
5
Totalrevenue($)
8 p x 1 12 p y 5 6160
5x 1 7y 5 3715 3 28 240x 2 56y 5 229,720
8x 1 12y 5 6160 3 5 40x 1 60y 5 30,800
4y 5 1080
y 5 270
When y 5 270:
5x 1 7y 5 3715
5x 1 7(270) 5 3715
5x 5 1825
x 5 365
The school sold 365 short sleeve T-shirts and 270 long sleeve T-shirts.
5. 12x 2 3y 5 29
24x 1 y 5 3 → y 5 4x 1 3
When y 5 4x 1 3:
12x 2 3y 5 29
12x 2 3(4x 1 3) 5 29
12x 2 12x 2 9 5 29
29 5 29
There are infi nitely many solutions.
6. 6x 1 15y 5 212 6x 1 15y 5 212
22x 2 5y 5 9 3 3 26x 2 15y 5 227
0 Þ 239
There is no solution.
7. 5x 1 3y 5 20 5x 1 3y 5 20
2x 2 3 } 5 y 5 24 3 5 25x 2 3y 5 220
0 5 0
There are infi nitely many solutions.
8. 12x 2 2y 5 21 12x 2 2y 5 21
3x 1 12y 5 24 3 (24) 212x 2 48y 5 16
250y 5 37
y 5 2 37
} 50
When y 5 2 37
} 50 :
12x 2 2y 5 21
12x 2 2(2 37
} 50 ) 5 21
12x 1 37
} 25 5 21
12x 5 488
} 25
x 5 122
} 75
The solution is 1 122 } 75 ,2
37 } 50 2 .
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124Algebra 2Worked-Out Solution Key
Chapter 3, continued
9. 8x 1 9y 5 15 3 5 40x 1 45y 5 75
5x 2 2y 5 17 3 (28) 240x 1 16y 5 2136
61y 5 261
y 5 21
When y 5 21:
8x 1 9y 5 15
8x 1 9(21) 5 15
8x 2 9 5 15
x 5 3
The solution is (3, 21).
10. 5x 1 5y 5 5 5x 1 5y 5 5
5x 1 3y 5 4.2 3 (21) 25x 2 3y 5 24.2
2y 5 0.8
y 5 0.4
When y 5 0.4:
5x 1 5y 5 5
5x 1 5(0.4) 5 5
5x 1 2 5 5
x 5 0.6
The solution is (0.6, 0.4).
3.2 Exercises (pp. 164–167)
Skill Practice
1. To solve a linear system where one of the coeffi cients is 1 or 21, it is usually easiest to use the substitution method.
2. Multiply one (or both) of the equations by a constant to obtain coeffi cients that differ only in sign for one of the variables. Add the revised equations together and solve for the remaining variable. Substitute this value into either of the original equations and solve for the other variable.
3. 2x 1 5y 5 7
x 1 4y 5 2 → x 5 2 2 4y
When x 5 2 2 4y: When y 5 21:
2x 1 5y 5 7 x 5 2 2 4y
2(2 2 4y) 1 5y 5 7 x 5 2 2 4(21)
4 2 8y 1 5y 5 7 x 5 6
23y 5 3
y 5 21
The solution is (6,21).
4. 3x 1 y 5 16 → y 5 23x 1 16
2x 2 3y 5 24
When y 5 23x 1 16: When x 5 4:
2x 2 3(23x 1 16) 5 24 3x 1 y 5 16
2x 1 9x 2 48 5 24 3(4) 1 y 5 16
11x 5 44 12 1 y 5 16
x 5 4 y 5 4
The solution is (4, 4).
5. 6x 2 2y 5 5
23x 1 y 5 7 → y 5 3x 1 7
When y 5 3x 1 7:
6x 2 2(3x 1 7) 5 5
6x 2 6x 2 14 5 5
214 Þ 5
There is no solution.
6. x 1 4y 5 1 → x 5 1 2 4y
3x 1 2y 5 212
When x 5 1 2 4y: When y 5 3 } 2 :
3(1 2 4y) 1 2y 5 212 x 1 4 1 3 } 2 2 5 1
3 2 12y 1 2y 5 212 x 1 6 5 1
210y 5 215 x 5 25
y 5 3 } 2
The solution is 1 25, 3 }
2 2 .
7. 3x 2 y 5 2 → y 5 3x 2 2
6x 1 3y 5 14
When y 5 3x 2 2: When x 5 4 } 3 :
6x 1 3(3x 2 2) 5 14 3 1 4 }
3 2 2 y 5 2
6x 1 9x 2 6 5 14 4 2 y 5 2
15x 5 20 y 5 2
x 5 4 } 3
The solution is 1 4 }
3 , 2 2 .
8. 3x 2 4y 5 25
2x 1 3y 5 25 → x 5 5 1 3y
When x 5 5 1 3y: When y 5 24:
3(5 1 3y) 2 4y 5 25 2x 1 3(24) 5 25
15 1 9y 2 4y 5 25 2x 2 12 5 25
5y 5 220 x 5 27
y 5 24
The solution is (27, 24).
9. 3x 1 2y 5 6
x 2 4y 5 212 → x 5 4y 2 12
When x 5 4y 2 12: When y 5 3:
3(4y 2 12) 1 2y 5 6 x 2 4(3) 5 212
12y 2 36 1 2y 5 6 x 2 12 5 212
14y 5 42 x 5 0
y 5 3
The solution is (0, 3).
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125Algebra 2
Worked-Out Solution Key
Chapter 3, continued
10. 6x 2 3y 5 15
22x 1 y 5 25 → y 5 2x 2 5
When y 5 2x 2 5:
6x 2 3(2x 2 5) 5 15
6x 2 6x 1 15 5 15
15 5 15
There are infi nitely many solutions.
11. 3x 1 y 5 21 → y 5 23x21
2x 1 3y 5 18
When y 5 23x 2 1: When x 5 23:
2x 1 3(23x 2 1) 5 18 3(23) 1 y 5 21
2x 2 9x 2 3 5 18 y 5 8
27x 5 21
x 5 23
The solution is (23, 8).
12. 2x 2 y 5 1 → y 5 21 1 2x
8x 1 4y 5 6
When y 5 21 1 2x: When x 5 5 } 8 :
8x 1 4(21 1 2x) 5 6 2 1 5 } 8 2 2 y 5 1
8x 2 4 1 8x 5 6 5 }
4 2 y 5 1
16x 5 10 y 5 1 } 4
x 5 5 } 8
The solution is 1 5 } 8
, 1
} 4 2 .
13. 3x 1 7y 5 13
x 1 3y 5 2 7 → x 5 27 2 3y
When x 5 27 2 3y: When y 5 217:
3(27 2 3y) 1 7y 5 13 x 1 3(217) 5 27
221 2 9y 1 7y 5 13 x 2 51 5 27
22y 5 34 x 5 44
y 5 217
The solution is (44, 217).
14. 2x 1 5y 5 10
23x 1 y 5 36 → y 5 3x 1 36
When y 5 3x 1 36: When x 5 210:
2x 1 5(3x 1 36) 5 10 23(210) 1 y 5 36
2x 1 15x 1 180 5 10 30 1 y 5 36
17x 5 2170 y 5 6
x 5 210
The solution is (210, 6).
15. 2x 1 6y 5 17 2x 1 6y 5 17
2x 2 10y 5 9 3 (21) 22x 1 10y 5 29
16y 5 8
When y 5 1 } 2 : y 5
1 } 2
2x 1 6 1 1 } 2 2 5 17
2x 1 3 5 17
x 5 7
The solution is 1 7, 1 }
2 2 .
16. 4x 2 2y 5 216 3 2 8x 2 4y 5 232
23x 1 4y 5 12 23x 1 4y 5 12
5x 5 220
When x 5 24: x 5 24
4(24) 2 2y 5 216
216 2 2y 5 216
y 5 0
The solution is (24, 0).
17. 3x 2 4y 5 210 3 (22) 26x 1 8y 5 20
6x 1 3y 5 242 6x 1 3y 5 242
11y 5 222
When y 5 22: y 5 22
3x 2 4(22) 5 210
3x 1 8 5 210
x 5 26
The solution is (26, 22).
18. 4x 2 3y 5 10 3 (22) 28x 1 6y 5 220
8x 2 6y 5 20 8x 2 6y 5 20
0 5 0
There are infi nitely many solutions.
19. 5x 2 3y 5 23 3 2 10x 2 6y 5 26
2x 1 6y 5 0 2x 1 6y 5 0
12x 5 26
When x 5 2 1 } 2 : x 5 2
1 } 2
5 1 2 1 } 2 2 2 3y 5 23
2 5 } 2 2 3y 5 23
23y 5 2 1 } 2
y 5 1 } 6
The solution is 1 2 1
} 2 , 1 }
6 2 .
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126Algebra 2Worked-Out Solution Key
20. 10x 2 2y 5 16 10x 2 2y 5 16
5x 1 3y 5 212 3 (22) 210x 2 6y 5 24
28y 5 40
When y 5 25: y 5 25
10x 2 2(25) 5 16
10x 1 10 5 16
x 5 3 } 5
The solution is 1 3 } 5 , 25 2 .21. 2x 1 5y 5 14 3 (23) 26x 2 15y 5 242
3x 2 2y 5 236 3 2 6x 2 4y 5 272
219y 5 2114
When y 5 6: y 5 6
2x 1 5(6) 5 14
2x 1 30 5 14
x 5 28
The solution is (28, 6).
22. 7x 1 2y 5 11 3 (23) 221x 2 6y 5 233
22x 1 3y 5 29 3 2 24x 1 6y 5 58
225x 5 25
When x 5 21: x 5 21
7(21) 1 2y 5 11
27 1 2y 5 11
y 5 9
The solution is (21, 9).
23. 3x 1 4y 5 18 3 (22) 26x 2 8y 5 236
6x 1 8y 5 18 6x 1 8y 5 18
0 Þ 218
There is no solution.
24. 2x 1 5y 5 13 3 (23) 26x 2 15y 5 239
6x 1 2y 5 213 6x 1 2y 5 213
213y 5 252
When y 5 4: y 5 4
2x 1 5(4) 5 13
2x 1 20 5 13
x 5 2 7 } 2
The solution is 1 2 7 } 2 , 4 2 .
25. 4x 2 5y 5 13 3 (23) 212x 1 15y 5 239
6x 1 2y 5 48 3 2 12x 1 4y 5 96
19y 5 57
When y 5 3: y 5 3
4x 2 5(3) 5 13
4x 2 15 5 13
x 5 7
The solution is (7, 3).
26. 6x 2 4y 5 14 3 2 12x 2 8y 5 28
2x 1 8y 5 21 2x 1 8y 5 21
14x 5 49
When x 5 7
} 2 : x 5 7 } 2
6 1 7 } 2
2 2 4y 5 14
21 2 4y 5 14
y 5 7 } 4
The solution is 1 7 } 2
, 7 }
4 2 .
27. The error was made when multiplying the fi rst equation by the constant 22. Not every term was multiplied by 22.
3x 1 2y 5 7 3 (22) 26x 2 4y 5 214
5x 1 4y 5 15 5x 1 4y 5 15
2x 5 21
x 5 1
28. 3x 1 2y 5 11
4x 1 y 5 22 → y 5 24x 2 2
When y 5 24x 2 2: When x 5 23:
3x 1 2(24x 2 2) 5 11 y 5 24(23) 2 2
3x 2 8x 2 4 5 11 y 5 10
25x 5 15
x 5 23
The solution is (23, 10).
29. 2x 2 3y 5 8 3 2 4x 2 6y 5 16
24x 1 5y 5 210 24x 1 5y 5 210
2y 5 6
When y 5 26: y 5 26
2x 2 3(26) 5 8
2x 1 18 5 8
x 5 25
The solution is (25, 26).
Chapter 3, continued
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127Algebra 2
Worked-Out Solution Key
30. 3x 1 7y 5 21 3 (22) 26x 2 14y 5 2
2x 1 3y 5 6 3 3 6x 1 9y 5 18
25y 5 20
When y 5 24: y 5 24
3x 1 7(24) 5 21
3x 2 28 5 21
x 5 9
The solution is (9, 24).
31. 4x 2 10y 5 18 4x 2 10y 5 18
22x 1 5y 5 29 3 2 24x 1 10y 5 218
0 5 0
There are infi nitely many solutions.
32. 3x 2 y 5 22 → y 5 3x 1 2
5x 1 2y 5 15
When y 5 3x 1 2: When x 5 1:
5x 1 2(3x 1 2) 5 15 y 5 3(1) 1 2
5x 1 6x 1 4 5 15 y 5 5
11x 5 11
x 5 1
The solution is (1, 5).
33. x 1 2y 5 28 → x 5 28 2 2y
3x 2 4y 5 224
When x 5 2822y: When y 5 0:
3(28 2 2y) 2 4y 5 224 x 5 28 2 2(0)
224 2 6y 2 4y 5 224 x 5 28
210y 5 0
y 5 0
The solution is (28, 0).
34. 2x 1 3y 5 26 3 (23) 26x 2 9y 5 18
3x 2 4y 5 25 3 2 6x 2 8y 5 50
217y 5 68
When y = 24: y 5 24
2x 1 3(24) 5 26
2x 2 12 5 26
x 5 3
The solution is (3, 24).
35. 3x 1 y 5 15 → y 5 23x 1 15
2x 1 2y 5 219
When y 5 23x 1 15: When x 5 7:
2x 1 2(23x 1 15) 5 219 y 5 23(7) 1 15
2x 2 6x 1 30 5 219 y 5 26
27x 5 249
x 5 7
The solution is (7, 26).
36. 4x 2 3y 5 8 3 2 8x 2 6y 5 16
28x 1 6y 5 16 28x 1 6y 5 16
0 Þ 32
There is no solution.
37. 4x 2 y 5 210 → y 5 4x 1 10
6x 1 2y 5 21
When y 5 4x 1 10: When x 5 2 3 } 2 :
6x 1 2(4x 1 10) 5 21 y 5 4 1 2 3 } 2 2 1 10
6x 1 8x 1 20 5 21 y 5 4
14x 5 221
x 5 2 3 } 2
The solution is 1 2 3 } 2 , 4 2 .
38. 7x 1 5y 5 212 3 4 28x 1 20y 5 248
3x 2 4y 5 1 3 5 15x 2 20y 5 5
43x 5 243
When x 5 21: x 5 21
7(21) 1 5y 5 212
5y 5 25
y 5 21
The solution is (21,21).
39. 2x 1 y 5 21 → y 5 22x 2 1
24x 1 6y 5 6
When y 5 22x 2 1: When x 5 2 3 } 4 :
24x 1 6(22x 2 1) 5 6 y 5 22 1 2 3 } 4 2 2 1
24x 2 12x 2 6 5 6 y 5 1 } 2
216x 5 12
x 5 2 3 } 4
The solution is 1 2 3 } 4 ,
1 }
2 2 .
40. B; 3x 1 2y 5 4 3 (22) 26x 2 4y 5 28
6x 2 3y 5 227 6x 2 3y 5 227
27y 5 235
When y 5 5: y 5 5
3x 1 2(5) 5 4
3x 5 26
x 5 22
The solution is (22, 5).
Chapter 3, continued
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128Algebra 2Worked-Out Solution Key
41. Let (x1, y1) 5 (1, 4) and (x2, y2) 5 (5, 0).
m 5 0 2 4
} 5 2 1 5 21
y 2 y1 5 m(x, 2 x1)
y 2 4 5 21(x 2 1)
y 5 2x 1 5
Let (x1, y1) 5 (0, 2) and (x2, y2) 5 (4, 4).
m 5 4 2 2
} 4 2 0 5
1 } 2
y 2 y1 5 m(x 2 x1)
y 2 2 5 1 } 2 (x 2 0)
y 5 1 } 2 x 1 2
System of equations:
y 5 2x 1 5
y 5 1 } 2 x 1 2
When y 5 2x 1 5: When x 5 2:
2x 1 5 5 1 } 2 x 1 2 y 5 2(2) 1 5
3 5 3 } 2 x y 5 3
2 5 x
The diagonals of the quadrilateral intersect at (2, 3).
42. Let (x1, y1) 5 (6, 1) and (x2, y2) 5 (3, 7).
m 5 7 2 1
} 3 2 6 5 22
y 2 y1 5 m(x 2 x1)
y 2 1 5 22(x2 6)
y 2 1 5 22x 1 12
y 5 22x 1 13
Let (x1, y1) 5 (1, 6) and (x2, y2) 5 (7, 4).
m 5 4 2 6
} 7 2 1 5 2 1 } 3
y 2 y1 5 m(x 2 x1)
y 2 6 5 2 1 } 3 (x 2 1)
y 2 6 5 2 1 } 3 x 1
1 } 3
y 5 2 1 } 3 x 1
19 } 3
System of equations:
y 5 22x 1 13
y 5 2 1 } 3 x 1
19 } 3
When y 5 22x 1 13: When x 5 4:
22x 1 13 5 2 1 } 3 x 1
19 } 3 y 5 22(4) 1 13
2 5 } 3 x 5 2
20 } 3 y 5 5
x 5 4
The diagonals of the quadrilateral intersect at (4, 5).
43. Let (x1, y1) 5 (7, 0) and (x2, y2) 5 (1, 3).
m 5 3 2 0
} 1 2 7 5 2 1 } 2
y 2 y1 5 m(x 2 x1)
y 2 0 5 2 1 } 2 (x 2 7)
y 5 2 1 } 2 x 1
7 } 2
Let (x1, y1) 5 (1, 21) and (x2, y2) 5 (5, 5).
m 5 5 1 1
} 5 2 1 5 3 } 2
y 2 y1 5 m(x 2 x1)
y 1 1 5 3 } 2 (x 2 1)
y 5 3 } 2 x 2
5 } 2
System of equations:
y 5 2 1 } 2 x 1
7 } 2
y 5 3 } 2 x 2
5 } 2
When y 5 2 1 } 2 x 1
7 } 2 : When x 5 3:
2 1 } 2 x 1
7 } 2 5
3 } 2 x 2
5 } 2 y 5 2
1 } 2 (3) 1
7 } 2
6 5 2x y 5 2 3 } 2 1
7 } 2
3 5 x y 5 2
The diagonals of the quadrilateral intersect at (3, 2).
44. 0.02x 2 0.05y 5 20.38 3 (2300) 26x 1 15y 5 114
0.03x 1 0.04y 5 1.04 3 200 6x 1 8y 5 208
23y 5 322
When y 5 14: y 5 14
0.02x 2 0.05(14) 5 20.38
0.02x 2 0.7 5 20.38
0.02x 5 0.32
x 5 16
The solution is (16, 14).
Chapter 3, continued
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129Algebra 2
Worked-Out Solution Key
45. 0.05x 2 0.03y 5 0.21 3 200 10x 2 6y 5 42
0.07x 1 0.02y 5 0.16 3 300 21x 1 6y 5 48
31x 5 90
When x 5 90
} 31 : x 5 90
} 31
10 1 90 }
31 2 2 6y 5 42
900
} 31
2 6y 5 42
26y 5 402
} 31
y 5 2 67
} 31
The solution is 1 90 }
31 ,2
67 } 31 2 .
46. 2 }
3 x 1 3y 5 234
x 2 1 } 2 y 5 21 → x 5
1 } 2 y 2 1
When x 5 1 } 2 y 2 1: When y 5 210:
2
} 3 1 1 }
2 y 2 1 2 1 3y 5 234 x 5
1 } 2 (210) 2 1
1 }
3 y 2
2 } 3 1 3y 5 234 x 5 26
10
} 3 y 5 2
100 } 3
y 5 210
The solution is (26,210).
47. 1 }
2 x 1
2 } 3 y 5
5 } 6 3 (230) 215x 2 20y 5 225
5 }
12 x 1
7 } 12 y 5
3 } 4 3 36 15x 1 21y 5 27
When y 5 2: y 5 2
1 }
2 x 1
2 } 3 (2) 5
5 } 6
1 }
2 x 1
4 } 3 5
5 } 6
x 5 21
The solution is (21, 2).
48. x 1 3
} 4 1
y 2 1 } 3 5 1
2x 2 y 5 12 → y 5 2x 2 12
When y 5 2x 2 12: When x 5 5:
12 1 x 1 3 }
4 1
(2x 2 12) 2 1 }} 3 2 5 1 p 12 y 5 2(5) 2 12
3(x 1 3) 1 4(2x 2 13) 5 12 y 5 22
3x 1 9 1 8x 2 52 5 12
11x 5 55
x 5 5
The solution is (5, 22).
49. x 2 1
} 2 1
y 1 2 } 3 5 4
x 2 2y 5 5 → x 5 2y 1 5
When x 5 2y 1 5:
6 1 2y 1 5 2 1 }
2 1
y 1 2 } 3 2 5 4 p 6
3(2y 1 4) 1 2(y 1 2) 5 24
6y 1 12 1 2y 1 4 5 24
8y 5 8
y 5 1
When y 5 1:
x 5 2(1) 1 5
x 5 7
The solution is (7, 1).
50. Sample answer: Two lines intersect at (21, 4). Choose another point in the plane to create one line. Then choose a point not on that line to create a different line.
x
y
1
1
(21, 4)
(0, 1)
(0, 5)
A B
Line A: (x1, y1) 5 (0, 1), (x2, y2) 5 (21, 4)
m 5 4 2 1
} 21 2 0 5 23
y 2 y1 5 m(x 2 x1)
y 2 1 5 23(x 2 0)
y 5 23x 1 1
Line B: (x1, y1) 5 (0, 5), (x2, y2) 5 (21, 4)
m 5 4 2 5
} 21 2 0 5 1
y 2 y1 5 m(x 2 x1)
y 2 5 5 1(x 2 0)
y 5 x 1 5
System of equations:
y 5 23x 1 1
y 5 x 1 5
Use substitution to check:
When y 5 x 1 5: When x 5 21:
x 1 5 5 23x 1 1 y 5 23(21) 1 1
4x 5 24 y 5 4
x 5 21
The lines intersect at (21, 4).
Chapter 3, continued
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130Algebra 2Worked-Out Solution Key
51. 7y 1 18xy 5 30
13y 2 18xy 5 90
20y 5 120
y 5 6
When y 5 6:
7(6) 1 18x(6) 5 30
108x 5 212
x 5 2 1 } 9
The solution is 1 2 1 } 9 , 6 2 .
52. xy 2 x 5 14 xy 2 x 5 14
5 2 xy 5 2x 2xy 2 2x 5 25
23x 5 9
When x 5 23: x 5 23
23y 2 (23) 5 14
23y 1 3 5 14
23y 5 11
y 5 2 11
} 3
The solution is 1 23,2 11
} 3 2 .
53. 2xy 1 y 5 44 2xy 1 y 5 44
32 2 xy 5 3y 3 2 64 2 2xy 5 6y
2xy 1 y 5 44
22xy 2 6y 5 264
25y 5 220
y 5 4
When y 5 4:
2x(4) 1 4 5 44
8x 5 40
x 5 5
The solution is (5, 4).
54. 23x 2 5y 5 9
rx 1 sy 5 t
a. The system will have no solution if the equations represent different lines with the same slope. Sample answer: r 5 23, s 5 25, and t Þ 9.
b. The system will have infi nitely many solutions if the equations represent the same line. Sample answer: r 5 23a, s 5 25a, and t 5 9a, where a is any real number.
c. The system will have a solution of (2,23) for any values or r, s, and t that satisfy the equation 2r 2 3s 5 t. Sample answer: r 5 1, s 5 1, and t 5 21.
Problem Solving
55. Let x 5 acoustic guitars and y 5 electric guitars.
Number of acoustics guitars 1
Number of electric guitars 5
Total number of guitars
x 1 y 5 9
Price peracousticguitar
p
Numberofacousticguitars
1
Price perelectricguitar
p
Numberofelectricguitars
5 Totalrevenue
339 p x 1 479 p y 5 3611
x 1 y 5 9 → y 5 9 2 x
339x 1 479y 5 3611
When y 5 9 2 x:
339x 1 479(9 2 x) 5 3611
339x 1 4311 2 479x 5 3611
2140x 5 2700
x 5 5
When x 5 5:
y 5 9 2 5
y 5 4
The music store sold 5 acoustic guitars and 4 electric guitars.
56. Let x 5 price of adult pass and y 5 price of children’s pass.
Adultprice 5
Childprice
1 2
x 5 y 1 2
Numberof adultpasses
p Adultprice
1
Numberof children’spasses
p Childprice
5 Total
revenue
378 p x 1 214 p y 5 2384
x 5 y 1 2
378x 1 214y 5 2384
When x 5 y 1 2:
378(y 1 2) 1 214y 5 2384
378y 1 756 1 214y 5 2384
592y 5 1628
y 5 2.75
When y 5 2.75:
x 5 2.75 1 2
x 5 4.75
The cost of an adult pass is $4.75.
Chapter 3, continued
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131Algebra 2
Worked-Out Solution Key
Chapter 3, continued
57. The company can fi ll its orders by operating Factory A for 5 weeks and Factory B for 3 weeks.
Let x 5 weeks of operation at Factory A, and y 5 weeks of operation at Factory B.
Factory A’s gasmowersper wk
p
Wks atFactoryA
1
Factory B’s gasmowersper wk
p
Wks atFactoryB
5
200 p x 1 400 p y 5 2200
Factory A’s electricmowersper wk
p
Wks atFactoryA
1
Factory B’s electricmowersper wk
p
Wks atFactoryB
5
Total #of electricmowers
100 p x 1 300 p y 5 1400
200x 1 400y 5 2200 200x 1 400y 5 2200
100x 1 300y 5 1400 3 (22) 2200x 2 600y 5 22800
2200y 5 2600
When y 5 3: y 5 3
200x 1 400(3) 5 2200
200x 5 1000
x 5 5
The solution is (5, 3).
58. B;
Let x 5 price of regular gasoline and y 5 price of premium gasoline.
Gallonsofregular
p
Regularprice
1
Gallonsofpremium
p
Premiumprice
5 Total
price
11 p x 1 16 p y 5 58.55
Premiumgas
5
Regularprice
1 0.2
y 5 x 1 0.2
11x 1 16y 5 58.55
y 5 x 1 0.2
When y 5 x 1 0.2:
11x 1 16(x 1 0.2) 5 58.55
11x 1 16x 1 3.2 5 58.55
27x 5 55.35
x 5 2.05
When x 5 2.05:
y 5 2.05 1 0.2
y 5 2.25
A gallon of premium gasoline costs $2.25.
59. Let x 5 doubles games in progress and y 5 singles games in progress.
Number of doublesgames
1
Number of singlesgames
5
Total number of games
x 1 y 5 26
Playerspergame
p
Doublegames 1
Playerspergame
p
Singlesgame
5 Totalnumber ofplayers
4 p x 1 2 p y 5 76
x 1 y 5 26 → y 5 26 2 x
4x 1 2y 5 76
When y 5 26 2 x: When x 5 12:
4x 1 2(26 2 x) 5 76 y 5 26 2 x
4x 1 52 2 2x 5 76 y 5 14
2x 5 24
x 5 12
There were 12 doubles games and 14 singles games in progress.
60. a. Let t 5 hours since 10:00 A.M. and d 5 distance traveled (miles).
Martha’sdistance
5 Martha’srate
p Martha’stime
d 5 4 p t
An equation for the distance Martha travels is d 5 4t.
b. Carol’sdistance
5 Carol’srate
p Carol’stime
d 5 6 p (t 2 2)
An equation for the distance Carol travels is d 5 6(t 2 2).
c. d 5 4t
d 5 6t 2 12
When d 5 4t:
4t 5 6t 2 12
t 5 6
Carol will catch up to Martha 6 hours after 10:00 A.M.,
or at 4:00 P.M.
d. Changing starting time:
Let h 5 the number of hours Carol starts after Martha.
d 5 4t
d 5 6(t 2 h) → d 5 6t 2 6h
When t 5 5: When d 5 20 and t 5 5:
d 5 4(5) 5 20 20 5 6(5) 2 6h
6h 5 10
h 5 5 } 3
Total #of gasmowers
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132Algebra 2Worked-Out Solution Key
Carol should start 5 }
3 hours, or 1 hour 40 minutes, after
Martha. It is reasonable that Carol could change her starting time.
Changing speed:
Let r 5 Carol’s speed (miles per hour).
d 5 4t
d 5 r(t 2 2)
When t 5 5: When d 5 20 and t 5 5:
d 5 4(5) 5 20 20 5 r (5 2 2)
r 5 20
} 3 5 6 2
} 3
Carol should run at a speed of 6 2
} 3 miles per hour.
This is a reasonable speed.
61. Let x 5 pounds of peanuts in one pound of mix and y 5 pounds of cashews in one pound mix.
Peanutpriceperpound
p
Amount of peanuts in 1lb of mix
1
Cashewpriceperpound
p
Amount ofcashewsin 1lb ofmix
5
Priceof 1lbof mix
2.80 p x 1 5.30 p y 5 3.30
Amount ofpeanuts in1lb of mix
1
Amount ofcashews in1lb of mix
5
Total weight of 1lb of mix
x 1 y 5 1
2.8x 1 5.3y 5 3.3
x 1 y 5 1 → y 5 1 2 x
When y 5 1 2 x: When x 5 0.8:
2.8x 1 5.3(1 2 x) 5 3.3 y 5 1 2 x
2.8x 1 5.3 2 5.3x 5 3.3 y 5 0.2
22.5x 5 22
x 5 0.8
One pound of mix should contain 0.8 pounds of peanuts and 0.2 pounds of cashews.
(0.8)100 5 80lb peanuts
(0.2)100 5 20lb cashews
The wholesaler should use 80 pounds of peanuts and20 pounds of cashews for 100 pounds of mix.
62. Let x 5 speed of the plane in calm air and y 5 speed of the wind.
Plane’sspeed incalm air
1
Speed oftailwind
5
Actual planespeed
x 1 y 5 1000
} 5 5 200
Plane’sspeed incalm air
2 Speed of
headwind
5
Actual planespeed
x 2 y 5 500
} 5 5 100
x 1 y 5 200 → y 5 200 2 x
x 2 y 5 100
When y 5 200 2 x: When x 5 150:
x 2 (200 2 x) 5 100 y 5 200 2 x
x 2 200 1 x 5 100 y 5 50
2x 5 300
x 5 150
The speed of the plane in calm air was 150 mi/h, and the speed of the wind was 50 mi/h.
63. Let x 5 hours the electrician worked and y 5 hours the apprentice worked.
Apprentice’shours
1 4 5
Electrician’shours
y 1 4 5 x
Electri-cian’spay rate
p
Electri-cian’shours
1
Appren-tice’spay rate
p Appren-
tice’shours
5
Totalearnings
50 p x 1 20 p y 5 550
y 1 4 5 x
50x 1 20y 5 550
When x 5 y 1 4:
50(y 1 4) 1 20y 5 550
50y 1 200 1 20y 5 550
70y 5 350
y 5 5
The apprentice earned 20(5) 5 $100.
When y 5 5:
x 5 y 1 4
x 5 9
The electrician earned 50(9) 5 $450.
Mixed Review
64. 25x 1 4 5 29
25x 5 25
x 5 25
65. 6(2a 2 3) 5 230
12a 2 18 5 230
12a 5 212
a 5 21
66. 1.2m 5 2.3m 2 2.2
21.1m 5 22.2
m 5 2
67. x 1 3 5 4
x 1 3 5 4 or x 1 3 5 24
x 5 1 or x 5 27
68. 2x 1 11 5 3
2x 1 11 5 3 or 2x 1 11 5 23
2x 5 28 or 2x 5 214
x 5 24 or x 5 27
Chapter 3, continued
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133Algebra 2
Worked-Out Solution Key
69. 2x 1 7 5 13
2x 1 7 5 13 or 2x 1 7 5 213
2x 5 6 or 2x 5 220
x 5 26 or x 5 20
70. Line 1: m1 5 10 2 5
} 2 2 1 5 5
Line 2: m2 5 27 2 (28)
} 3 2 8 5 2 1 } 5
Because m2 5 2 1 } m1 , the lines are perpendicular.
71. Line 1: m1 5 5 1 2
} 4 2 9 5 2 7 } 5
Line 2: m2 5 26 1 1
} 6 1 2 5 2 5 } 8
Because m1 Þ m2 and m2 Þ 2 1 } m1 , the lines are neither
parallel nor perpendicular.
72. (x1, y1) 5 (2, 4), (x2, y2) 5 (5, 1)
m 5 y2 2 y1
} x2 2 x1 5
1 2 4 } 5 2 2 5 21
y 2 y1 5 m(x 2 x1)
y 2 4 5 21(x 2 2)
y 5 2x 1 6
73. (x1, y1) 5 (3, 1), (x2, y2) 5 (21, 27)
m 5 y2 2 y1
} x2 2 x1 5
27 2 1 }
21 2 3 5 2
y 2 y1 5 m(x 2 x1)
y 2 1 5 2(x 2 3)
y 5 2x 2 5
74. (x1, y1) 5 (2, 4), (x2, y2) 5 (22, 22)
m 5 y2 2 y1
} x2 2 x1 5
22 2 4 }
22 2 2 5 3 } 2
y 2 y1 5 m(x 2 x1)
y 2 4 5 3 } 2 (x 2 2)
y 5 3 } 2 x 1 1
75. x < 23 76. y ≥ 2
1
x
y
21
1
x
y
21
77. 2x 1 y > 1 78. y ≤ 2x 1 4
1
x
y
21
1
x
y
21
79. 4x 2 y ≥ 5 80. y < 23x 1 2
1
x
y
21
1
x
y
21
Quiz 3.1–3.2 (p. 167)
1.
1
x
y
21
From the graph, the lines appear to intersect at (2, 5).
The solution appears to be (2, 5).
3x 1 y 5 11 x 2 2y 5 28
3(2) 1 5 0 11 2 2 2(5) 0 28
6 1 5 0 11 2 2 10 0 28
11 5 11 ✓ 28 5 28 ✓
2.
1
x
y
21
From the graph, the lines appear to intersect at (23, 1).
The solution appears to be (23, 1).
2x 1 y 5 25 2x 1 3y 5 6
2(23) 1 1 0 25 2(23) 1 3(1) 0 6
26 1 1 0 25 3 1 3 0 6
25 5 25 ✓ 6 5 6 ✓
3. 2
x
y
21
From the graph, the lines appear to intersect at (26, 22).
The solution appears to be (26, 22).
x 2 2y 5 22 3x 1 y 5 220
26 2 2(22) 0 22 3(26) 1 (22) 0 220
26 1 4 0 22 218 2 2 0 220
22 5 22 ✓ 220 5 220 ✓
4.
x
y
2
1
4x 1 8y 5 8
x 1 2y 5 6
The graphs of the equations are parallel lines. There is no solution. The system is inconsistent.
Chapter 3, continued
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134Algebra 2Worked-Out Solution Key
5.
x
y
2
2
25x 1 3y 5 25
y 5 x 1 153
The graphs of the equations are parallel lines. There is no solution. The system is inconsistent.
6.
x
y1
21
x 2 2y 5 2
2x 2 y 5 25
(24, 23)
The solution is (24, 23). The system is consistent and independent.
7. 3x 2 y 5 24
x 1 3y 5 228 → x 5 228 2 3y
When x 5 228 2 3y: When y 5 28:
3(228 2 3y) 2 y 5 24 x 5 228 2 3(28)
284 2 9y 2 y 5 24 x 5 24
210y 5 80
y 5 28
The solution is (24, 28).
8. x 1 5y 5 1 → x 5 1 2 5y
23x 1 4y 5 16
When x 5 1 2 5y: When y 5 1:
23(1 2 5y) 1 4y 5 16 x 5 1 2 5(1)
23 1 15y 1 4y 5 16 x 5 24
19y 5 19
y 5 1
The solution is (24, 1).
9. 6x 1 y 5 26 → y 5 26x 2 6
4x 1 3y 5 17
When y 5 26x 2 6: When x 5 2 5 } 2 :
4x 1 3(26x 2 6) 5 17 y 5 26 1 2 5 } 2 2 2 6
4x 2 18x 2 18 5 17 y 5 9
214x 5 35
x 5 2 5 } 2
The solution is 1 2 5 } 2 , 9 2 .
10. 2x 2 3y 5 21
2x 1 3y 5 219
4x 5 220
x 5 25
When x 5 25:
2(25) 2 3y 5 21
23y 5 9
y 5 23
The solution is (25, 23).
11. 3x 2 2y 5 10 3 2 6x 2 4y 5 20
26x 1 4y 5 220 26x 1 4y 5 220
0 5 0
There are infi nitely many solutions.
12. 2x 1 3y 5 17 3 (25) 210x 2 15y 5 285
5x 1 8y 5 20 3 2 10x 1 16y 5 40
y 5 245
When y 5 245:
2x 1 3(245) 5 17
2x 5 152
x 5 76
The solution is (76, 245).
13. Let x 5 cost per foot of cable and y 5 cost per connector.
6 p Cost per
foot of cable
1 2 p
Cost perconnector
5
Cost of 6 footcable with
connectors
6 p x 1 2 p y 5 15.50
3 p Cost per
foot of cable
1 2 p
Cost perconnector
5
Cost of 3 footcable with
connectors
3 p x 1 2 p y 5 10.25
6x 1 2y 5 15.50 6x 1 2y 5 15.50
3x 1 2y 5 10.25 3 (21) 23x 2 2y 5 210.25
3x 5 5.25
When x 5 1.75: x 5 1.75
6(1.75) 1 2y 5 15.50
2y 5 5
y 5 2.5
4 - foot cable 5 4x 1 2y
5 4(1.75) 1 2(2.50)
5 12.00
A 4 - foot cable should cost $12.00.
Chapter 3, continued
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135Algebra 2
Worked-Out Solution Key
Lesson 3.3
3.3 Guided Practice (pp. 169–170)
1. y ≤ 3x 2 2
1
x21
y
y > 2x 1 4
2. 2x 2 1 } 2 y ≥ 4
22
21
y
x
4x 2 y ≤ 5
3. x 1 y > 23 4. y ≤ 4 26x 1 y < 1 y ≥ x 2 5
1
x21
y
1
x
y
1
5. y > 22 6. y ≥ 2x 1 1 y ≤ 2x 1 2 y < x 1 1
1
x1
y
x
y
2
1
7. Range of discounts: 20%–50%
Regular prices: $40– $100
a. Let x 5 regular footwear price and y 5 sale footwear price.
x ≥ 40
00 20 40 60 10080
20
40
80
60
x
y
Regular prices (dollars)
Sal
e p
rice
s (d
olla
rs)
y 5 0.8x
x 5 40
y 5 0.5x
x 5 100 x ≤ 100
y ≥ 0.5x
y ≤ 0.8x
b. When x 5 60:
0.5(60) ≤ y ≤ 0.8(60)
30 ≤ y ≤ 48
Footwear regularly priced at $60 sells for between $30 and $48, inclusive, during the sale.
3.3 Exercises (pp. 171–173)
Skill Practice
1. The ordered pair must be in the area of the graph that is common to all of the inequalities in the system.
2. Graph each inequality in the system. Then identify the region common to all the graphs of the inequalities.
3. D
4. x > 21 5. x ≤ 2 x < 3 y ≤ 5
1
x22
y
1
x
y
21
6. y ≥ 5 7. 2x 1 y < 23
y ≤ 1 2x 1 y > 4
x
y
211
x
y
1
1
8. y < 10 9. 4x 2 4y ≥ 216
y > x 2x 1 2y ≥ 24
1
x21
y
1
x
y
21
10. 2x ≥ y 11. y > x 2 4
2x 1 y ≥ 25 3y < 22x 1 9
1
x21
y
1 x
21
y
Chapter 3, continued
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136Algebra 2Worked-Out Solution Key
12. x 1 y ≥ 23 13. 2y < 25x 2 10
26x 1 4y < 14 5x 1 2y > 22
1
21
y
x
x
y
1
1
14. 3x 2 y > 12 15. x 2 4y ≤ 210
2x 1 8y > 24 y ≤ 3x 2 1
1
3
y
x
1
y
x21
16. The graph of y ≤ 2x 2 2 should be shaded to the right of the line y 5 2x 2 2.
21
1
y
x
17. x < 6 18. x ≥ 28
y > 21 y ≤ 21
y < x y < 22x 2 4
1
x21
y
1
y
x23
19. 3x 1 2y > 26 20. x 1 y < 5
25x 1 2y > 22 2x 2 y > 0
y < 5 2x 1 5y > 220
1
x
y
23
1
21
y
x
21. x ≥ 2 22. y ≥ x 23x 1 y < 21 x 1 3y < 5
4x 1 3y < 12 2x 1 y ≥ 23
1
21 x
y
1
22 x
y
23. y ≥ 0 24. x 1 y < 5
x > 3 x 1 y > 25
x 1 y ≥ 22 x 2 y < 4
y < 4x x 2 y > 22
2
1 x
y
1
21 x
y
25. x ≤ 10
24 x
y6
x ≥ 22
3x 1 2y < 6
6x 1 4y > 212
26. B; y
x
2
2
Q III Q IV
Q IQ II
y ≤ 2x 2 3 1 2
4x 2 5y ≤ 20
27. Sample answer: x ≥ 0 y ≤ 21
28. y < x 29. y ≤ x 2 2 y > 2x y ≥ x 2 2
1
21 x
y
1
x
y
22
Chapter 3, continued
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137Algebra 2
Worked-Out Solution Key
30. y ≤ 2x 2 3 1 2
1
21 x
y
y > x 2 3 2 1
31. y ≤ 3 y ≥ 22
x ≥ 23
x ≤ 432. Line through points (22, 2) and (2, 4):
m 5 4 2 2
} 2 2 (22)
5 1 } 2
y 2 y1 5 m(x 2 x1)
y 2 4 5 1 } 2 (x 2 2)
y 5 1 } 2 x 1 3 (boundary)
Line through points (22, 22) and (2, 24):
m 5 24 2 (22)
} 2 2 (22)
5 2 1 } 2
y 2 y1 5 m(x 2 x1)
y 1 4 5 2 1 } 2 (x 2 2)
y 5 2 1 } 2 x 2 3 (boundary)
System:
y ≤ 1 } 2 x 1 3
y ≥ 2 1 } 2 x 2 3
x ≥ 22
x ≤ 2 33. Line through points (22, 23) and (1, 3):
m 5 3 2 (23)
} 1 2 (22)
5 2
y 2 y1 5 m(x 2 x1)
y 2 3 5 2(x 2 1)
y 5 2x 1 1
Line through points (22, 23) and (2, 22):
m 5 22 2 (23)
} 2 2 (22)
5 1 } 4
y 2 y1 5 m(x 2 x1)
y 1 2 5 1 } 4 (x 2 2)
y 5 1 } 4 x 2
5 } 2
Line through points (1, 3) and (2, 22):
m 5 22 2 3
} 2 2 1 5 25
y 2 y1 5 m(x 2 x1)
y 2 3 5 25(x 2 1)
y 5 25x 1 8
System:
y ≤ 2x 1 1
y ≥ 1 } 4 x 2
5 } 2
y ≤ 25x 1 8
34. Let x 5 the hours dog walking and y 5 the hours car washing.
x 1 y ≤ 20
x ≥ 0 y ≥ 0 7.5x 1 6y ≥ 92
35. Discount: 30% –70%
Regular prices: $20 –$50
Let x 5 the regular price and y 5 the sale price.
x ≥ 20
100 20 30 40 50
10
0
20
30
40
Regular price (dollars)
Sal
e p
rice
(d
olla
rs)
x
y
y 5 0.3x
y 5 0.7x
x 5 50x 5 20 x ≤ 50
y ≥ 0.3x
y ≤ 0.7x
When x 5 20:
0.3(20) ≤ y ≤ 0.7(20)
6 ≤ y ≤ 14
Games regularly priced at $20 sell between $6 and $14, inclusive, during the sale.
36. x > 8.0
00 8.0 8.1 8.2 8.3
75
77
81
79
x
y
pH level
Tem
per
atu
re (
8F) x < 8.3
y > 76
y < 80
In degrees Celsius:
C 5 5 } 9 (F 2 32) C 5
5 } 9 (F 2 32)
C 5 5 } 9 (80 2 32) C 5
5 } 9 (76 2 32)
C 5 5 } 9 (48) C 5
5 } 9 (44)
C 5 26. } 6 C 5 24. } 4
The graphs would look similar; the y axis would be incremented differently to show 24.44 < y < 26.67.
37. a. x ≥ 2 b.
10
23456
x
y
10 2 3 4 5 6 7 8
Number of juniors
Nu
mb
er o
f se
nio
rs
y ≥ 2 x 1 y ≤ 8 x 1 y ≥ 5
c. Sample answer:
3 juniors, 2 seniors
6 juniors, 2 seniors
Chapter 3, continued
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138Algebra 2Worked-Out Solution Key
38. Let x 5 the horizontal distance (in inches) from the left side of home plate and y 5 the distance above the ground (in inches).
x ≥ 0
00 4 8 12 16
10
20
40
30
x
y
Horizontal distance (in.)
Ver
tica
l dis
tan
ce (
in.) x ≤ 17
y ≥ 20
y ≤ 42
39. a. x ≤ 65 b.
Hea
rt r
ate
(bea
ts p
er m
inu
te)
Age (years)
0100 20 30 40 50 70 80 9060
255075
100125
200225
175150
x
y
x ≥ 20
y ≤ 0.75(220 2 x)
y ≥ 0.5(220 2 x)
c. No; the target zone for a 40-year-old is between 0.5(220 2 40) 5 90 and 0.75(220 2 40) 5 135. A heart rate of 158 is outside of this range.
40.
5000 1000 1500 2000
500
0
1000
1500
2000
Your guess
You
fri
end
’s g
ues
s
x
y
x ≥ 500
y ≥ 500
y 2 1000 > x 2 1000 When y < 1000:
y 2 1000 5 1000 2 y > x 2 1000 2y > x 2 1000 2 1000
y < 1000 2 x 2 1000 When y ≥ 1000:
y 2 1000 5 y 2 1000 > x 2 1000 y > 1000 1 x 2 1000 System of inequalities:
x ≥ 500
y ≥ 500
y 2 1000 > x 2 1000
Mixed Review
41. 6x 2 8y 5 6(4) 2 8(21)
5 24 1 8
5 32
42. 12x 1 3y 5 12(4) 1 3(25)
5 48 2 15
5 33
43. x2 2 2xy 1 3y 5 (22)2 2 2(22)(3) 1 3(3)
5 4 1 12 1 9
5 25
44. 4x2y2 2 xy 5 4(5)2(26)2 2 (5)(26)
5 4(25)(36) 1 30
5 3600 1 30
5 3630
45. x 2 8 ≤ 25
x ≤ 3
21 0 1 2 3 4
46. 5x 2 11 > 2x 1 7
6x > 18
x > 3
21 0 1 2 3 4
47. 9x 1 2 ≥ 23x 2 13
12x ≥ 215
x ≥ 2 5 } 4
0 2 424 22
542
48. 25x 1 y 5 211
4x 2 y 5 7 → y 5 4x 2 7
When y 5 4x 2 7: When x 5 4:
25x 1 4x 2 7 5 211 y 5 4(4) 2 7
2x 5 24 y 5 9
x 5 4
The solution is (4, 9).
49. 9x 1 4y 5 27 9x 1 4y 5 27
3x 2 5y 5 234 3 (23) 29x 1 15y 5 102
19y 5 95
When y 5 5: y 5 5
9x 1 4(5) 5 27
9x 5 227
x 5 23
The solution is (23, 5).
Chapter 3, continued
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139Algebra 2
Worked-Out Solution Key
50. 4x 1 9y 5 210 3 (22) 28x 2 18y 5 20
8x 1 18y 5 20 8x 1 18y 5 20
0 Þ 40
There is no solution.
51. x 2 5y 5 18 → x 5 5y 1 18
2x 1 3y 5 10
When x 5 5y 1 18: When y 5 22:
2(5y 1 18) 1 3y 5 10 x 5 5(22) 1 18
10y 1 36 1 3y 5 10 x 5 8
13y 5 226
y 5 22
The solution is (8, 22).
52. 16x 2 12y 5 28 16x 2 12y 5 28
8x 2 6y 5 24 3 (22) 216x 1 12y 5 8
0 5 0
There are infi nitely many solutions.
53. 16x 1 5y 5 24 16x 1 5y 5 24
8x 2 2y 5 7 3 (22) 216x 1 4y 5 214
9y 5 218
When y 5 22: y 5 22
8x 2 2(22) 5 7
8x 5 3
x 5 3 } 8
The solution is 1 3 } 8 , 22 2 .
3.3 Extension (p. 176)
1. At (0, 7): C 5 0 1 2(7) 5 14
At (1, 0): C 5 1 1 2(0) 5 1
At (8, 0): C 5 8 1 2(0) 5 8
Minimum value: 1; maximum value: 14
2. At (28, 4): C 5 4(28) 2 2(4) 5 240
At (28, 28): C 5 4(28) 2 2(28) 5 216
At (2, 28): C 5 4(2) 2 2(28) 5 24
At (6, 22): C 5 4(6) 2 2(22) 5 28
Minimum value: 240; maximum value: 28
3. At (20, 60): C 5 3(20) 1 5(60) 5 360
At (40, 10): C 5 3(40) 1 5(10) 5 170
At (80, 0): C 5 3(80) 1 5(0) 5 240
At (100, 40): C 5 3(100) 1 5(40) 5 500
At (60, 80): C 5 3(60) 1 5(80) 5 580
Minimum value: 170; maximum value: 580
4.
x
y
1
1
(0, 5)
(5, 0)(0, 0)
At (0, 0): C 5 3(0) 1 4(0) 5 0
At (5, 0): C 5 3(5) 1 4(0) 5 15
At (0, 5): C 5 3(0) 1 4(5) 5 20
Minimum value: 0; maximum value: 20
5.
x
y
2
2
(5, 9)
(25, 3) (5, 3)
At (25, 3): C 5 2(25) 1 5(3) 5 5
At (5, 3): C 5 2(5) 1 5(3) 5 25
At (5, 9): C 5 2(5) 1 5(9) 5 55
Minimum value: 5; maximum value: 55
6.
x
y
1
1
(0, 4)
(2, 22)(0, 0)
At (0, 4): C 5 3(0) 1 4 5 4
At (0, 0): C 5 3(0) 1 0 5 0
At (2, 22): C 5 3(2) 1 (22) 5 4
Minimun value: 0; no maximum value
7. x 5 the number of mini piñatas
y 5 the number of regular-sized piñatas
2x 1 3y ≤ 30
x
y
3
3
(6, 6)
(15, 0)(12, 0)
x 1 y ≥ 12
x ≥ 0
y ≥ 0
P 5 12x 1 24y
At (12, 0): P 5 12(12) 1 24(0) 5 144
At (15, 0): P 5 12(15) 1 24(0) 5 180
At (6, 6): P 5 12(6) 1 24(6) 5 216
The owner should make 6 mini piñatas and 6 regular-sized piñatas to maximize profi t.
Chapter 3, continued
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140Algebra 2Worked-Out Solution Key
8. x 5 the number of inkjet printers
y 5 the number of laser printers
x 1 y ≤ 60
x
y
10
10 (60, 0)
(30, 30)(0, 40)
(0, 0)
x 1 3y ≤ 120
x ≥ 0
y ≥ 0
P 5 40x 1 60y
At (0, 0): P 5 40(0) 1 60(0) 5 0
At (60, 0): P 5 40(60) 1 60(0) 5 2400
At (30, 30): P 5 40(30) 1 60(30) 5 3000
At (0, 40): P 5 40(0) 1 60(40) 5 2400
The company should make 30 inkjet printers and30 laser printers.
9. t 5 the number of jars of tomato sauce
s 5 the number of jars of salsa
10t 1 5s ≤ 180
s
t
6
3
(0, 15)
(0, 0)
6013
18013( (,
t 1 0.25s ≤ 15
t ≥ 3s
s ≥ 0
t ≥ 0
P 5 2t 1 1.5s
At (0, 0): P 5 2(0) 1 1.5(0) 5 0
At (0, 15): P 5 2(15) 1 1.5(0) 5 30
At 1 60 }
13 ,
180 }
13 2 : P 5 2 1 180
} 13
2 1 1.5 1 60 }
13 2 5
450 } 13 ø 34.62
The maximum value of P occurs when t 5 180
} 13 ø 13.85
and s 5 60
} 13 ø 4.62.
You should make 13 jars of tomato sauce and 4 jarsof salsa.
10. a. Sample answer: b. Sample answer:
1
x
y
21
(5, 7)
(7, 2)
1
x
y
21
(2, 6)
(6, 1)(0, 0)
Lesson 3.4
Investigating Algebra Activity 3.4 (p. 177)
1. 4x 1 3y 1 2z 5 12 2. 2x 1 2y 1 3z 5 6
(3, 0, 0)
(0, 4, 0)
(0, 0, 6)
y
x
z
(3, 0, 0)
(0, 3, 0)
(0, 0, 2)
y
x
z
3. x 1 5y 1 3z 5 15
(15, 0, 0)
(0, 3, 0)
(0, 0, 5)
y
x
z
4. 5x 2 y 2 2z 5 10
(0, 210, 0)
(0, 0, 25)
(2, 0, 0)
y
x
z
5. 27x 1 7y 1 2z 5 14
(22, 0, 0)
(0, 2, 0)
(0, 0, 7)
y
x
z
6. 2x 1 9y 2 3z 5 218
(29, 0, 0)
(0, 22, 0)
(0, 0, 6)
y
x
z
Chapter 3, continued
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141Algebra 2
Worked-Out Solution Key
7. Sample answer:
The planes of three linear equations in three variables can intersect in three different ways. They could intersect in exactly one point, intersect in a line, or never intersectat all.
3.4 Guided Practice (pp. 180–181)
1. 3x 1 y 2 2z 5 10 3 (24) 212x 2 4y 1 8z 5 240
x 1 4y 1 3z 5 7 x 1 4y 1 3z 5 7
211x 1 11z 5 233
2x 1 z 5 23
6x 2 2y 1 z 5 22 3 2 12x 2 4y 1 2z 5 24
x 1 4y 1 3z 5 7 x 1 4y 1 3z 5 7
13x 1 5z 5 3
2x 1 z 5 23 3 (25) 5x 2 5z 5 15
13x 1 5z 5 3 13x 1 5z 5 3
18x 5 18
x 5 1
21 1 z 5 23 → z 5 22
3(1) 1 y 2 2(22) 5 10 → y 5 3
The solution is (1, 3, 22).
2. x 1 y 2 z 5 2 3 (22) 22x 2 2y 1 2z 5 24
2x 1 2y 2 2z 5 6 2x 1 2y 2 2z 5 6
0 5 2
Because this is a false equation, there is no solution.
3. x 1 y 1 z 5 3 x 1 y 2 z 5 3
x 1 y 2 z 5 3 2x 1 2y 1 z 5 6
2x 1 2y 5 6 3x 1 3y 5 9
2x 1 2y 5 6 3 3 6x 1 6y 5 18
3x 1 3y 5 9 3 (22) 26x 2 6y 5 218
0 5 0
Because you obtain the identity 0 5 0, the system has infi nitely many solutions.
4. x 1 y 1 z 5 60
x 1 (x 1 z) 1 z 5 60
2x 1 2z 5 60
1000x 1 200y 1 500z 5 25,000
1000x 1 200(x 1 z) 1 500z 5 25,000
1200x 1 700z 5 25,000
2x 1 2z 5 60 3 (2600) 21200x 2 1200z 5 236,000
1200x 1 700z 5 25,000 1200x 1 700z 5 25,000
2500z 5 211,000
z 5 22
2x 1 2(22) 5 60 → x 5 8
8 1 y 1 22 5 60 → y 5 30
The department should run 8 TV ads, 30 radio ads, and22 newspaper ads each month.
3.4 Exercises (pp. 182–185)
Skill Practice
1. A linear equation in three variables has the formax 1 by 1 cz 5 d. The graph of this equation will bea plane.
2. Sample answer: If one variable is expressed in terms of the other two, substitute that expression into the remaining two equations, and solve the new linear system in two variables.
3. 2x 2 y 1 z 5 25 x 2 3y 1 z 5 25
2(1) 2 4 1 (23) 0 25 1 2 3(4) 1 (23) 0 25
25 5 25 ✓ 214 Þ 25
5x 1 2y 2 2z 5 19
5(1) 1 2(4) 2 2(23) 0 19
19 5 19 ✓
(1, 4, 23) is not a solution.
4. 4x 2 y 1 3z 5 13
4(21) 2 (22) 1 3(5) 0 13
13 5 13 ✓
x 1 y 1 z 5 2
21 1 (22) 1 5 0 2
2 5 2 ✓
x 1 3y 2 2z 5 217
21 1 3(22) 2 2(5) 0 217
217 5 217 ✓
(21, 22, 5) is a solution.
5. x 1 4y 2 2z 5 12
6 1 4(0) 2 2(23) 0 12
12 5 12 ✓
3x 2 y 1 4z 5 6
3(6) 2 0 1 4(23) 0 6
6 5 6 ✓
2x 1 3y 1 z 5 29
2(6) 1 3(0) 1 (23) 0 29
29 5 29 ✓
(6, 0, 23) is a solution.
Chapter 3, continued
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142Algebra 2Worked-Out Solution Key
6. 3x 1 4y 2 2z 5 211
3(25) 1 4(1) 2 2(0) 0 211
211 5 211 ✓
2x 1 y 2 z 5 11
2(25) 1 1 2 (0) 0 11
29 Þ 11
x 1 4y 1 3z 5 21
25 1 4(1) 1 3(0) 0 21
21 5 21 ✓
(25, 1, 0) is not a solution.
7. 3x 2 y 1 5z 5 34
3(2) 2 (8) 1 5(4) 0 34
18 Þ 34
x 1 3y 2 6z 5 2
2 1 3(8) 2 6(4) 0 2
2 5 2 ✓
23x 1 y 2 2z 5 26
23(2) 1 8 2 2(4) 0 26
26 5 26 ✓
(2, 8, 4) is not a solution.
8. 2x 1 4y 2 z 5 223
2(0) 1 4(24) 2 7 0 223
223 5 223 ✓
x 2 5y 2 3z 5 21
0 2 5(24) 2 3(7) 0 21
21 5 21 ✓
2x 1 y 1 4z 5 24
2(0) 1 (24) 1 4(7) 0 24
24 5 24 ✓
(0, 24, 7) is a solution.
9. 3x 1 y 1 z 5 14
5x 2 y 1 5z 5 30
8x 1 6z 5 44
2x 1 2y 2 3z 5 29 2x 1 2y 2 3z 5 29
5x 2 y 1 5z 5 30 3 2 10x 2 2y 1 10z 5 60
9x 1 7z 5 51
8x 1 6z 5 44 3 (27) 256x 2 42z 5 2308
9x 1 7z 5 51 3 6 54x 1 42z 5 306
22x 5 22
x 5 1
9(1) 1 7z 5 51 → z 5 6
3(1) 1 y 1 6 5 14 → y 5 5
The solution is (1, 5, 6).
10. 2x 2 y 1 2z 5 27 2x 2 y 1 2z 5 27
x 1 4y 2 6z 5 21 3 (22) 22x 2 8y 1 12z 5 2
29y 1 14z 5 25
2x 1 2y 2 4z 5 5
x 1 4y 2 6z 5 21
6y 2 10z 5 4
29y 1 14z 5 25 3 5 245y 1 70z 5 225
6y 2 10z 5 4 3 7 42y 2 70z 5 28
23y 5 3
y 5 21
29(21) 1 14z 5 25 → z 5 21
x 1 4(21) 2 6(21) 5 21 → x 5 23
The solution is (23, 21, 21).
11. 3x 2 y 1 2z 5 4 3 (22) 26x 1 2y 2 4z 5 28
6x 2 2y 1 4z 5 28 6x 2 2y 1 4z 5 28
0 5 216
No solution.
12. 4x 2 y 1 2z 5 218 3 2 8x 2 2y 1 4z 5 236
3x 1 3y 2 4z 5 44 3x 1 3y 2 4z 5 44
11x 1 y 5 8
2x 1 2y 1 z 5 11 3 4 24x 1 8y 1 4z 5 44
3x 1 3y 2 4z 5 44 3x 1 3y 2 4z 5 44
2x 1 11y 5 88
11x 1 y 5 8 11x 1 y 5 8
2x 1 11y 5 88 3 11 211x 1 121y 5 968
122y 5 976
y 5 8
2x 1 11(8) 5 88 → x 5 0
2(0) 1 2(8) 1 z 5 11 → z 5 25
The solution is (0, 8, 25).
13. 5x 1 y 2 z 5 6
x 1 y 1 z 5 2
6x 1 2y 5 8
6x 1 2y 5 8 6x 1 2y 5 8
3x 1 y 5 4 3 (22) 26x 2 2y 5 28
0 5 0
Infi nitely many solutions.
Chapter 3, continued
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143Algebra 2
Worked-Out Solution Key
14. 2x 1 y 2 z 5 9
5x 1 7y 1 z 5 4
7x 1 8y 5 13
2x 1 6y 1 2z 5 217 2x 1 6y 1 2z 5 217
5x 1 7y 1 z 5 4 3 (22) 210x 2 14y 2 2z 5 28
211x 2 8y 5 225
7x 1 8y 5 13
211x 2 8y 5 225
24x 5 212
x 5 3
7(3) 1 8y 5 13 → y 5 21
5(3) 1 7(21) 1 z 5 4 → z 5 24
The solution is (3, 21, 24).
15. x 1 y 2 z 5 4
3x 1 2y 1 4z 5 17
2x 1 5y 1 z 5 8 → z 5 x 2 5y 1 8
x 1 y 2 (x 2 5y 1 8) 5 4 → y 5 2
z 5 x 2 5(2) 1 8 → z 5 x 2 2
3x 1 2(2) 1 4(x 2 2) 5 17 → x 5 3
z 5 3 2 2 → z 5 1
The solution is (3, 2, 1).
16. 2x 2 y 2 z 5 15 → z 5 2x 2 y 2 15
4x 1 5y 1 2z 5 10
2x 2 4y 1 3z 5 220
4x 1 5y 1 2(2x 2 y 2 15) 5 10
8x 1 3y 5 40
2x 2 4y 1 3(2x 2 y 2 15) 5 220
5x 2 7y 5 25
8x 1 3y 5 40 3 7 56x 1 21y 5 280
5x 2 7y 5 25 3 3 15x 2 21y 5 75
71x 5 355
x 5 5
8(5) 1 3y 5 40 → y 5 0
2(5) 2 0 2 z 5 15 → z 5 25
The solution is (5, 0, 25).
17. 4x 1 y 1 5z 5 240
23x 1 2y 1 4z 5 10
x 2 y 2 2z 5 22 → x 5 y 1 2z 2 2
4(y 1 2z 2 2) 1 y 1 5z 5 240
5y 1 13z 5 232
23(y 1 2z 2 2) 1 2y 1 4z 5 10
2y 2 2z 5 4
5y 1 13z 5 232 5y 1 13z 5 232
2y 2 2z 5 4 3 5 25y 2 10z 5 20
3z 5 212
z 5 24
2y 2 2(24) 5 4 → y 5 4
x 2 4 2 2(24) 5 22 → x 5 26
The solution is (26, 4, 24).
18. x 1 3y 2 z 5 12
2x 1 4y 2 2z 5 6
2x 2 2y 1 z 5 26 → z 5 x 1 2y 2 6
2x 1 4y 2 2(x 1 2y 2 6) 5 6
12 5 6
No solution.
19. 2x 2 y 1 z 5 22 → z 5 22x 1 y 2 2
6x 1 3y 2 4z 5 8
23x 1 2y 5 3z 5 26
6x 1 3y 2 4(22x 1 y 2 2) 5 8
14x 2 y 5 0
23x 1 2y 1 3(22x 1 y 2 2 5 26
29x 1 5y 5 0
14x 2 y 5 0 3 5 70x 2 5y 5 0
29x 1 5y 5 0 29x 1 5y 5 0
61x 5 0
x 5 0
14(0) 2 y 5 0 → y 5 0
2(0) 2 0 1 z 5 22 → z 5 22
The solution is (0, 0, 22).
Chapter 3, continued
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144Algebra 2Worked-Out Solution Key
20. 3x 1 5y 2 z 5 12
x 1 y 1 z 5 0 → x 5 2y 2 z
2x 1 2y 1 2z 5 227
3(2y 2 z) 1 5y 2 z 5 12
2y 2 4z 5 12
2(2y 2 z) 1 2y 1 2z 5 227
3y 1 3z 5 227
2y 2 4z 5 12 3 (23) 26y 1 12z 5 236
3y 1 3z 5 227 3 2 6y 1 6z 5 254
18z 5 290
z 5 25
2y 2 4(25) 5 12 → y 5 24
x 1 (24) 1 (25) 5 0 → x 5 9
The solution is (9, 24, 25).
21. The error is in line 2. Equation 2 is 3x 1 2y 1 z 5 11. So 2 times Equation 2 is 6x 1 4y 1 2z 5 22, not6x 1 2y 1 2z 5 22.
22. The error is in line 1. Equation 2 is 3x 1 2y 1 z 5 11. After you solve for z, you should get z 5 11 2 3x 2 2y, not z 5 11 1 3x 1 2y.
23. A;
2x 1 5y 1 3z 5 10
2(7) 1 5(1) 1 3(23) 0 10
10 5 10 ✓
3x 2 y 1 4z 5 8
3(7) 2 (1) 1 4(23) 0 8
8 5 8 ✓
5x 2 2y 1 7z 5 12
5(7) 2 2(1) 1 7(23) 0 12
12 5 12 ✓
Because all three equations are true, (7, 1, 23) is the solution of the system.
24. B;
2x 2 2y 2 z 5 6
2(x) 2 2(x 2 3) 2 0 0 6
6 5 6 ✓
2x 1 y 1 3z 5 23
2x 1 (x 2 3) 1 0 0 23
23 5 23
3x 2 3y 1 2z 5 9
3x 2 3(x 2 3) 1 0 0 9
9 5 9 ✓
Because all three equations are true, (x, x 2 3, 0) describes all of the solutions of the system.
25. x 1 5y 2 2z 5 21
2x 2 2y 1 z 5 6
3y 2 z 5 5
2x 2 2y 1 z 5 6 3 (22) 2x 1 4y 2 2z 5 212
22x 2 7y 1 3z 5 7 22x 2 7y 1 3z 5 7
23y 1 z 5 25
3y 2 z 5 5
23y 1 z 5 25
0 5 0
Infi nitely many solutions.
26. 4x 1 5y 1 3z 5 15
x 2 3y 1 2z 5 26 → x 5 3y 2 2z 2 6
2x 1 2y 2 z 5 3
4(3y 2 2z 2 6) 1 5y 1 3z 5 15
17y 2 5z 5 39
2(3y 2 2z 2 6) 1 2y 2 z 5 3
2y 1 z 5 23
17y 2 5z 5 39 17y 2 5z 5 39
2y 1 z 5 23 3 5 25y 1 5z 5 215
12y 5 24
y 5 2
22 1 z 5 23 → z 5 21
2x 1 2(2) 2 (21) 5 3 → x 5 2
The solution is (2, 2, 21).
27. 6x 1 y 2 z 5 22 6x 1 y 2 z 5 22
2x 1 y 1 2z 5 5 3 6 26x 1 6y 1 12z 5 30
7y 1 11z 5 28
x 1 6y 1 3z 5 23
2x 1 y 1 2z 5 5
7y 1 5z 5 28
7y 1 11z 5 28 7y 1 11z 5 28
7y 1 5z 5 28 3 (21) 27y 2 5z 5 228
6z 5 0
z 5 0
7y 1 5(0) 5 28 → y 5 4
x 1 6(4) 1 3(0) 5 23 → x 5 21
The solution is (21, 4, 0).
Chapter 3, continued
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145Algebra 2
Worked-Out Solution Key
28. x 1 2y 5 21 → x 5 22y 2 1
3x 2 y 1 4z 5 17
24x 1 2y 2 3z 5 230
3(22y 2 1) 2 y 1 4z 5 17
27y 1 4z 5 20
24(22y 2 1) 1 2y 2 3z 5 230
10y 2 3z 5 234
27y 1 4z 5 20 3 3 221y 1 12z 5 60
10y 2 3z 5 234 3 4 40y 2 12z 5 2136
19y 5 276
y 5 24
27(24) 1 4z 5 20 → z 5 22
x 1 2(24) 5 21 → x 5 7
The solution is (7, 24, 22).
29. 2x 2 y 1 2z 5 221 3 (22) 24x 1 2y 2 4z 5 42
23x 1 2y 1 4z 5 6 23x 1 2y 1 4z 5 6
27x 1 4y 5 48
x 1 5y 2 z 5 25 3 4 4x 1 20y 2 4z 5 100
23x 1 2y 1 4z 5 6 23x 1 2y 1 4z 5 6
x 1 22y 5 106
27x 1 4y 5 48 27x 1 4y 5 48
x 1 22y 5 106 3 7 7x 1 154y 5 742
158y 5 790
y 5 5
27x 1 4(5) 5 48 → x 5 24
2(24) 2 5 1 2z 5 221 → z 5 24
The solution is (24, 5, 24).
30. 4x 2 8y 1 2z 5 10 4x 2 8y 1 2z 5 10
2x 2 4y 1 z 5 8 3 (22) 24x 1 8y 2 2z 5 216
0 5 26
No solution.
31. 2x 1 5y 2 z 5 216
x 1 y 2 z 5 28
6y 2 2z 5 224
2x 1 3y 1 4z 5 18 2x 1 3y 1 4z 5 18
x 1 y 2 z 5 28 3 (22) 22x 2 2y 1 2z 5 16
y 1 6z 5 34
6y 2 2z 5 224 3 3 18y 2 6z 5 272
y 1 6z 5 34 y 1 6z 5 34
19y 5 238
y 5 22
6(22) 2 2z 5 224 → z 5 6
x 1 (22) 2 6 5 28 → x 5 0
The solution is (0, 22, 6).
32. 2x 2 y 1 4z 5 19 3 (22) 24x 1 2y 2 8z 5 238
4x 1 2y 1 3z 5 37 4x 1 2y 1 3z 5 37
4y 2 5z 5 21
2x 1 3y 2 2z 5 27 3 4 24x 1 12y 2 8z 5 228
4x 1 2y 1 3z 5 37 4x 1 2y 1 3z 5 37
14y 2 5z 5 9
4y 2 5z 5 21 3 (21) 24y 1 5z 5 1
14y 2 5z 5 9 14y 2 5z 5 9
10y 5 10
y 5 1
4(1) 2 5z 5 21 → z 5 1
2x 1 3(1) 2 2(1) 5 27 → x 5 8
The solution is (8, 1, 1).
33. x 1 y 1 z 5 3 → x 5 2y 2 z 1 3
3x 2 4y 1 2z 5 228
2x 1 5y 1 z 5 23
3(2y 2 z 1 3) 2 4y 1 2z 5 228
27y 2 z 5 237
2(2y 2 z 1 3) 1 5y 1 z 5 23
6y 1 2z 5 26
27y 2 z 5 237 3 2 214y 2 2z 5 274
6y 1 2z 5 26 6y 1 2z 5 26
28y 5 248
y 5 6
27(6) 2 z 5 237 → z 5 25
x 1 6 1 (25) 5 3 → x 5 2
The solution is (2, 6, 25).
34. a. Sample answer: b. Sample answer:
x 1 6y 2 2z 5 7 3x 2 2y 1 z 5 10
3x 2 y 1 z 5 4 8x 2 y 1 4z 5 24
x 2 12y 1 10z 5 7 26x 1 4y 2 2z 5 3
Solution is (1, 2, 3). No Solution
c. Sample answer:
4x 2 6y 1 2z 5 72
x 1 y 2 z 5 26
24x 2 4y 1 4z 5 24 Infi nitely many solutions.
35. x 1 1 } 2 y 1
1 } 2 z 5
5 } 2 3 2 2x 1 y 1 z 5 5
1 }
3 x 1
3 } 2 y 1
2 } 3 z 5
13 } 6 3 (26) 22x 2 9y 2 4z 5 213
28y 2 3z 5 28
3 }
4 x 1
1 } 4 y 1
3 } 2 z 5
7 } 4 3 (28) 26x 2 2y 2 12z 5 214
1 }
3 x 1
3 } 2 y 1
2 } 3 z 5
13 } 6 3 18 6x 1 27y 1 12z 5 39
25y 5 25
y 5 1
28(1) 2 3z 5 28 → z 5 0
x 1 1 } 2 (1) 1
1 } 2 (0) 5
5 } 2 → x 5 2
The solution is (2, 1, 0).
Chapter 3, continued
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146Algebra 2Worked-Out Solution Key
36. 1 }
3 x 1
5 } 6 y 1
2 } 3 z 5
4 } 3 3 (212) 24x 2 10y 2 8z 5 216
2 }
3 x 1
1 } 6 y 1
3 } 2 z 5
4 } 3 3 6 4x 1 y 1 9z 5 8
29y 1 z 5 28
1 }
6 x 1
2 } 3 y 1
1 } 4 z 5
5 } 6 3 24 4x 1 16y 1 6z 5 20
2 }
3 x 1
1 } 6 y 1
3 } 2 z 5
4 } 3 3 (26) 24x 2 y 2 9z 5 28
15y 2 3z 5 12
29y 1 z 5 28 3 3 227y 1 3z 5 224
15y 2 3z 5 12 15y 2 3z 5 12
212y 5 212
y 5 1
29(1) 1 z 5 28 → z 5 1
1 }
3 x 1
5 } 6 (1) 1
2 } 3 (1) 5
4 } 3 → x 5 2
1 } 2
The solution is 1 2 1 } 2 , 1, 1 2 .
37. Substitute 21 for x, 2 for y, and 23 for z.
21 1 2(2) 2 3(23) 5 a → a 5 12
2(21) 2 2 1 (23) 5 b → b 5 24
2(21) 1 3(2) 2 2(23) 5 c → c 5 10
When a 5 12, b 5 24, and c 5 10, the system has(21, 2, 23) as its solution.
38. w 1 x 1 y 1 z 5 2 Add Equation 1
3w 1 x 1 y 2 z 5 25 to Equation 4.
4w 1 2x 1 y 5 23 New Equation 1
22w 1 x 2 2y 1 z 5 21 Add 21 times Equation 2
3w 1 x 1 y 2 z 5 25 to Equation 4.
w 1 2x 2 y 5 26 New Equation 2
2w 1 2x 2 y 1 2z 5 22 Add Equation 3
6w 1 2x 1 2y 2 2z 5 210 to 2 times Equation 4.
5w 1 4x 1 y 5 212 New Equation 3
4w 1 2x 1 2y 5 23 Add new Equation 1
210w 2 8x 2 2y 5 24 to 22 times new Eq. 3.
26w 2 6x 5 21 Call this Equation 5.
w 1 2x 2 y 5 26 Add new Equation 2
5w 1 4x 1 y 5 212 to new Equation 3.
6w 1 6x 5 218 Call this Equation 6.
26w 2 6x 5 21 Add Equation 5
6w 1 6x 5 218 to Equation 6.
0 5 3
Because 0 5 3 is a false equation, the system has no solution.
39. 2w 1 x 2 3y 1 z 5 4 Add Equation 1
22w 2 2x 1 2y 2 6z 5 228 to 22 times Eq. 4.
2x 2 y 2 5z 5 224 New Equation 1
w 2 3x 1 y 1 z 5 32 Add Equation 2
2w 2 x 1 y 2 3z 5 214 to 21 times Equation 4.
24x 1 2y 2 2z 5 18 New Equation 2
2w 1 2x 1 2y 2 z 5 210 Add Equation 3
w 1 x 2 y 1 3z 5 14 to Equation 4.
3x 1 y 1 2z 5 4 New Equation 3
2x 2 y 2 5z 5 224 Add new Equation 1
3x 1 y 1 2z 5 4 to new Equation 3.
2x 2 3z 5 220 Call this Equation 4.
24x 1 2y 2 2z 5 18 Add new Equation 2
26x 2 2y 2 4z 5 28 to 22 times new Eq. 3.
210x 2 6z 5 10 Call this Equation 5.
24x 1 6z 5 40 Add 22 times Eq. 4
210x 2 6z 5 10 to Equation 5
214x 5 50
x 5 2 25
} 7 Solve for x.
z 5 30
} 7 Substitute into Equation 4 or 5 to fi nd z.
y 5 43
} 7 Substitute into a new equation to fi nd y.
w 5 76
} 7 Substitute into an original equation to fi nd w.
The solution is 1 76 } 7 , 2
25 } 7 ,
43 } 7 ,
30 } 7 2 .
40. 210w 1 5x 1 20y 1 10z 5 235 Add 5 times Eq. 2
22w 2 4x 1 4y 2 10z 5 26 to 22 times Eq. 3.
212w 1 x 1 24y 5 241 New Eq. 2
24w 1 48x 1 120y 5 264 Add 24 times Eq. 1
60w 2 5x 2 120y 5 205 to 25 times new Eq. 2.
84w 1 43x 5 469 New Eq. 1
x 5 3w 2 1 Rewrite Eq 4.
84w 1 43(3w 2 1) 5 469 Substitute 3w 2 1 for x in new Eq. 1.
213w 5 512
w 5 512
} 213 Solve for w.
x 5 441
} 71 Substitute into Eq. 4 to fi nd x.
y 5 2 163
} 213 Substitute into new Eq. 2 to fi nd y.
z 5 2 569
} 213 Substitute into an original equation to fi nd z.
The solution is 1 512 }
213 ,
441 }
71 , 2
163 } 213 , 2
569 } 273 2 .
Chapter 3, continued
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147Algebra 2
Worked-Out Solution Key
41. 2w 1 7x 2 3y 5 41 Add Equation 1
22w 2 2x 1 2y 5 216 to 2 times Equation 4.
5x 2 y 5 25 New Equation 1
2w 2 2x 1 y 5 213 Add Equation 2
w 1 x 2 y 5 8 to 21 times Equation 4.
2x 5 25 New Equation 2
x 5 5 Solve for x.
y 5 0 Substitute into new Equation 1 to fi nd y.
w 5 3 Substitute into an original equation for fi nd w.
z 5 22 Substitute into original Equation 3 to fi nd z.
The solution is (3, 5, 0, 22).
Problem Solving
42. p 5 cost of a small pizza
x 5 cost of a liter of soda
y 5 cost of a salad
Equation 1: 2p 1 x 1 y 5 14
Equation 2: p 1 x 1 3y 5 15
Equation 3: 3p 1 x 5 16 → x 5 16 2 3p
2p 1 (16 2 3p) 1 y 5 14
2p 1 y 5 22
p 1 (16 2 3p) 1 3y 5 15
22p 1 3y 5 21
2p 1 y 5 22 3 (22) 2p 2 2y 5 4
22p 1 3y 5 21 22p 1 3y 5 21
y 5 3
2p 1 y 5 22 → p 5 5
2(5) 1 x 1 3 5 14 → x 5 1
A small pizza costs $5, a liter of soda costs $1, and a salad costs $3.
43. x 5 gallons received in 1st delivery
y 5 gallons received in 2nd delivery
z 5 gallons received in 3rd delivery
Equation 1: 0.7x 1 0.5y 1 0.3z 5 1200
Equation 2: 0.2x 1 0.3y 1 0.3z 5 900
Equation 3: 0.1x 1 0.2y 1 0.4z 5 1000
21.4x 2 y 2 0.6z 5 22400
1.4x 1 2.1y 1 2.1z 5 6300
1.1y 1 1.5z 5 3900
0.7x 1 0.5y 1 0.3z 5 1200
20.7x 2 1.4y 2 2.8z 5 27000
20.9y 2 2.5z 5 25800
1.1y 1 1.5z 5 3900 3 5
20.9y 2 2.5z 5 25800 3 3
5.5y 1 7.5z 5 19,500
22.7y 2 7.5z 5 217,400
2.8y 5 2100
y 5 750
1.1(750) 1 1.5z 5 3900 → z 5 2050
0.7x 1 0.5(750) 1 0.3(2050) 5 1200 → x 5 300
The health club received 300 gallons in the fi rst delivery, 750 gallons in the second delivery, and 2050 gallons in the third delivery.
44. a. c 5 comedy shows
d 5 dramas
r 5 reality shows
30c 1 60d 1 60r 5 360
c 1 d 1 r 5 7
2c 5 d
b. 30c 1 60d 1 60r 5 360 Add Eq. 1
260c 2 60d 2 60r 5 2420 to 260 times Eq. 2.
230c 5 260
c 5 2 Solve for c.
d 5 4 Substitute into Eq. 3 to fi nd d.
r 5 1 Substitute to fi nd r.
There are two comedies, four dramas, and one reality show on the tape.
c. New system of equations:
30c 1 60d 1 60r 5 360
c 1 d 1 r 5 5
2c 5 d
30c 1 60d 1 60r 5 360 Add Equation 1
260c 2 60d 2 60r 5 2300 to 260 times Eq. 2.
230c 5 60
c 5 22 Solve for c.
Because a negative number of shows cannot be recorded, this situation is impossible.
45. a. x 5 athletes who fi nished in 1st place
y 5 athletes who fi nished in 2nd place
z 5 athletes who fi nished in 3rd place
Equation 1: x 1 y 1 z 5 20
Equation 2: 5x 1 3y 1 z 5 68
Equation 3: y 5 x 1 z
x 1 (x 1 z) 1 z 5 20 5x 1 3(x 1 z) 1 z 5 68
2x 1 2z 5 20 8x 1 4z 5 68
2x 1 2z 5 20 3 (24) 28x 2 8z 5 280
8x 1 4z 5 68 8x 1 4z 5 68
24z 5 212
z 5 3
2x 1 2(3) 5 20 → x 5 7
7 1 y 1 3 5 20 → y 5 10
Seven athletes fi nished in fi rst place, ten fi nished second, and three fi nished third.
Chapter 3, continued
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148Algebra 2Worked-Out Solution Key
b. The new Equation 2 would be 5x 1 3y 1 z 5 70.
x 1 (x 1 z) 1 z 5 20 5x 1 3(x 1 z) 1 z 5 70
2x 1 2z 5 20 8x 1 4z 5 70
2x 1 2z 5 20 3 (24) 28x 2 8z 5 280
8x 1 4z 5 70 8x 1 4z 5 70
24z 5 210
z 5 2.5
This claim must be false because you cannot have 2.5 athletes.
46. m 5 cost of mixed nuts
g 5 cost of granola
d 5 cost of dried fruit
Equation 1: m 1 0.5g 1 d 5 8
Equation 2: 2m 1 0.5g 1 0.5d 5 9
Equation 3: m 1 2g 1 0.5d 5 9
m 1 0.5g 1 d 5 8 m 1 0.5g 1 d 5 8
m 1 2g 1 0.5d 5 9 3 (22) 22m 2 4g 2 d 5 218
2 m 2 3.5g 5 210
2m 1 0.5g 1 0.5d 5 9 2m 1 0.5g 1 0.5d 5 9
m 1 2g 1 0.5d 5 9 3 (21) 2m 2 2g 2 0.5d 5 29
m 2 1.5g 5 0
2m 2 3.5g 5 210
m 2 1.5g 5 0
25g 5 210
g 5 2
m 2 1.5(2) 5 0 → m 5 3
3 1 0.5(2) 1 d 5 8 → d 5 4
Mixed nuts cost $3 per pound, granola costs $2 per pound, and dried fruit costs $4 per pound.
47. a. r 5 number of roses
l 5 number of lilies
i 5 number of irises
Equation 1: r 1 l 1 i 5 12
Equation 2: 2.5r 1 4l 1 2i 5 32
Equation 3: r 5 2(l 1 i) or r 5 2l 1 2i
b. (2l 1 2i) 1 l 1 i 5 12
3l 1 3i 5 12
2.5(2l 1 2i) 1 4l 1 2i 5 32
9l 1 7i 5 32
3l 1 3i 5 12 3 (23) 29l 2 9i 5 236
9l 1 7i 5 32 9l 1 7i 5 32
22i 5 24
i 5 2
3l 1 3(2) 5 12 → l 5 2
r 1 2 1 2 5 12 → r 5 8
Each bouquet should contain 8 roses, 2 lilies, and 2 irises.
c. r 1 l 1 i 5 12 r 1 l 1 i 5 12
(2l 1 2i) 1 l 1 i 5 12 r 1 4 5 12
3l 1 3i 5 12 r 5 8
3(l 1 i) 5 12
l 1 i 5 4
No; Three possible solutions are:(8 roses, 2 lilies, 2 irises), (8 roses, 3 lilies, 1 iris), (8 roses, 1 lily, 3 irises).
48. t 5 weight of one tangerine
b 5 weight of one banana
g 5 weight of one grapefruit
t 1 a 5 g Equation 1
t 1 b 5 a Equation 2
2g 5 3b Equation 3
t 1 (t 1 b) 5 g Substitute t 1 b for a in Eq. 1.
2t 1 b 5 g New Equation 1
g 5 3 } 2 b Rewrite Equation 3.
2t 1 b 5 3 } 2 b Substitute
3 }
2 b for g in new Eq. 1.
2 1 } 2 b 5 22t
b 5 4t Solve for b.
t 1 4t 5 a Substitute 4t for b in Eq. 2
5t 5 a
Because 5t 5 a, fi ve tangerines will balance the applein the fi nal picture.
Mixed Review
49. 15 1 (28) 5 7
50. 24 2 (213) 5 24 1 13 5 9
51. 15 p 7 5 105 52. 24(28) 5 4 p 8 5 32
53. (1, 24), (2, 6)
m 5 6 2 (24)
} 2 2 1 5 10
The slope is positive, so the line rises.
54. (4, 2), (218, 1)
m 5 1 2 2
} 218 2 4 5
1 } 22
The slope is positive, so the line rises.
55. (6, 26), (26, 6)
m 5 6 2 (26)
} 26 2 6 5 21
The slope is negative, so the line falls.
56. (25, 2), (25, 10)
m 5 10 2 2
} 25 2 (25)
5 8 } 0
The slope is undefi ned, so the line is vertical.
Chapter 3, continued
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149Algebra 2
Worked-Out Solution Key
57. (22, 4), (26, 8)
m 5 8 2 4 }
26 2 (22) 5 21
The slope is negative, so the line falls.
58. (27, 3), (5, 3)
m 5 3 2 3
} 5 2 (27)
5 0
The slope is 0, so the line is horizontal.
59. 3x 2 y 5 27 3 3 9x 2 3y 5 221
2x 1 3y 5 21 2x 1 3y 5 21
11x 5 0
x 5 0
3(0) 2 y 5 27 → y 5 7
The solution is (0, 7).
60. 3x 1 2y 5 23 3 3 9x 1 6y 5 29
4x 2 3y 5 238 3 2 8x 2 6y 5 276
17x 5 285
x 5 25
3(25) 1 2y 5 23 → y 5 6
The solution is (25, 6).
61. 5x 1 y 5 11 → y 5 11 2 5x
2x 1 3y 5 219
2x 1 3(11 2 5x) 5 219
213x 5 252
x 5 4
5(4) 1 y 5 11 → y 5 29
The solution is (4, 29).
Mixed Review of Problem Solving (p. 186)
1. a. n 5 the number of necklaces
b 5 the number of bracelets
3.5n 1 2.5b 5 121 Equation 1
9.0n 1 7.5b 5 324 Equation 2
b. 210.5n 2 7.5b 5 2363 Add 23 times Eq. 1
9.0n 1 7.5b 5 324 to Equation 2.
21.5n 5 239
n 5 26 Solve for n.
b 5 12 Substitute to fi nd b.
At the craft fair, 26 necklaces and 12 bracelets were sold.
2. a. t 5 number of tapers
p 5 number of pillars
j 5 number of jar candles
t 1 4p 1 6j 5 24 Equation 1
t 1 p 1 j 5 8 Equation 2
t 5 p 1 j Equation 3
b. (p 1 j) 1 4p 1 6j 5 24 Substitute p 1 j for t in Equation1.
5p 1 7j 5 24 New Equation 1
(p 1 j) 1 p 1 j 5 8 Substitute p 1 j for t in Equation 2.
2p 1 2j 5 8 New Equation 2
10p 1 14j 5 48 Add 2 times new Eq. 1.
210p 2 10j 5 240 to 25 times new Eq. 2.
4j 5 8
j 5 2 Solve for j.
p 5 2 Substitute to fi nd p.
t 5 4 Substitute to fi nd t.
There are four tapers, two pillars, and two jar candles in each basket.
c. t 1 p 1 j 5 8 t 1 p 1 j 5 8
(p 1 j) 1 p 1 j 5 8 t 1 4 5 8
2p 1 2j 5 8 t 5 4
2(p 1 j) 5 8
p 1 j 5 4
No; three possible solutions are: (4 tapers, 2 pillars, 2 jar candles), (4 tapers, 3 pillars, 1 jar candle), (4 tapers, 1 pillar, 3 jar candles).
3. Sample answer: y > 1, x > 2, y < 4 2 x
4. a. s 5 the number of small tables
l 5 the number of large tables
s 1 l 5 20 Equation 1
4s 1 6l 5 90 Equation 2
24s 2 4l 5 280 Add 24 times Equation. 1
4s 1 6l 5 90 to Equation 2.
2l 5 10
l 5 5 Solve for l.
s 5 15 Substitute to solve for s.
The restaurant has fi fteen 4-seat tables and fi ve 6-seat tables.
b. Sample answer:
1. The restaurant could buy fi ve more 6-seat tables and fi ve more 4-seat tables.
2. The restaurant could buy one additional 6-seat table and eleven more 4-seat tables.
3. The restaurant could buy three more 6-seat tables and 8 more 4-seat tables.
Chapter 3, continued
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150Algebra 2Worked-Out Solution Key
5. s 5 the price of a soda
p 5 the price of a pretzel
h 5 the price of a hot dog
s 1 p 1 2h 5 7 Equation 1
2s 1 p 1 2h 5 8 Equation 2
s 1 4h 5 10 Equation 3
s 1 p 1 2h 5 7 Add Equation 1
22s 2 p 2 2h 5 28 to 21 times Equation 2.
2s 5 21
s 5 1 Solve for s.
h 5 2.25 Substitute into original Equation. 3 to fi nd h.
The price of one hotdog is $2.25.
6. h 5 the number of hours skating
b 5 the number of hours bicycling
6h 1 8b 5 34 Equation 1
b 1 h 5 5 Equation 2
b 5 5 2 h Rewrite Equation 2.
6h 1 8(5 2 h) 5 34 Substitute 5 2 h for b in Eq. 1.
22h 5 26
h 5 3 Solve for h.
The student spends three hours skating.
7. a. y 5 0.29x 1 73.14
b. y 5 0.14x 1 79.12
c.
50 10 15 20 25 30 35 40 45
720
7476788082848688
Years since 1996
Life
sp
an (
year
s)
x
y
(40, 84.7)
The lines intersect at about (40, 84.7). This means that in the year 2036, men and women will both have a life span of about 84.7 years.
8. a. s 5 the number of small chairs
l 5 the number of large chairs
45s 1 70l ≤ 2000
80s 1 110l ≥ 2750
Number of small chairs
Nu
mb
er o
f la
rge
chai
rs
0 10 20 30 40 s
l
0
10
20
30
b. Sample answer:
30 small chairs and 5 large chairs
20 small chairs and 13 large chairs
10 small chairs and 20 large chairs
Lesson 3.5
3.5 Guided Practice (pp. 188–190)
1. F 22 5 11 4 26 8G1 F 23 1 25
22 28 4G 5 F 22 1 (23) 5 1 1 11 1 (25)
4 1 (22) 26 1 (28) 8 1 4G 5 F 25 6 6
2 214 12G 2. F 24 0
7 22
23 1G 2 F 2 2
23 0
5 214G
5 F 24 2 2 0 2 2
7 2 (23) 22 2 0
23 2 5 1 2 (214)G
5 F 26 22
10 22
28 15G
3. 24 F 2 21 23
27 6 1
22 0 25G
5 F 24(2) 24(21) 24(23)
24(27) 24(6) 24(1)
24(22) 24(0) 24(25)G
5 F28 4 12
28 224 24
8 0 20G
4. 3 F 4 21 23 25G 1 F 22 22
0 6G 5 F 3(4) 1 (22) 3(21) 1 (22)
3(23) 1 0 3(25) 1 6G 5 F 10 25 29 29G
5. B 2 A 5 F 95 114
316 215
205 300G 2 F 125 100
278 251
225 270G
5 F 95 2 125 114 2 100
316 2 278 215 2 251
205 2 225 300 2 270G 5 F 230 14
38 236
220 30G
This matrix represents the change in the number of DVD racks sold from last month to this month.
Chapter 3, continued
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151Algebra 2
Worked-Out Solution Key
6. 22 (F 23x 21 4 yG 1 F 9 24
25 3G) 5 F 12 10
2 218G 22 F23x 1 9 25
21 y 1 3G 5 F 12 10
2 218G F22(23x 1 9) 22(25)
22(21) 22(y 1 3)G 5 F 12 10
2 218G F6x 2 18 10
2 22y 2 6G 5 F 12 10
2 218G 6x 2 18 5 12 22y 2 6 5 218
x 5 5 y 5 6
The solution is x 5 5 and y 5 6.
3.5 Exercises (pp. 190–193)
Skill Practice
1. The dimensions of a matrix with 3 rows and 4 columns are 3 3 4.
2. To determine if two matrices are equal, fi rst compare the dimensions and then compare the corresponding elements. If they are the same, then the matrices are equal.
3. The fi nal matrix has the wrong dimensions.
It should be F 13.1
21.2 G.
4. F 5 2
21 8G 1 F 28 10
26 3 G 5 F 5 1 (28) 2 1 10 21 1 (26) 8 1 3G
5 F 23 12 27 11G
5. F 10 28 5 23G 2 F 12 23
3 24G 5 F10 2 12 28 2 (23)
5 2 3 23 2 (24)G 5 F 22 25 2 1G
6. This operation is not possible. You cannot subtract a 2 3 1 matrix from a 2 3 2 matrix.
7. F1.2 5.3
0.1 4.4
6.2 0.7G 1 F2.4 20.6
6.1 3.1
8.1 21.9G
5 F 1.2 1 2.4 5.3 1 (20.6)
0.1 1 6.1 4.4 1 3.1
6.2 1 8.1 0.7 1 (21.9)G 5 F 3.6 4.7
6.2 7.5
14.3 21.2G
8. This operation is not possible. You cannot add a 3 3 3 matrix to a 3 3 2 matrix.
9. F 7 23
12 5
24 11G 2 F 9 2
22 6
6 5G
5 F 7 2 9 23 2 2
12 2 (22) 5 2 6
24 2 6 11 2 5G 5 F 22 25
14 21
210 6G
10. 2 F 21 4 3 26 G 5 F 2(21) 2(4)
2(3) 2(26)G
5 F 22 8 6 212G
11. 23 F 2 0 25 4 7 23G 5 F 23(2) 23(0) 23(25)
23(4) 23(7) 23(23)G 5 F 26 0 15
212 221 9G 12. 24 F 2 23 22
2 5 } 8 11 }
2 7 }
4 G
5 F 24(2) 24(23) 24(22)
24 1 2 5 } 8 2 24 1 11 } 2 2 24 1 7 }
4 2 G
5 F 28 12 8
5 } 2 222 27G 13. 1.5 F 22 3.4 1.6
5.4 0 23G 5 F 1.5(22) 1.5(3.4) 1.5(1.6)
1.5(5.4) 1.5(0) 1.5(23)G 5 F 23 5.1 2.4
8.1 0 24.5G 14.
1 }
2 F 22 8 12
20 21 0
28 10 2G
5 F 1 } 2 (22) 1 }
2 (8) 1 } 2 (12)
1 } 2 (20) 1 }
2 (21) 1 } 2 (0)
1 } 2 (28) 1 }
2 (10) 1 } 2 (2)
G 5 F 21 4 6
10 2 1 } 2 0
24 5 1G
Chapter 3, continued
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152Algebra 2Worked-Out Solution Key
15. 22.2 F 6 3.1 4.5
21 0 2.5
5.5 21.8 6.4G
5 F 22.2(6) 22.2(3.1) 22.2(4.5)
22.2(21) 22.2(0) 22.2(2.5)
22.2(5.5) 22.2(21.8) 22.2(6.4)G
5 F 213.2 26.82 29.9
2.2 0 25.5
212.1 3.96 214.08G
16. A 1 B 5 F 5 24
3 21G 1 F 18 212
26 0G 5 F 5 1 18 24 1 (212)
3 1 (26) 21 1 0G 5 F 23 216
23 21G17. B 2 A 5 F 18 212
26 0G 2 F 5 24
3 21G 5 F 18 2 5 212 2 (24)
26 2 3 0 2 (21)G 5 F 13 28
29 1G18. 4A 2 B 5 4F 5 24
3 21G 2 F 18 212
26 0G 5 F 4(5) 2 18 4(24) 2 (212)
4(3) 2 (26) 4(21) 2 0G 5 F 2 24
18 24G19.
2 }
3 B 5
2 } 3 F 18 212
26 0G 5 F 2 } 3 (18) 2 }
3 (212)
2 } 3 (26) 2 }
3 (0)G
5 F 12 28
24 0G20. C 1 D 5 F 1.8 21.5 10.6
28.8 3.4 0G 1 F 7.2 0 25.4
2.1 21.9 3.3G 5 F 1.8 1 7.2 21.5 1 0 10.6 1 (25.4)
28.8 1 2.1 3.4 1 (21.9) 0 1 3.3G 5 F 9 21.5 5.2
26.7 1.5 3.3G21. C 1 3D 5 F 1.8 21.5 10.6
28.8 3.4 0G 1 3F 7.2 0 25.4
2.1 21.9 3.3G 5 F 1.8 1 3(7.2) 21.5 1 3(0) 10.6 1 3(25.4)
28.8 1 3(2.1) 3.4 1 3(21.9) 0 1 3(3.3)G 5 F 23.4 21.5 25.6
22.5 22.3 9.9G
22. D 2 2C 5 F 7.2 0 25.4
2.1 21.9 3.3G 2 2F 1.8 21.5 10.6
28.8 3.4 0G 5 F 7.2 2 2(1.8) 0 2 2(21.5) 25.4 2 2(10.6)
2.1 2 2(28.8) 21.9 2 2(3.4) 3.3 2 2(0)G 5 F 3.6 3 226.6
19.7 28.7 3.3G 23. 0.5C 2 D 5 0.5F 1.8 21.5 10.6
28.8 3.4 0G 2 F 7.2 0 25.4
2.1 21.9 3.3G5 F 0.5(1.8) 2 7.2 0.5(21.5) 2 0 0.5(10.6) 2 (25.4)
0.5(28.8) 2 2.1 0.5(3.4) 2 (21.9) 0.5(0) 2 3.3G5 F 26.3 20.75 10.7
26.5 3.6 23.3G 24. F 21 3x
24 5G 5 F 21 218
2y 5G 3x 5 218 2y 5 24
x 5 26 y 5 22
The solution is x 5 26 and y 5 22.
25. F 22x 6
1 28G 1 2 F 5 21
27 6G 5 F 29 4
213 yG F 22x 1 2(5) 6 1 2(21)
1 1 2(27) 28 1 2(6)G 5 F 29 4
213 yG F 22x 1 10 4
213 4G 5 F 29 4
213 yG 22x 1 10 5 29 y 5 4
x 5 19
} 2
The solution is x 5 19
} 2 and y 5 4.
26. 2F 8 2x
5 6G 2 F 3 29
10 24yG 5 F 13 4
0 16G F 2(8) 2 3 2(2x) 2 (29)
2(5) 2 10 2(6) 2 (24y)G 5 F 13 4
0 16G F 13 22x 1 9
0 12 1 4yG 5 F 13 4
0 16G 22x 1 9 5 4 12 1 4y 5 16
x 5 5 } 2 y 5 1
The solution is x 5 5 } 2 and y 5 1.
Chapter 3, continued
n2ws-03-a.indd 152 6/27/06 9:51:55 AM
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153Algebra 2
Worked-Out Solution Key
27. 4xF 21 2
3 6G 5 F 8 216
224 3yG F 4x(21) 4x(2)
4x(3) 4x(6)G 5 F 8 216
224 3y G F 24x 8x
12x 24xG 5 F 8 216
224 3yG 24x 5 8 3y 5 24x
x 5 22 3y 5 24(22)
y 5 216
The solution is x 5 22 and y 5 216.
28. C; 2x 5 6.4 3y 5 20.75
x 5 3.2 y 5 20.25
3x 2 2y 5 3(3.2) 2 2(20.25)
5 9.6 1 0.50
5 10.1
29. Sample answer:
A 5 F 10 3
25 4 G, B 5 F 5 2
23 2 G30. a. X 1 F 25 0
4 23G 5 F 7 28
23 5 G F a 1 (25) b 1 0
c 1 4 d 1 (23)G 5 F 7 28
23 5 G a 5 12, b 5 28, c 5 27, and d 5 8.
X 5 F 12 28
27 8 G b. X 2 F 2 3
5 0 G 5 F 8 6
21 3 G F a 2 2 b 2 3
c 2 5 d 2 0G 5 F 8 6
21 3G a 5 10, b 5 9, c 5 4, and d 5 3.
X 5 F 10 9
4 3 G c. 2X 1 F 23 1
4 7 G 5 F 8 29
0 10 G F 2a 1 (23) 2b 1 1
2c 1 4 2d 1 7G 5 F 8 29
0 10 G a 5 211, b 5 10, c 5 4, and d 5 23.
X 5 F 211 10
4 23 G
d. 3X 2 F 11 26
2 1 G 5 F 213 15
219 2 G F 3a 2 11 3b 2 (26)
3c 2 2 3d 2 1G 5 F 213 15
219 2 G a 5 2
2 } 3 , b 5 3, c 5 2
17 } 3 , and d 5 1.
X 5 F 2 2 } 3 3
2 17
} 3 1G
Problem Solving
31. Change in sales 5 Sales for 2004 2 Sales for 2003
5 F 32 47 30 19
5 16 20 14
29 39 36 31G
2 F 32 42 29 20
12 17 25 16
28 40 32 21G
5 F 32 2 32 47 2 42 30 2 29 19 2 20
5 2 12 16 2 17 20 2 25 14 2 16
29 2 28 39 2 40 36 2 32 31 2 21G
5 F 0 5 1 21
27 21 25 22
1 21 4 10G
32. City mpg Highway mpg
Economy
Mid-size
Mini-van
Suv
F 32 40
24 34
18 25
19 22
G After an 8% increase: 1.08F 32 40
24 34
18 25
19 22
G 5 F 34.56 43.2
25.92 36.72
19.44 27
20.52 23.76
G 33. a. May(M) June (J)
A B C A B C
Downtown
Mall
F 31 42 18
22 25 11 GF 25 36 12
38 32 15 G b. M 1 J 5 F 31 42 18
22 25 11 G 1 F 25 36 12
38 32 15 G 5 F 31 1 25 42 1 36 18 1 12
22 1 38 25 1 32 11 1 15 G 5 F 56 78 30
60 57 26 G This matrix represents the total sales for May and June.
Chapter 3, continued
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154Algebra 2Worked-Out Solution Key
c. 1 }
2 (M 1 J) 5
1 } 2 F 56 78 30
60 57 26G 5 F 28 39 15
30 28.5 13 G34. No, the matrix A 1 B does not give meaningful
information. The team size in each matrix is an average, but the sum of the two averages is not an average.
35. A 5 F 1 1 5 5
1 4 1 4G
2
x
y
2
3A 5 F 3 3 15 15
3 12 3 12 G The height of the large rectangle is three times the
height of the small rectangle, and the width of the large rectangle is three times the width of the small rectangle. Therefore, the large rectangle is nine times the size of the small rectangle.
Mixed Review
36. 5 1 (28) 5 23 37. 27 1 6 5 21
38. 8(27) 5 256
39. 26 2 (215) 5 26 1 15 5 9
40. 25(29) 5 45
41. 6 2 (218) 5 6 1 18 5 24
42. y 5 x 1 1 when x ≤ 0.
y 5 2x 1 1 when x ≥ 0.
y 5 x 1 1
43. vertex: (1, 1) → y 5 ax 2 1 1 1
passes through (3, 22) → 22 5 a(3 2 1) 1 1
a 5 2 3 } 2
y 5 2 3 } 2 x 2 1 1 1
44. vertex: (3, 21) → y 5 ax 2 3 2 1
passes through (2, 2) → 2 5 a2 2 3 2 1
a 5 3
y 5 3x 2 3 2 1
45. 0 1 2(3) ≤? 23 25 1 2(1) ≤? 23
6 µ 23 23 ≤ 23 ✓
(25, 1) is a solution of the inequality, but (0, 3) is not.
46. 5(25) 2 0 >?2 5(5) 2 23 >?2
225 ò 2 2 ò 2
Neither ordered pair is a solution.
47. 28(21) 2 3(1) <? 5 28(3) 2 3(29) <? 5
5ñ 5 3 < 5 ✓
(3, 29) is a solution of the inequality, but (21, 1) is not.
48. 21(2) 2 10(3) >?4 21(21) 2 10(0) >?4
12 > 4 ✓ 221ò 4
(2, 3) is a solution of the inequality, but (21, 0) is not.
Quiz 3.3–3.5 (p. 193)
1.
1
x
y
22
2.
1
x
y
1
3.
x
y
2
1
4. 1
x
y
21
5.
8
x
y
22
6.
1
x
y
21
7. 2x 2 y 2 3z 5 5
x 1 2y 2 5z 5 211
2x 2 3y 5 10 → x 5 210 2 3y
2(210 2 3y) 2 y 2 3z 5 5
27y 2 3z 5 25
(210 2 3y) 1 2y 2 5z 5 211
2y 2 5z 5 21
27y 2 3z 5 25 27y 2 3z 5 25
2y 2 5z 5 21 3 (27) 7y 1 35z 5 7
32z 5 32
z 5 1
27y 2 3(1) 5 25 → y 5 24
x 1 2(24) 2 5(1) 5 211 → x 5 2
The solution is (2, 24, 1).
8. x 1 y 1 z 5 23 3 (22) 22x 2 2y 2 2z 5 6
4x 2 5y 1 2z 5 16 4x 2 5y 1 2z 5 16
2x 2 7y 5 22
2x 2 3y 1 z 5 9 3 (22) 24x 1 6y 2 2z 5 218
4x 2 5y 1 2z 5 16 4x 2 5y 1 2z 5 16
y 5 22
2x 2 7(22) 5 22 → x 5 4
4 1 (22) 1 z 5 23 → z 5 25
The solution is (4, 22, 25).
Chapter 3, continued
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155Algebra 2
Worked-Out Solution Key
9. 2x 2 4y 1 3z 5 1
22x 1 5y 2 2z 5 2
y 1 z 5 3
6x 1 2y 1 10z 5 19 6x 1 2y 1 10z 5 19
22x 1 5y 2 2z 5 2 3 3 26x 1 15y 2 6z 5 6
17y 1 4z 5 25
y 1 z 5 3 3 (24) 24y 2 4z 5 212
17y 1 4z 5 25 17y 1 4z 5 25
13y 5 13
y 5 1
1 1 z 5 3 → z 5 2
2x 2 4(1) 1 3(2) 5 1 → x 5 2 1 } 2
The solution is 1 2 1 } 2 , 1, 2 2 .
10. A 1 B 5 F 2 25
3 21 G 1 F 24 3
8 10 G 5 F 2 1 (24) 25 1 3
3 1 8 21 1 10 G 5 F 22 22
11 9 G 11. B 2 2A 5 F 24 3
8 10 G 2 2 F 2 25
3 21 G 5 F 24 2 2(2) 3 2 2(25)
8 2 2(3) 10 2 2(21) G 5 F 28 13
2 12G 12. The sum 3A 1 C is not possible. You cannot add a 2 3 2
matrix to a 2 3 3 matrix.
13. 2 } 3 C 5 F 2 }
3 (26) 2 }
3 (22) 2 }
3 (9)
2 } 3 (1) 2 }
3 (24) 2 }
3 (21)G
5 F24 2 4 } 3 6
2 } 3 2 8 } 3 2 2 } 3 G
14. e 5 pounds of Empire apples
r 5 pounds of Red Delicious apples
g 5 pounds of Golden Delicious apples
Equation 1: e 1 r 1 g 5 21
Equation 2: 1.4e 1 1.1r 1 1.3g 5 25
Equation 3: r 5 2(g 1 e), or 22e 1 r 2 2g 5 0
e 1 r 1 g 5 21 3 (21.4)
21.4e 2 1.4r 2 1.4g 5 229.4
1.4e 1 1.1r 1 1.3g 5 25
1.4e 1 1.1r 1 1.3g 5 25
20.3r 2 0.1g 5 24.4
e 1 r 1 g 5 21 3 2 2e 1 2r 1 2g 5 42
22e 1 r 2 2g 5 0 22e 1 r 2 2g 5 0
3r 5 42
r 5 14
20.3(14) 2 0.1g 5 24.4 → g 5 2
e 1 14 1 2 5 21 → e 5 5
You should buy 5 pounds of Empire, 14 pounds of Red Delicious, and 2 pounds of Golden Delicious apples.
Graphing Calculator Activity 3.5 (p. 194)
1. F 19 25
8 8G 2. F 32.24 17.68 23.12
22.08 14.56 28.32G 3. F 26 11 25
21 11 7
18 20 28G 4. F 22 225
38 239
211 12G
5. R M S C Hardcover
Paperback
F 44 36 38 21
76 44 22 50 GLesson 3.6
3.6 Guided Practice (pp. 195–198)
1. AB is defi ned and has dimensions 5 3 2.
2. AB is not defi ned because the number of columns in A does not equal the number of rows in B.
3. AB 5 F 23 3
1 22 GF 1 5
23 22 G 5 F 23(1) 1 3(23) 23(5) 1 3(22)
1(1) 1 (22)(23) 1(5) 1 (22)(22)G
5 F 212 221
7 9 G 4. A(B 2 C) 5 F 21 2
23 0
4 1G (F 3 2
22 21G 2 F 24 5
1 0G) 5 F 21 2
23 0
4 1G F 7 23
23 21G 5 F 21(7) 1 2(23) 21(23) 1 2(21)
23(7) 1 0(23) 23(23) 1 0(21)
4(7) 1 1(23) 4(23) 1 1(21)G
5 F 213 1
221 9
25 213G
Chapter 3, continued
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156Algebra 2Worked-Out Solution Key
5. AB 2 AC 5 (F 21 2
23 0
4 1G F 3 2
22 21 G) 2 (F 21 2
23 0
4 1G F 24 5
1 0 G) 5 F 21(3) 1 2(22) 21(2) 1 2(21)
23(3) 1 0(22) 23(2) 1 0(21)
4(3) 1 1(22) 14(2) 1 1(21)G
2 F 21(24) 1 2(1) 21(5) 1 2(0)
23(24) 1 0(1) 23(5) 1 0(0)
4(24) 1 1(1) 4(5) 1 1(0)G
5 F 27 24
29 26
10 7G 2 F 6 25
12 215
215 20G
5 F 213 1
221 9
25 213G
6. 2 1 } 2 (AB) 5 2
1 } 2 F 27 24
29 26
10 7G
5 F 2 1 } 2 (27) 2 1 } 2 (24)
2 1 } 2 (29) 2 1 } 2 (26)
2 1 } 2 (10) 2 1 } 2 (7)G 5 F 7
} 2 2
9 } 2 3
25 2 7 } 2 G
7. F 14 30 18
16 25 20 G F 75
1
45G
5 F 14(75) 1 30(1) 1 18(45)
16(75) 1 25(1) 1 20(45)G 5 F 1890
2125 G The total cost of equipment for the women’s team is
$1890 and the total cost of equipment for the men’s team is $2125.
3.6 Exercises (pp. 199–202)
Skill Practice
1. The product of matrices A and B is defi ned provided the number of columns in A is equal to the number of rowsin B.
2. Sample answer: To fi nd the element in the fi rst row and fi rst column of AB, multiply each element in the fi rst row of A by the corresponding element in the fi rst column of B, then add the products.
3. AB is defi ned and has dimensions 2 3 2.
4. AB is defi ned and has dimensions 3 3 2.
5. AB is not defi ned. The number of columns in A does not equal the number of rows in B.
6. AB is defi ned and has dimensions 1 3 3.
7. AB is not defi ned. The number of columns in A does not equal the number of rows in B.
8. AB is defi ned and has dimensions 2 3 5.
9. A; Matrix A has 2 rows and matrix B has 2 columns, so AB has dimensions 2 3 2.
10. F 3 21G F 5
7 G 5 F 3(5) 1 (21)(7) G 5 F 8 G
11. F 1
4 G F 22 1 G 5 F 1(22) 1(1)
4(22) 4(1) G 5 F 22 1
28 4 G 12. Not defi ned; The number of columns in the fi rst matrix
does not equal the number of rows in the second matrix.
13.
F 9 23
0 2 G F 0 1
4 22 G 5 F 9(0) 1 (23)(4) 9(1) 1 (23)(22)
0(0) 1 2(4) 0(1) 1 2(22)G
5 F 212 15
8 24 G 14. F 5 0
24 1 G F 23 2
6 2 G 5 F 5(23) 1 0(6) 5(2) 1 0(2)
24(23) 1 1(6) 24(2) 1 1(2)G
5 F 215 10
18 26 G 15. F 5 2
0 24
1 6G F 3 7
22 0 G 5 F 5(3) 1 2(22) 5(7) 1 2(0)
0(3) 1 (24)(22) 0(7) 1 (24)(0)
1(3) 1 6(22) 1(7) 1 6(0)G
5 F 11 35
8 0
29 7G
16. Not defi ned; The number of columns in the fi rst matrix does not equal the number of rows in the second matrix.
17. F 1 3 0
2 12 24 G F 9 1
4 23
22 4G
5 F 1(9) 1 3(4) 1 0(22) 1(1) 1 3(23) 1 0(4)
2(9) 1 12(4) 1 (24)(22) 2(1) 1 12(23) 1 (24)(4)G
5F 21 28
74 250G
Chapter 3, continued
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157Algebra 2
Worked-Out Solution Key
18. F 2 5
21 4
3 27G F 0 1 5
23 10 24 G 5 F 2(0) 1 5(23) 2(1) 1 5(10) 2(5) 1 5(24)
21(0) 1 4(23) 21(1) 1 4(10) 21(5) 1 4(24)
3(0) 1 (27)(23) 3(1) 1 (27)(10) 3(5) 1 (27)(24)G
5 F 215 52 210
212 39 221
21 267 43G
19. The multiplication should be row 1 of the left matrix by column 1 of the right matrix.
3(7) 1 (21)(1) 5 20
20. The multiplication should be row 1 of the left matrix by column 1 of the right matrix.
2(4) 1 5(3) 5 23
21. B;
F 1 24
3 22 G F 4 21
0 23 G 5 F 1(4) 1 (24)(0) 1(21) 1 (24)(23)
3(4) 1 (22)(0) 3(21) 1 (22)(23)G
5 F 4 11
12 3G22. 3AB 5 3F 5 23
22 4GF 0 1
4 22G
5 F 15 29
26 12GF 0 1
4 22G
5 F 15(0) 1 (29)(4) 15(1) 1 (29)(22)
26(0) 1 12(4) 26(1) 1 12(22)G
5 F 236 33
48 230G
23. 2 1 } 2 AC 5 2
1 } 2 F 5 23
22 4G F 26 3
4 1G 5 2
1 } 2 F 5(26) 1 (23)(4) 5(3) 1 (23)(1)
22(26) 1 4(4) 22(3) 1 4(1)G
5 F 2 1 } 2 (242) 2 1 } 2 (12)
2 1 } 2 (28) 2
1 } 2 (22)
G 5 F 21 26
214 1G
24. AB 1 AC 5 F 5 23
22 4G F 0 1
4 22 G 1 F 5 23
22 4G F 26 3
4 1G 5 F 5(0) 1 (23)(4) 5(1) 1 (23)(22)
22(0) 1 4(4) 22(1) 1 4(22)G
1 F 5(26) 1 (23)(4) 5(3) 1 (23)(1)
22(26) 1 4(4) 22(3) 1 4(1)G
5 F 212 11
16 210 G 1 F 242 12
28 22 G 5 F 254 23
44 212G 25. AB 2 BA 5 F 5 23
22 4G F 0 1
4 22 G 2 F 0 1
4 22 G F 5 23
22 4G 5 F 5(0) 1 (23)(4) 5(1) 1 (23)(22)
22(0) 1 4(4) 22(1) 1 4(22)G
2 F 0(5) 1 1(22) 0(23) 1 1(4)
4(5) 1 (22)(22) 4(23) 1 (22)(4)G
5 F 212 11
16 210G 2 F 22 4
24 220G 5 F 210 7
28 10G 26. E(D 1 E)
5 F 23 1 4
7 0 22
3 4 21G (F 1 3 2
23 1 4
2 1 22G 1 F 23 1 4
7 0 22
3 4 21G)
5 F 23 1 4
7 0 22
3 4 21GF 22 4 6
4 1 2
5 5 23G
5 F 23(22) 1 1(4) 1 4(5) 23(4) 1 1(1) 1 4(5)
7(22) 1 0(4) 1 (22)(5) 7(4) 1 0(1) 1 (22)(5)
3(22) 1 4(4) 1 (21)(5) 3(4) 1 4(1) 1 (21)(5)
23(6) 1 1(2) 1 4(23)
7(6) 1 0(2) 1 (22)(23)
3(6) 1 4(2) 1 (21)(23)G
5 F 30 9 228
224 18 48
5 11 29 G
Chapter 3, continued
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158Algebra 2Worked-Out Solution Key
27. (D 1 E)D
5 (F 1 3 2
23 1 4
2 1 22G 1 F 23 1 4
7 0 22
3 4 21G) F 1 3 2
23 1 4
2 1 22G
5 F 22 4 6
4 1 2
5 5 23GF 1 3 2
23 1 4
2 1 22G
5 F 22(1) 1 4(23) 1 6(2) 22(3) 1 4(1) 1 6(1)
4(1) 1 1(23) 1 2(2) 4(3) 1 1(1) 1 2(1)
5(1) 1 5(23) 1 (23)(2) 5(3) 1 5(1) 1 (23)(1)
22(2) 1 4(4) 1 6(22)
4(2) 1 1(4) 1 2(22)
5(2) 1 5(4) 1 (23)(22)G
5 F 22 4 0
5 15 8
216 17 36 G
28. 22(BC) 5 22 F 0 1
4 22G F 26 3
4 1 G 5 22 F 0(26) 1 1(4) 0(3) 1 1(1)
4(26) 1 (22)(4) 4(3) 1 (22)(1)G
5 F 22(4) 22(1)
22(232) 22(10)G 5 F 28 22
64 220G29. 4AC 1 3AB 5 4F 5 23
22 4G F 26 3
4 1 G 1 3F 5 23
22 4G F 0 1
4 22G 5 4F 5(26) 1 (23)(4) 5(3) 1 (23)(1)
(22)(26) 1 4(4) (22)(3) 1 4(1)G
1 3F 5(0) 1 (23)(4) 5(1) 1 (23)(22)
(22)(0) 1 4(4) (22)(1) 1 4(22)G
5 F 4(242) 4(12)
4(28) 4(22)G
1 F 3(212) 3(11)
3(16) 3(210)G
5 F 2168 2 36 48 1 33
112 1 48 28 1 (230)G
5 F 2204 81
160 238G
30. 3(1) 1 2(x) 1 4(3) 5 19
2x 5 4
x 5 2
0(1) 1 (22)(x) 1 4(3) 5 y
0(1) 1 (22)(2) 1 4(3) 5 y
8 5 y
The solution is x 5 2 and y 5 8.
31. 22(9) 1 x(2) 1 1(21) 5 213
2x 5 6
x 5 3
4(9) 1 1(2) 1 3(21) 5 y
35 5 y
The solution is x 5 3 and y 5 35.
32. A2 5 F 1 21
0 2G F 1 21
0 2G 5 F 1(1) 1 (21)(0) 1(21) 1 (21)(2)
0(1) 1 2(0) 0(21) 1 2(2)G
5 F 1 23
0 4G A3 5 F 1 21
0 2G F 1 21
0 2G F 1 21
0 2G 5 F 1 23
0 4G F 1 21
0 2G 5 F 1(1) 1 (23)(0) 1(21) 1 (23)(2)
0(1) 1 4(0) 0(21) 1 4(2)G
5 F 1 27
0 8G 33. A2 5 F 24 1
2 21G F 24 1
2 21G 5 F (24)(24) 1 1(2) (24)(1) 1 1(21)
2(24) 1 (21)(2) 2(1) 1 (21)(21)G
5 F 18 25
210 3G A3 5 F 24 1
2 21G F 24 1
2 21G F 24 1
2 21G 5 F 18 25
210 3G F 24 1
2 21G 5 F 18(24) 1 (25)(2) 18(1) 1 (25)(21)
210(24) 1 3(2) 210(1) 1 3(21)G
5 F 282 23
46 213G
Chapter 3, continued
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159Algebra 2
Worked-Out Solution Key
34. A2 5 F 2 0 21
1 3 2
22 21 0G F 2 0 21
1 3 2
22 21 0G
5 F 6 1 22
1 7 5
25 23 0G
A3 5 F 2 0 21
1 3 2
22 21 0G F 2 0 21
1 3 2
22 21 0G F 2 0 21
1 3 2
22 21 0G
5 F 6 1 22
1 7 5
25 23 0GF 2 0 21
1 3 2
22 21 0G
5 F 17 5 24
21 16 13
213 29 21G
35. Sample answer:
A 5 F 2 6
4 9G, B 5 F 1 0
0 1G36. k(AB) 5 k(F a b
c dGF e f
g hG)
5 kF ae 1 bg af 1 bh
ce 1 dg cf 1 dhG 5 F kae 1 kbg kaf 1 kbh
kce 1 kdg kcf 1 kdhG (kA)B 5 (kF a b
c dG) F e f
g hG
5 F ka kb
kc kdG F e f g hG
5 F kae 1 kbg kaf 1 kbh
kce 1 kdg kcf 1 kdhG A(kB) 5 F a b
c dG (kF e f g hG)
5 F a b c dG F ke kf
kg khG 5 F kae 1 kbg kaf 1 kbh
kce 1 kdg kcf 1 kdhG All of the matrices are equal, so k(AB) 5 (kA)B 5 A(kB).
Problem Solving
37. Equipment Cost (dollars)
Bats Balls Uniforms Bats
Balls
Uniforms
F 21
4
30G F 12 45 15 G
Total cost 5 F 12 45 15 G F 21
4
30G
5 F 12(21) 1 45(4) 1 15(30) G 5 F 882 G The total cost of equipment for the softball team is $882.
38. Art Supplies
Paint Brushes Canvases
Class 1
Class 2
F 24 12 17
20 14 15G Cost (dollars)
Paint
Brushes
Canvases
F 3.35
1.75
4.50G
Total cost 5 F 24 12 17
20 14 15G F 3.35
1.75
4.50G
5 F 24(3.35) 1 12(1.75) 1 17(4.50) 20(3.35) 1 14(1.75) 1 15(4.50)G 5 F 177.9
159.00G The supplies for class 1 cost $177.90 and the supplies for
class 2 cost $159.00.
39. Attendance
Students Adults Seniors
Friday
Saturday
F 120 150 40
192 215 54G Ticket cost (dollars)
Students
Adults
Seniors
F 2
5
4G
F 120 150 40
192 215 54G F 2
5
4G
5 F 120(2) 1 150(5) 1 40(4) 192(2) 1 215(5) 1 54(4)G 5 F 1150
1675G The income from Friday night’s play was $1150 and the
income from Saturday night’s play was $1675.
Chapter 3 continued
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160Algebra 2Worked-Out Solution Key
40. Medals Won
Gold Silver Bronze
USA
China
Russia
F 35 39 29
32 17 14
27 27 38G
Points
Gold
Silver
Bronze
F 3
2
1G
F 35 39 29
32 17 14
27 27 38G F 3
2
1G
5 F 35(3) 1 39(2) 1 29(1)
32(3) 1 17(2) 1 14(1)
27(3) 1 27(2) 1 38(1)G 5 F 212
144
173G
USA scored 212 points, China scored 144 points, and Russia scored 173 points.
41. SP: (3 3 2)(1 3 3) PS: (1 3 3)(3 3 2) not equal equal
The matrix PS is defi ned.
PS 5 F 650 825 1050G F 21 16
40 33
15 19G 5
F 650(21)1825(40)11050(15) 650(16)1825(33)11050(19)G 5 F 62,400 57,575 G This matrix shows that dealer A made a profi t of $62,400
and dealer B made a profi t of $57,575.
42. Grades (G)
Homework Quizzes Tests Jean
Ted
Pat
Al
Matt
F 82 88 86
92 88 90
82 73 81
74 75 78
88 92 90
G Weights (W)
Homework
Quizzes
Tests
F 0.2
0.3
0.5G
GW 5 F 82(0.2) 1 88(0.3) 1 86(0.5)
92(0.2) 1 88(0.3) 1 90(0.5)
82(0.2) 1 73(0.3) 1 81(0.5)
74(0.2) 1 75(0.3) 1 78(0.5)
88(0.2) 1 92(0.3) 1 90(0.5)
G 5 F 85.8
89.8
78.8
76.3
90.2
G The students received the following overall grades:
Jean, 85.8; Ted, 89.8; Pat, 78.8; Al, 76.3; and Matt, 90.2.
43. a. T 5 F 1 2 p q
p 1 2 qG 5 F 1 2 0.2 0.05
0.2 1 2 0.05G 5 F 0.8 0.05
0.2 0.95G b. M1 5 F 0.8 0.05
0.2 0.95G F 5000 8000G
5 F 0.8(5000) 1 0.05(8000) 0.2(5000) 1 0.95(8000)G 5 F 4400
8600G This matrix represents the number of commuters who
will drive and use public transportation after one year.
c. M2 5 F 0.8 0.05
0.2 0.95G F 4400 8600G
5 F 0.8(4400) 1 0.05(8600) 0.2(4400) 1 0.95(8600)G 5 F 3950
9050G M3 5 F 0.8 0.05
0.2 0.95G F 3950 9050G
5 F 0.8(3950) 1 0.05(9050) 0.2(3950) 1 0.95(9050)G 5 F 3612.5
9387.5G M4 5 F 0.8 0.05
0.2 0.95G F 3612.5 9387.5G
5 F 0.8(3612.5) 1 0.05(9387.5) 0.2(3612.5) 1 0.95(9387.5)G 5 F 3359.375
9640.625G M2, M3, and M4 represent the number of commuters
after 2, 3, and 4 years, respectively.
44. a. Cost (C) Price (P)
Plain
Class year
School name
Mascot
F 10
15
20
20
G Plain
Class year
School name
Mascot
F 15
20
25
30
G b. Sales (S)
Plain Class year School name Mascot
Year 1
Year 2
Year 3
F 0 20 100 0
10 100 50 30
20 300 100 50G
Chapter 3, continued
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161Algebra 2
Worked-Out Solution Key
c. SC 5 F 0 20 100 0
10 100 50 30
20 300 100 50G F 10
15
20
20
G 5 F 0(10) 1 20(15) 1 100(20) 1 0(20)
10(10) 1 100(15) 1 50(20) 1 30(20)
20(10) 1 300(15) 1 100(20) 1 50(20)G
5 F 2300
3200
7700G
SC represents the cost of making the scarves each year.
SP 5 F 0 20 100 0
10 100 50 30
20 300 100 50G F 15
20
25
30
G 5 F 0(15) 1 20(20) 1 100(25) 1 0(30)
10(15) 1 100(20) 1 50(25) 1 30(30)
20(15) 1 300(20) 1 100(25) 1 50(30)G
5 F 2900
4300
10,300G
SP represents the total price received for the scarveseach year.
d. SP 2 SC 5 F 2900
4300
10,300G 2 F 2300
3200
7700G
5 F 2900 2 2300
4300 2 3200
10,300 2 7700G 5 F 600
1100
2600G
This matrix represents the profi t that is made each year.
45. a. AB 5 F 0 21
1 0G F 27 24 24
4 8 2G 5 F 24 28 22
27 24 24G
1
x
y
21
(24, 27)
(22, 24)(28, 24)
b. For the triangle rotated 1808:
F 0 21
1 0G F 0 21
1 0G F 27 24 24
4 8 2G 5 F 21 0
0 21G F 27 24 24
4 8 2G 5 F 7 4 4
24 28 22G The vertices of this triangle are (7, 24), (4, 28),
and (4, 22).
For the triangle rotated 2708:
F 0 21
1 0G F 0 21
1 0G F 0 21
1 0G F 27 24 24
4 8 2G 5 F 21 0
0 21G F 0 21
1 0GF 27 24 24
4 8 2G 5 F 0 21
1 0GF 27 24 24
4 8 2G 5 F 4 8 2
7 4 4G The vertices of this triangle are (4, 7), (8, 4), and (2, 4).
Mixed Review
46.
1
x
y
21
47. 1
x
y
21
48.
1
x
y
21
49. 1
x
y
21
50. 1
x
y
21
51.
1
x
y
21
52. y 2 (24) 5 2(x 2 0) 53. y 2 2 5 23(x 2 5)
y 1 4 5 2x y 2 2 5 23x 1 15
y 5 2x 2 4 y 5 23x 1 17
Chapter 3, continued
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162Algebra 2Worked-Out Solution Key
54. y 2 (21) 5 2 2 } 3 (x 2 0) 55. y 2 3 5
3 } 4 (x 2 0)
y 1 1 5 2 2 } 3 x y 2 3 5
3 } 4 x
y 5 2 2 } 3 x 2 1 y 5
3 } 4 x 1 3
56. m 5 2 2 8
} 1 2 4 5 2
y 2 2 5 2(x 2 1)
y 2 2 5 2x 2 2
y 5 2x
57. m 5 1 2 8
} 0 2 (28)
5 2 7 } 8
y 2 1 5 2 7 } 8 (x 2 0)
y 2 1 5 2 7 } 8 x
y 5 2 7 } 8 x 1 1
58. 3x 1 2y 5 5
2x 1 3y 5 13 → x 5 3y 2 13
3(3y 2 13) 1 2y 5 5 → y 5 4
2x 1 3(4) 5 13 → x 5 21
The solution is (21, 4).
59. 3x 2 5y 5 11
2x 1 5y 5 24
5x 5 35
x 5 7
3(7) 2 5y 5 11 → y 5 2
The solution is (7, 2).
60. 3x 2 y 5 4 3 3 9x 2 3y 5 12
22x 1 3y 5 226 22x 1 3y 5 226
7x 5 214
x 5 22
3(22) 2 y 5 4 → y 5 210
The solution is (22, 210).
61. 4x 2 3y 5 17 4x 2 3y 5 17
2x 1 5y 5 15 3 (22) 24x 2 10y 5 230
213y 5 213
y 5 1
4x 2 3(1) 5 17 → x 5 5
The solution is (5, 1).
62. 4x 2 2y 5 14 4x 2 2y 5 14
22x 1 y 5 27 3 2 24x 1 2y 5 214
0 5 0
Infi nitely many solutions.
63. x 1 4y 5 4 x 1 4y 5 4
3x 2 2y 5 19 3 2 6x 2 4y 5 38
7x 5 42
x 5 6
6 1 4y 5 4 → y 5 2 1 } 2
The solution is 1 6, 2 1 } 2 2 .
Lesson 3.7
3.7 Guided Practice (pp. 204–207)
1. 3 2 6 1 5 3(1) 2 6(22) 5 15
2. 4 21 2 23 22 21
0 5 1 4 21
23 22
0 5
5 (28 1 0 2 30) 2 (0 2 20 1 3)
5 238 2 (217) 5 221
3. 10 22 3 2 212 4
0 27 22 10 22
2 212
0 27
5 (240 1 0 2 42) 2 (0 2 280 1 8)
5 198 1 272 5 470
4. Area 5 6 1 } 2
5 11 1 9 2 1
1 3 1 5 11
9 2
1 3
5 6 1 } 2 [(10 1 11 1 27) 2 (2 1 15 1 99)]
5 6 1 } 2 (268) 5 34
The area of the triangle is 34 square units.
5. 3 24 2 5 5 15 2 (28) 5 23
x 5
215 24 13 5
} 23
5 275 2 (252)
}} 23 5 21
y 5
3 215 2 13
} 23
5 39 2 (230)
} 23 5 3
The solution is (21, 3).
6. 4 7 23 22 5 28 2 (221) 5 13
x 5
2 7 28 22
} 13
5 24 2 (256)
} 13 5 4
y 5
4 2 23 28
} 13
5 232 2 (26)
} 13 5 22
The solution is (4, 22).
Chapter 3, continued
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163Algebra 2
Worked-Out Solution Key
7. 3 24 2 4 1 25
2 23 1 3 24
4 1
2 23
5 (3 1 40 1 (224)) 2 (4 1 45 1 (216)) 5 214
x 5
18 24 2
213 1 25
11 23 1 18 24
213 1
11 23
}}} 214
5 (18 1 220 1 78) 2 (22 1 270 1 52)
}}} 214 5 2
y 5
3 18 2 4 213 25
2 11 1 3 18
4 213
2 11
}} 214
5 (239 2 180 1 88) 2 (252 2 165 1 72)
}}}} 214 5 21
z 5
3 24 18 4 1 213
2 23 11 3 24
4 1
2 23
}} 214
5 (33 1 104 2 216) 2 (36 1 117 2 176)
}}} 214 5 4
The solution is (2, 21, 4).
3.7 Exercises (pp. 207–209)
Skill Practice
1. The determinant of a 2 3 2 matrix is the difference of the products of the elements of the diagonals.
2. Cramer’s rule is used to solve a system of linear equations. The numerators for x and y are the determinants of the matrices formed by replacing the coeffi cients of x and y, respectively, with the column of constants. The denominator of x and y is the determinant of the coeffi cient matrix.
3. 2 21 4 25 5 2(25) 2 4(21) 5 26
4. 7 1 0 3 5 7(3) 2 0(1) 5 21
5. 24 3 1 27 5 24(27) 2 1(3) 5 25
6. 1 23 2 6 5 1(6) 2 2(23) 5 12
7. 10 26 27 5 5 10(5) 2 (26)(27) 5 8
8. 0 3 5 23 5 0(23) 2 5(3) 5 215
9. 9 23 7 2 5 9(2) 2 7(23) 5 39
10. 25 12 4 6 5 25(6) 2 4(12) 5 278
11. 21 12 4 0 2 25
3 0 15 21 12 4 0 2 25
3 0 1 21 12
0 2
3 0
5 (22 2 180 1 0) 2 (24 1 0 1 0)
5 2206
12. 1 2 3 5 28 1
2 4 35 1 2 3 5 28 1
2 4 3 1 2
5 28
2 4
5 (224 1 4 1 60) 2 (248 1 4 1 30)
5 54
13. 5 0 2 23 9 22
1 24 0 5 5 0 2 23 9 22
1 24 0 5 0
23 9
1 24
5 (0 1 0 1 24) 2 (18 1 40 1 0)
5 234
14. 27 4 5 1 2 24
210 1 65 27 4 5 1 2 24
210 1 6 27 4
1 2
210 1
5 (284 1 160 1 5) 2 (2100 1 28 1 24)
5 129
15. 12 5 8 0 6 28
1 10 45 12 5 8 0 6 28
1 10 4 12 5
0 6
1 10
5 (288 2 40 1 0) 2 (48 2 960 1 0)
5 1160
16. 24 3 29 12 6 0
8 212 0 5 24 3 29 12 6 0
8 212 0 24 3
12 6
8 212
5 (0 1 0 1 1296) 2 (2432 1 0 1 0)
5 1728
17. 22 6 0 8 15 3
4 21 7 5 22 6 0 8 15 3
4 21 7 22 6
8 15
4 21
5 (2210 1 72 1 0) 2 (0 1 6 1 336)
5 2480
18. 5 7 6
24 0 8
1 8 7 5 5 7 6
24 0 8
1 8 7 5 7
24 0
1 8
5 (0 1 56 2 192) 2 (0 1 320 2 196)
5 2260
Chapter 3, continued
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164Algebra 2Worked-Out Solution Key
19. The values in the upward diagonals should be subtracted from the values in the downward diagonals.
2 0 21 4 1 6
23 2 5 2 0
4 1
23 2
5 (10 1 0 2 8) 2 (3 1 24 1 0)
5 2 2 27 5 225
The determinant is 225.
20. The column extensions should be on the right sideof the matrix.
3 0 1 2 2 23
23 5 0 3 0
2 2
25 5
5 (0 1 0 1 10) 2 (26 2 45 1 0)
5 10 2 (251) 5 61
21. D;
det A 5 212 2 6 5 218
det B 5 8 2 18 5 210
det C 5 25 2 (221) 5 16
det D 5 25 2 (22) 5 27
The greatest determinant is det D.
22. Area 5 6 1 } 2
1 5 1 4 6 1
7 3 1 1 5
4 6
7 3
5 6 1 } 2 [(6 1 35 1 12) 2 (42 1 3 1 20)]
5 6
The area of the triangle is 6 square units.
23. Area 5 6 1 } 2
4 2 1 4 8 1
8 5 1 4 2
4 8
8 5
5 6 1 } 2 [(32 1 16 1 20) 2 (64 1 20 1 8)]
5 12
The area of the triangle is 12 square units.
24. Area 5 6 1 } 2
24 6 1 0 3 1
6 6 1 24 6
0 3
6 6
5 6 1 } 2 [(212 1 36 1 0) 2 (18 2 24 1 0)]
5 15
The area of the triangle is 15 square units.
25. Area 5 6 1 } 2
24 24 1 21 2 1
2 26 1 24 24
21 2
2 26
5 6 1 } 2 [(28 2 8 1 6) 2 (4 1 24 1 4)]
5 21
The area of the triangle is 21 square units.
26. Area 5 6 1 } 2
5 24 1 6 3 1
8 21 1 5 24
6 3
8 21
5 6 1 } 2 [(15 2 32 2 6) 2 (24 2 5 2 24)]
5 9
The area of the triangle is 9 square units.
27. Area 5 6 1 } 2
26 1 1 22 26 1
0 3 1 26 1
22 26
0 3
5 6 1 } 2 [(36 1 0 2 6) 2 (0 2 18 2 2)]
5 25
The area of the triangle is 25 square units.
28. A;
Area 5 6 1 } 2
23 4 1 6 3 1
2 21 1 23 4
6 3
2 21
5 6 1 } 2 [(29 1 8 2 6) 2 (6 1 3 1 24)]
5 20
The area of the triangle is 20 square units.
29. 3 5
21 2 5 6 2 (25) 5 11
x 5
3 5
10 2 }
11 5
6 2 50 } 11 5 24
y 5
3 3 21 10
} 11
5 30 2 (23)
} 11 5 3
The solution is (24, 3).
30. 2 21
1 2 5 4 2 (21) 5 5
x 5
22 21
14 2 }}
5 5
24 2 (214) } 5 5 2
y 5
2 22
1 14 }
5 5
28 2 (22) } 5 5 6
The solution is (2, 6).
31. 5 1
2 25 5 225 2 2 5 227
x 5
240 1
11 25 }}
227 5
200 2 11 }
227 5 27
y 5
5 240
2 11 }}
227 5
55 2 (280) }
227 5 25
The solution is (27, 25).
Chapter 3, continued
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165Algebra 2
Worked-Out Solution Key
32. 21 1 1 4 21 4
1 2 21 21 1
4 21
1 2
5 (21 1 4 1 8) 2 (21 2 8 2 4) 5 24
x 5
23 1 1 214 21 4
9 2 21 23 1
214 21
9 2
}}} 24
5 (23 1 36 2 28) 2 (29 2 24 1 14)
}}} 24 5 1
y 5
21 23 1 4 214 4
1 9 21 21 23
24 214
1 9
}}} 24
5 (214 2 12 1 36) 2 (214 2 36 1 12)
}}} 24 5 2
z 5
21 1 23 4 21 214
1 2 9 21 1
24 21
1 2
}}} 24
5 (9 2 14 2 24) 2 (3 1 28 1 36)
}}} 24 5 24
The solution is (1, 2, 24).
33. 21 22 4 1 1 2
2 1 23 21 22
1 1
2 1
5 (3 2 8 1 4) 2 (8 2 2 1 6) 5 213
x 5
228 22 4
211 1 2
30 1 23 228 22
211 1
30 1 }}} 213
5 (84 2 120 2 44) 2 (120 2 56 2 66)
}}} 213 5 6
y 5
21 228 4
1 211 2
2 30 23 21 228
1 211
2 30 }}} 13
5 (233 2 112 1 120) 2 (288 2 60 1 84)
}}}} 213 5 23
z 5
21 22 228 1 1 211
2 1 30 21 22
1 1
2 1 }}} 213
5 (230 1 44 2 28) 2 (256 1 11 2 60)
}}} 213 5 27
The solution is (6, 23, 27).
34. 4 1 3 2 25 4
1 21 2 4 1
2 25
1 21
5 (240 1 4 2 6) 2 (215 2 16 1 4) 5 215
x 5
7 1 3
219 25 4
22 21 2 7 1
219 25
22 21
}}} 215
5 (270 2 8 1 57) 2 (30 2 28 2 38)
}}} 215 5 21
y 5
4 7 3
2 219 4
1 22 2 4 7
2 219
1 22
}}} 215
5 (2152 1 28 2 12) 2 (257 2 32 1 28)
}}} 215 5 5
z 5
4 1 7
2 25 219
1 21 22 4 1
2 25
1 21
}} 215
5 (40 2 19 2 14) 2 (235 1 76 2 4)
}}} 215 5 2
The solution is (21, 5, 2).
35. 5 21 22
1 3 4
2 24 1 5 21
1 3
2 24
5 (15 2 8 1 8) 2 (212 2 80 2 1) 5 108
x 5
26 21 22
16 3 4
215 24 1 26 21
16 3
215 24
}}} 108
5 (218 1 60 1 128) 2 (90 1 96 2 16)
}}} 108 5 0
y 5
5 26 22
1 16 4
2 215 1 5 26
1 16
2 215
}}} 108
5 (80 2 48 1 30) 2 (264 2 300 2 6)
}}} 108 5 4
z 5
5 21 26
1 3 16
2 24 215 5 21
1 3
2 24
}}} 108
5 (2225 2 32 1 24) 2 (236 2 320 1 15)
}}}} 108 5 1
The solution is (0, 4, 1).
Chapter 3, continued
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166Algebra 2Worked-Out Solution Key
36. 1 1 1
3 23 2
21 2 22 1 1
3 23
21 2
5 (6 2 2 1 6) 2 (3 1 4 2 6) 5 9
x 5
28 1 1
221 23 2
11 2 22 28 1
221 23
11 2 }} 9
5 (248 1 22 2 42) 2 (233 2 32 1 42)
}}} 9 5 25
y 5
1 28 1
3 221 2
21 11 22 1 28
3 221
21 11 }}} 9
5 (42 1 16 1 33) 2 (21 1 22 1 48)
}}} 9 5 0
z 5
1 1 28
3 23 221
21 2 11 1 1
3 23
21 2 }}} 9
5 (233 1 21 2 48) 2 (224 2 42 1 33)
}}} 9 5 23
The solution is (25, 0, 23).
37. 3 21 1
21 2 23
1 1 1 3 21
21 2
1 1
5 (6 1 3 2 1) 2 (2 2 9 1 1) 5 14
x 5
25 21 1
217 2 23
21 1 1 25 21
217 2
21 1 }} 14
5 (50 1 63 2 17) 2 (42 2 75 1 17)
}}} 14 5 8
y 5
3 25 1
21 217 23
1 21 1 3 25
21 217
1 21 }}} 14
5 (251 2 75 2 21) 2 (217 2 189 2 25)
}}} 14 5 6
z 5
3 21 25
21 2 217
1 1 21 3 21
21 2
1 1 }}} 14
5 (126 1 17 2 25) 2 (50 2 51 1 21)
}}} 14 5 7
The solution is (8, 6, 7).
38. Sample answer:
F 10 17 5 9G
39. a. AB 5 F 2 21 1 2GF 3 5
22 24G 5 F 2(3) 1 (21)(22) 2(5) 1 (21)(24)
1(3) 1 2(22) 1(5) 1 2(24)G 5 F 8 14
21 23G det AB 5 224 2 (214) 5 210
det A 5 4 2 (21) 5 5
det B 5 212 2 (210) 5 22
The determinant of AB is equal to the product of det A and det B.
b. det kA 5 2k 2k k 2k 5 4k2 2 (2k2) 5 5k2
The determinant of kA is equal to k2 times the determinant of A.
Problem Solving
40. Area 5 6 1 } 2
938 454 1 900 2518 1
0 0 1 938 454
900 2518
0 0
5 6 1 } 2 F (2485,884 1 0 1 0) 2 (0 1 0 1 408,600) G
5 447,242
The area of the Bermuda Triangle is 447,242 square miles.
41. Area 5 6 1 } 2
0 0 1 5 2 1
3 6 1 0 0
5 2
3 6
5 6 1 } 2 [(0 1 0 1 30) 2 (6 1 0 1 0)]
5 12
The area of the garden will be 12 square feet.
Chapter 3, continued
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167Algebra 2
Worked-Out Solution Key
42. f 5 number of fl oor seats
n 5 number of other seats
Equation 1: 40f 1 25n 5 185,500
Equation 2: f 1 n 5 6700
Method 1: Cramer’s rule
40 25 1 15 40 2 25 5 15
f 5
185,500 25 6700 1
}} 15
5 185,500(1) 2 6700(25)
}} 15 5 1200
n 5
40 185,500 1 6700
}} 15
5 40(6700) 2 1(185,500)
}} 15 5 5500
Method 2: Substitution
n 5 6700 2 f
40f 1 25(6700 2 f ) 5 185,500
15f 5 18,000
f 5 1200
n 5 6700 2 1200 5 5500
Method 3: Elimination
40f 1 25n 5 185,500 40f 1 25n 5 185,500
f 1 n 5 6700 3 (240) 240f 2 40n 5 2268,000
215n 5 282,500
n 5 5500
f 1 5500 5 6700 → f 5 1200
Each method used above produces the same outcome. 1200 fl oor seats and 5500 other seats were sold.
43. a. s 5 number of single scoop cones
d 5 number of double scoop cones
t 5 number of triple scoop cones
Equation 1: 0.90s 1 1.20d 1 1.60t 5 134
Equation 2: s 1 d 1 t 5 120
Equation 3: s 5 d 1 t, or s 2 d 2 t 5 0
0.90 1.20 0.60 1 1 1
1 21 21 0.90 1.20
1 1
1 21
5 (20.90 1 1.20 2 1.60) 2 (1.60 2 0.90 2 1.20)
5 20.80
s 5
134 1.20 1.60 120 1 1
0 21 21 134 1.20
120 1
0 21 }}} 20.80
5 2(134 1 0 2 192) 2 (0 2 134 2 144)
}}} 20.80 5 60
d 5
0.90 134 1.60 1 120 1
1 0 21 0.90 134
1 1
1 0
}}} 20.80
5 (2108 1 134 1 0) 2 (192 1 0 2 134)
}}} 20.80 5 40
t 5
0.90 1.20 134 1 1 120
1 21 0 0.90 1.20
1 1
1 21 }}} 20.80
5 (0 1 144 2 134) 2 (134 2 108 1 0)
}}} 20.80 5 20
The ice cream shop sells 60 single scoop, 40 double scoop, and 20 triple scoop cones.
b. New prices:
Single scoop: 0.90(1.10) 5 0.99
Double scoop: 1.20(1.10) 5 1.32
Triple scoop: 1.60(1.10) 5 1.76
Number of cones sold:
Single scoop: 60(0.95) 5 57
Double scoop: 40(0.95) 5 38
Triple scoop 20(0.95) 5 19
Total revenue 5 0.99(57) 1 1.32(38) 1 1.76(19)
5 140.03
The total revenue from the ice cream cones is $140.03.
44. F 1 Na 5 42
Na 1 Cl 5 58.5
5F 1 Cl 5 130.5
1 1 0 0 1 1
5 0 1 1 1
0 1
5 0
5 (1 1 5 1 0) 2 (0 1 0 1 0) 5 6
F 5
42 1 0 58.5 1 1
130.5 0 1 42 1
58.5 1
130.5 0 }} 6
5 (42 1 130.5 1 0) 2 (0 1 0 1 58.5)
}}} 6 5 19
Na 5
1 42 0 0 58.5 1
5 130.5 1 1 42
0 58.5
5 130.5 }} 6
5 (58.5 1 210 1 0) 2 (0 1 130.5 1 0)
}}} 6 5 23
Chapter 3, continued
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168Algebra 2Worked-Out Solution Key
Cl 5
1 1 42 0 1 58.5
5 0 130.5 1 1
0 1
5 0 }}} 6
5 (130.5 1 292.5 1 0) 2 (210 1 0 1 0)
}}} 6 5 35.5
The atomic weights of fl ourine, sodium, and chlorine are 19, 23, and 35.5, respectively.
45. a. Area 5 6 1 } 2
70 128 1 0 70 1
124 36 1 70 128
0 70
124 36
5 6 1 } 2 [(4900 1 15872 1 0) 2 (8680 1 2520 1 0)]
5 4786
The area of the top triangular region is 4786 mi2.
b. Area 5 6 1 } 2
0 70 1 67 0 1
124 36 1 0 70
67 0
124 36
5 6 1 } 2 [(0 1 8680 1 2412) 2 (0 1 0 1 4690)]
5 3201
The area of the bottom triangular region is 3201 mi2.
c. Total area 5 4786 1 3201 5 7987
The total of the Dinosaur Diamond is 7987 mi2.
d. You could connect Vernal, UT, to Moab, UT, to create a left and right triangle.
46. 6 1 } 2
x 120 1 100 50 1
0 0 1 x 120
100 50
0 0
5 5000
6 1 } 2 F (50x 1 0 1 0) 2 (0 1 0 1 12,000) G 5 5000
6 1 } 2 (50x 2 12,000) 5 5000
Two possibilities:
1. 1 }
2 (50x 2 12,000) 5 5000
25x 5 11,000
x 5 440
2. 2 1 } 2 (50x 2 12,000) 5 5000
225x 5 21000
x 5 40
The farmer could put the fi nal post at (40, 120) or(440, 120).
Mixed Review
47. f (8) 5 8 2 12 5 24 48. f (7) 5 4(7) 1 8 5 36
49. f (25) 5 (25)2 2 10 5 15
50. f (3) 5 2(3)2 1 2(3) 5 23
51. f (4) 5 2(4)2 2 4 1 5 5 215
52. f (22) 5 (22)2 2 2(22) 1 4 5 12
53.
1
x
y
21
54. 1
x
y
21
55.
8
x
y
4
56. 1
x
y
21
57. F 2 24 6 1GF 23 0
1 7G 5 F 2(23) 1 (24)(1) 2(0) 1 (24)(7)
6(23) 1 1(1) 6(0) 1 1(7)G 5 F 210 228
217 7G 58. F 26 28
2 24GF 0 5 7 1G
5 F 26(0) 1 (28)(7) 26(5) 1 (28)(1) 2(0) 1 (24)(7) 2(5) 1 (24)(1)G
5 F 256 238 228 6G
59. F 1 0 3 22GF 25 10
2 0 G 5 F 1(25) 1 0(2) 1(10) 1 0(0)
3(25) 1 (22)(2) 3(10) 1 (22)(0)G 5 F 25 10
219 30GLesson 3.8
3.8 Guided Practice (pp. 210–213)
1. 1 }
24 2 2 F 4 21
22 6G 5 1 } 22
F 4 21
22 6G 5 F 2
} 11
2 1 } 22
2 1 } 11 3 } 11 G
2. 1 }
28 2 (220) F 8 25
4 21G 5 1 } 12
F 8 25
4 21G 5 F 2 } 3 2 5 }
12
1 } 3 2 1 } 12
G
Chapter 3, continued
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169Algebra 2
Worked-Out Solution Key
3. 1 }
6 2 4 F 22 4
1 23G 5 1 } 2 F 22 4
1 23G 5 F 21 2
1 } 2 2 3 } 2 G4. 2
1 } 24 F 6 21
0 24G 5 F 2 1 } 4 1 } 24
0 1 } 6 G
F 2 1 } 4 1 } 24
0 1 } 6 GF 24 1
0 6GX 5 F 2 1 } 4 1 } 24
0 1 } 6 GF 8 9
24 6G F 1 0
0 1GX 5 F 22 1 1 2 9 } 4 1 1 } 4
0 1 4 0 1 1G
X 5 F 21 22
4 1G5. A 5 F 2 22 0
2 0 22
12 24 26G; A
21 5 F 21 21.5 0.5
21.5 21.5 0.5
21 22 0.5G
Check:
AA21 5 F 2 22 0
2 0 22
12 24 26GF 21 21.5 0.5
21.5 21.5 0.5
21 22 0.5G
5 F 22 1 3 1 0 23 1 3 1 0 1 2 1 1 0
22 1 0 1 2 23 1 0 1 4 1 1 0 2 1
212 1 6 1 6 218 1 6 1 12 6 2 2 2 3G
5 F 1 0 0
0 1 0
0 0 1G 5 I
A21A 5 F 21 21.5 0.5
21.5 21.5 0.5
21 22 0.5GF 2 22 0
2 0 22
12 24 26G
5 F 22 2 3 1 6 2 1 0 2 2 0 1 3 2 3
23 2 3 1 6 3 1 0 2 2 0 1 3 2 3
22 2 4 1 6 2 1 0 2 2 0 1 4 2 3G
5 F 1 0 0
0 1 0
0 0 1G 5 I
6. A 5 F 23 4 5
1 5 0
5 2 2G
A21 ø F 20.0654 20.0131 0.1634
0.01307 0.2026 20.0327
0.1503 20.1699 0.1242G
Check:
AA21 ø F 23 4 5
1 5 0
5 2 2GF 20.0654 20.0131 0.1634
0.01307 0.2026 20.0327
0.1503 20.1699 0.1242G
5 F 0.1962 1 0.05228 1 0.7515
20.0654 1 0.06535 1 0
20.327 1 0.02614 1 0.3006
0.0393 1 0.8104 2 0.8495
20.0131 1 1.0130 1 0
20.0655 1 0.4052 2 0.3398
20.4902 2 0.1308 1 0.621
0.1634 2 0.1635 1 0
0.817 2 0.0654 1 0.2482G
5 F 0.99998 2 3 1024 0
25 3 1025 0.9999 21 3 1024
22.6 3 1024 21 3 1024 0.9998G
ø F 1 0 0
0 1 0
0 0 1G 5 I
A21A 5 F 20.0654 20.0131 0.1634
0.01307 0.2026 20.0327
0.1503 20.1699 0.1242GF 23 4 5
1 5 0
5 2 2G
5 F 0.1962 2 0.0131 1 0.817
20.03921 1 0.2026 2 0.1635
20.4509 2 0.1699 1 0.621
20.2616 2 0.0655 1 0.3268
0.05228 1 1.013 2 0.0654
0.6012 2 0.8495 1 0.2484
20.327 1 0 1 0.3268
0.06535 1 0 2 0.0654
0.7515 1 0 1 0.2484G
5 F 1.0001 23 3 1024 22 3 1024
21.1 3 1024 0.99988 25 3 1025
2 3 1024 1 3 1024 0.9999G
ø F 1 0 0
0 1 0
0 0 1G 5 I
Chapter 3, continued
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170Algebra 2Worked-Out Solution Key
7. A 5 F 2 1 22
5 3 0
4 3 8G; A21 5 F 12 27 3
220 12 25
1.5 21 0.5G
Check:
AA21 5 F 2 1 22
5 3 0
4 3 8GF 12 27 3
220 12 25
1.5 21 0.5G
5F 24 2 20 2 3 214 1 12 1 2 6 2 5 2 1
60 2 60 1 0 235 1 36 1 0 15 2 15 1 0
48 2 60 1 12 228 1 36 2 8 12 2 15 1 4G
5 F 1 0 0
0 1 0
0 0 1G 5 I
A21A 5 F 12 27 3
220 12 25
1.5 21 0.5GF 2 1 22
5 3 0
4 3 8G
5 F 24 2 35 1 12 12 2 21 1 9
240 1 60 2 20 220 1 36 2 15
3 2 5 1 2 1.5 2 3 1 1.5
224 1 0 1 24
40 1 0 2 40
23 1 0 1 4G
5 F 1 0 0
0 1 0
0 0 1G 5 I
8. F 4 1 3 5GF x
y G 5 F 10 21G
A21 5 1 } 17 F 5 21
23 4G 5 F 5 } 17
2 1 } 17
2 3 } 17 4 } 17
G X 5 A21B 5 F 5 }
17 2 1 } 17
2 3 } 17 4 } 17
GF 10 21G 5 F 3
22G The solution of the system is (3, 22).
9. F 2 21 6 23GF x
y G 5 F 26 218G
A21 5 1 } 0 F 23 1
26 2G Because A 5 0,
1 }
A is undefi ned and A does not
have an inverse. The system has infi nitely many solutions.
10. F 3 21 24 2GF x
y G 5 F 25 8 G
A21 5 1 } 2 F 2 1
4 3G 5 F 1 1 } 2
2 3 } 2 G X 5 A21B 5 F 1 1 } 2
2 3 } 2 GF 25 8 G 5 F 21
2G The solution of the system is (21, 2).
11. F 2 1 0
2 2 1
4 3 2GF m
p
dG 5 F 17
35
69G
A21B 5 F 8
1
17G
A movie pass costs $8, a package of popcorn costs $1, and a DVD costs $17.
3.8 Exercises (pp. 214–217)
Skill Practice
1. Matrix of variables: F x y G
Matrix of constants:F 4 22G
2. First, you fi nd the determinant of A. Then, switch the elements on the downward diagonal and negate the elements on the upward diagonal. Finally, you
multiply 1 }
det A and the new matrix.
3. 1 }
4 2 5 F 4 5
1 1G 5 21F 4 5 1 1G 5 F 24 25
21 21G 4. 1 }
28 2 (29) F 4 23
3 22G 5 1F 4 23
3 22G 5 F 4 23
3 22G 5.
1 }
12 2 10 F 2 22
25 6G 5 1 } 2 F 2 22
25 6G 5 F 1 21
2 5 } 2 3 G 6.
1 }}
221 2 (218) F 3 9
22 27G 5 2 1 } 3 F 3 9
22 27G 5 F 21 23
2 } 3 7 }
3 G
Chapter 3, continued
n2ws-03-b.indd 170 6/27/06 9:56:19 AM
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171Algebra 2
Worked-Out Solution Key
7. 1 }}
228 2 (224) F 7 6
24 24G 5 2 1 } 4 F 7 6
24 24G 5 F 2 7 } 4 2 3 } 2
1 1G
8.1 }
120 2 264 F 20 22
12 6G 5 2 1 } 144
F 20 22 12 6G
5 F 2 5 } 36 2 11 } 72
2 1 } 12 2 1 } 24 G
9. 1 }}
2720 2 (2360) F 30 260
6 224 G 5 2 1 } 360
F 30 260 6 224 G
5 F 2 1 } 12 1 } 6
2 1 } 60 1 } 15
G10.
1 }
2 4 } 3 2 1 2
10 } 3 2 F 21 2 5 } 6
4 4 } 3 G 5
1 } 2 F 21 2 5 } 6
4 4 } 3 G
5 F 2 1 } 2 2 5 } 12
2 2 } 3 G
11. The new matrix should have been multiplied by 1 }
det A ,
not by det A.
F 2 4 1 5G21
5 1 } 6 F 5 24
21 2G 5 F 5 }
6 2 2 } 3
2 1 } 6 1 } 3 G
12. C;
1 }
210 2 (29) F 21 3
23 10G 5 F 1 23 3 210G
13. 1 }
1 F 5 21
24 1G 5 F 5 21 24 1G
F 5 21 24 1GF 1 1
4 5GX 5 F 5 21 24 1GF 2 3
21 6G F 1 0
0 1GX 5 F 10 1 1 15 2 6 28 2 1 212 1 6G
X 5 F 11 9 29 26G
14. 1 }
2 F 3 28
22 6G 5 F 3 } 2 24
21 3G
F 3 } 2 24
21 3GF 6 8
2 3GX 5 F 3 } 2 24
21 3GF 4 3
0 22G F 1 0
0 1GX 5 F 6 1 0 9 } 2 1 8
24 1 0 23 2 6G
X 5 F 6 25 }
2
24 29G
15. 2 1 } 4 F 4 0
26 21G 5 F 21 0
3 } 2 1 }
4 G
F 21 0
3 } 2 1 }
4 GF 21 0
6 4GX 5 F 21 0
3 } 2 1 }
4 GF 3 21
4 5G F 1 0
0 1GX 5 F 23 1 0 1 1 0
9
} 2 1 1 2 3 } 2 1 5 } 4 G
X 5 F 23 1
11 }
2 2 1 }
4 G
16. 2 1 } 12
F 2 26 21 23G 5 F 2 1 } 6 1 } 2
1 } 12
1 } 4 G F 2 1 } 6 1 } 2
1 } 12
1 } 4 GF 23 6 1 2GX 5 F 2 1 } 6 1 } 2
1 } 12
1 } 4 GF 5 21 8 2G
F 1 0 0 1GX 5 F 2
5 } 6 1 4
1 }
6 1 1
5 } 12
1 2 2 1 } 12 1 1 } 2 G X 5 F 19
} 6 7 }
6
29 }
12 5 } 12 G
17. 2 1 } 2 F 22 25
0 1G 5 F 1 5 } 2
0 2 1 } 2 G F 1 5 }
2
0 2 1 } 2 GF 1 5 0 22GX 5 F 1 5 }
2
0 2 1 } 2 GF 3 21 0 6 8 4G
F 1 0 0 1GX 5 F 3 1 15 21 1 20 0 1 10
0 2 3 0 2 4 0 2 2G X 5 F 18 19 10
23 24 22G
Chapter 3, continued
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172Algebra 2Worked-Out Solution Key
18.1 } 3 F 3 22
9 25G 5 F 1 2 2 } 3
3 2 5 } 3 G F 1 2 2 } 3
3 2 5 } 3 GF 25 2 29 3GX 5 F 1 2 2 } 3
3 2 5 } 3 GF 4 5 0 3 1 6G
F 1 0 0 1GX 5 F 4 2 2 5 2 2 } 3 0 2 4
12 2 5 15 2 5 } 3 0 2 10G
X 5 F 2 13 }
3 24
7 40 } 3 210
G19. A21 5 F 20.3 20.2 0.3
0.9 0.6 0.1
20.2 0.2 0.2G
Check:
AA21 5 F 1 1 22
22 0 3
3 1 0GF 20.3 20.2 0.3
0.9 0.6 0.1
20.2 0.2 0.2G
5 F 20.3 1 0.9 1 0.4 20.2 1 0.6 2 0.4
0.6 1 0 2 0.6 0.4 1 0 1 0.6
20.9 1 0.9 1 0 20.6 1 0.6 1 0
0.3 1 0.1 2 0.4
20.6 1 0 1 0.6
0.9 1 0.1 1 0G
5 F 1 0 0
0 1 0
0 0 1G 5 I
A21A 5 F 20.3 20.2 0.3
0.9 0.6 0.1
20.2 0.2 0.2GF 1 1 22
22 0 3
3 1 0G
5 F 20.3 1 0.4 1 0.9 20.3 1 0 1 0.3
0.9 2 1.2 1 0.3 0.9 1 0 1 0.1
20.2 2 0.4 1 0.6 20.2 1 0 1 0.2
0.6 2 0.6 1 0
21.8 1 1.8 1 0
0.4 1 0.6 1 0G
5 F 1 0 0
0 1 0
0 0 1G 5 I
20. A21 5 F 21. } 3 1. } 3 20. } 3 20.8 } 3 0. } 3 0.1 } 6 1.1 } 6 20. } 6 0.1 } 6
G Check:
AA21 5 F 1 0 2
2 1 3
1 4 4GF 21. } 3 1. } 3 20. } 3
20.8 } 3 0. } 3 0.1 } 6 1.1 } 6 20. } 6 0.1 } 6
G 5 F 1 0 2
2 1 3
1 4 4GF 2 4 } 3 4 }
3 2 1 } 3
2 5 } 6 1 } 3 1 }
6
7 } 6 2 2 } 3 1 } 6 G
5 F 2 4 } 3 1 0 1 7 } 3 4 }
3 1 0 2 4 }
3 2 1 } 3 1 0 1 1 }
3
2 8 } 3 2 5 } 6 1 7 }
2 8 }
3 1 1 }
3 2 2 2 2 } 3 1 1 }
6 1 1 }
2
2 4 } 3 2 10 }
3 1 14
} 3 4 }
3 1 4 }
3 2 8 }
3 2 1 } 3 1 2 }
3 1 2 }
3 G
5 F 1 0 0
0 1 0
0 0 1G 5 I
A21A 5 F 21. } 3 1. } 3 20. } 3 20.8 } 3 0. } 3 0.1 } 6 21.1 } 6 20. } 6 0.1 } 6
GF 1 0 2
2 1 3
1 4 4G
5F 2 4 } 3 4 } 3 2 1 } 3
2 5 } 6 1 } 3 1 }
6
7 } 6 2 2 } 3 1 } 6 G F 1 0 2
2 1 3
1 4 4G
5 F 2 4 } 3 1 8 } 3 2 1 }
3 0 1 4 }
3 2 4 }
3 2 8 } 3 1 4 2 4 }
3
2 5 } 6 1 2 } 3 1 1 }
6 0 1 1 }
3 1 2 }
3 2 5 } 3 1 1 1 2 }
3
7 } 6 2 4 } 3 1 1 }
6 0 2 2 }
3 1 2 }
3 7 }
3 2 2 1 2 }
3 G
5 F 1 0 0
0 1 0
0 0 1G 5 I
Chapter 3, continued
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173Algebra 2
Worked-Out Solution Key
21. A21 5 F 20.5 0 0.5
21.0625 0.125 0.4375
0.21875 0.0625 20.03125G
Check:
AA21 5 F 1 21 2
22 3 10
3 21 2GF 20.5 0 0.5
21.0625 0.125 0.4375
0.21875 0.0625 20.03125G
5 F 20.5 1 1.0625 1 0.4375 0 2 0.125 1 0.125
1 2 3.1875 1 2.1875 0 1 0.375 1 0.625
21.5 1 1.0625 1 0.4375 0 2 0.125 1 0.125
0.5 2 0.4375 2 0.0625
21 1 1.3125 2 0.3125
1.5 2 0.4375 2 0.0625G
5 F 1 0 0
0 1 0
0 0 1G 5 I
A21A 5 F 20.5 0 0.5
21.0625 0.125 0.4375
0.21875 0.0625 20.03125GF 1 21 2
22 3 10
3 21 2G
5 F 20.5 1 0 1 1.5
21.0625 2 0.25 1 1.3125
0.21875 2 0.125 2 0.09375
0.5 1 0 2 0.5
1.0625 1 0.375 2 0.4375
20.21875 1 0.1875 1 0.03125
21 1 0 1 1
22.125 1 1.25 1 0.875
0.4375 1 0.625 2 0.0625G
5 F 1 0 0
0 1 0
0 0 1G 5 I
22. A21 ø F 0.03425 0.03425 0.08904
0.08219 0.08219 0.01370
20.65753 0.34247 20.10960G
Check:
AA21 5 F 22 5 21
0 8 1
12 25 0G
F 0.03425 0.03425 0.08904
0.08219 0.08219 0.01370
20.65753 0.34247 20.10960G
5 F 20.0685 1 0.41095 1 0.65753
0 1 0.65752 2 0.65753
0.411 2 0.41095 1 0
20.0685 1 0.41095 2 0.34247
0 1 0.65752 1 0.34247
0.411 2 0.41095 1 0
20.17808 1 0.0685 1 0.10960
0 1 0.1096 2 0.10960
1.06848 2 0.0685 1 0G
5 F 0.99998 22.0 3 1025 2.0 3 1025
21.0 3 1025 0.99999 0
5 3 1025 5 3 1025 0.99998G
ø F 1 0 0
0 1 0
0 0 1G 5 I
A21A ø F 0.03425 0.03425 0.08904
0.08219 0.08219 0.01370
20.65753 0.34247 20.10960G
F 22 5 21
0 8 1
12 25 0G
5 F 20.0685 1 0 1 1.06848
20.16438 1 0 1 0.1644
1.31506 1 0 2 1.3152
0.17125 1 0.274 2 0.4452
0.41095 1 0.65752 2 0.0685
23.28765 1 2.73976 1 0.548
20.03425 1 0.03425 1 0
20.08219 1 0.08219 1 0
0.65753 1 0.34247 1 0G
5 F 0.99998 5 3 1025 0
2 3 1025 0.99997 0
21.4 3 1024 1.1 3 1024 1G
ø F 1 0 0
0 1 0
0 0 1G 5 I
Chapter 3, continued
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174Algebra 2Worked-Out Solution Key
23. A21 5 F 0.15 0.3 0.05
20.06875 0.1125 0.01875
20.025 20.05 20.175G
Check:
AA21 5 F 3 28 0
2 4 1
21 0 26GF 0.15 0.3 0.05
20.06875 0.1125 0.01875
20.025 20.05 20.175G
5 F 0.45 1 0.55 1 0 0.9 2 0.9 1 0
0.3 2 0.275 2 0.025 0.6 1 0.45 2 0.05
20.15 1 0 1 0.15 20.3 1 0 1 0.3
0.15 2 0.15 1 0
0.1 1 0.075 2 0.175
20.05 1 0 1 1.05G
5 F 1 0 0
0 1 0
0 0 1G 5 I
A21A 5 F 0.15 0.3 0.05
20.06875 0.1125 0.01875
20.025 20.05 20.175GF 3 28 0
2 4 1
21 0 26G
5 F 0.45 1 0.6 2 0.05
20.20625 1 0.225 2 0.01875
20.075 2 0.1 1 0.175
21.2 1 1.2 1 0 0 1 0.3 2 0.3
0.55 1 0.45 1 0 0 1 0.1125 2 0.1125
0.2 2 0.2 1 0 0 2 0.05 1 1.05G
5 F 1 0 0
0 1 0
0 0 1G 5 I
24. A21 ø F 0.2766 20.2340 20.1915
0.3191 0.1915 20.2979
20.0851 0.1489 0.2128G
AA21 5 F 4 1 5
22 2 1
3 21 6GF 0.2766 20.2340 20.1915
0.3191 0.1915 20.2979
20.0851 0.1489 0.2128G
5 F 1.1064 1 0.3191 2 0.4255
20.5532 1 0.6382 2 0.0851
0.8298 2 0.3191 2 0.5106
20.936 1 0.1915 1 0.7445
0.468 1 0.383 1 0.1489
20.702 2 0.1915 1 0.8934
20.766 2 0.2979 1 1.064
0.383 2 0.5958 1 0.2128
20.5745 1 0.2979 1 1.2768G
5 F 1 0 1 3 1024
21 3 1024 0.9999 0
1 3 1024 21 3 1024 1.0002G
ø F 1 0 0
0 1 0
0 0 1G 5 I
A21A ø F 0.2766 20.2340 20.1915
0.3191 0.1915 20.2979
20.0851 0.1489 0.2128GF 4 1 5
22 2 1
3 21 6G
5 F 1.1064 1 0.468 2 0.5745
1.2764 2 0.383 2 0.8937
20.3404 2 0.2978 1 0.6384
0.2766 2 0.468 1 0.1915
0.3191 1 0.383 1 0.2979
20.0851 1 0.2978 2 0.2128
1.383 2 0.234 2 1.149
1.5955 1 0.1915 2 1.7874
20.4255 1 0.1489 1 1.2768G
5 F 0.9999 1 3 1024 0
23 3 1024 1 24 3 1024
2 3 1024 21 3 1024 1.0002G
ø F 1 0 0
0 1 0
0 0 1G 5 I
25. F 4 21 27 22GF x
yG 5 F 10 225G
A21 5 2 1 } 15 F 22 1
7 4G 5 F 2 } 15
2 1 } 15
2 7 } 15 2 4 } 15 G X 5 A21B 5 F 2 }
15 2 1 } 15
2 7 } 15 2 4 } 15 GF 10 225G 5 F 3
2G The solution is (3, 2).
26. F 4 7 2 3GF x
yG 5 F 216 24G
A21 5 2 1 } 2 F 3 27
22 4G 5 F 2 3 } 2 7 } 2
1 22G
X 5 A21B 5 F 2 3 } 2 7 } 2
1 22GF 216
24G 5 F 10
28G The solution (10, 28).
Chapter 3, continued
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175Algebra 2
Worked-Out Solution Key
Chapter 3, continued
27. F 3 22 6 25GF x
yG 5 F 5 14G
A21 5 2 1 } 3 F 25 2
26 3G 5 F 5 } 3 2 2 } 3
2 21G
X 5 A21B 5 F 5 } 3 2 2 } 3
2 21GF 5
14G 5 F 21 24G
The solution is (21, 24).
28. F 1 21 9 210GF x
yG 5 F 4 45G
A21 5 21F 210 1 29 1G 5 F 10 21
9 21G X 5 A21B 5 F 10 21
9 21GF 4 45G 5 F 25
29G The solution is (25, 29).
29. F 22 29 4 16GF x
yG 5 F 22 8G
A21 5 1 } 4 F 16 9
24 22G 5 F 4 9 } 4
21 2 1 } 2 G
X 5 A21B 5 F 4 9 } 4
21 2 1 } 2 GF 22
8G 5 F 10 22G
The solution is (10, 22).
30. F 2 27 21 5GF x
yG 5 F 26 3G
A21 5 1 } 3 F 5 7
1 2G 5 F 5 } 3 7 }
3
1 } 3
2 } 3 G
X 5 A21B 5 F 5 } 3 7 }
3
1 } 3
2 } 3 GF 26
3G 5 F 23 0G
The solution is (23, 0).
31. F 6 1 21 3GF x
yG 5 F 22 225G
A21 5 1 } 19 F 3 21
1 6G 5 F 3 } 19
2 1 } 19
1 } 19
6 } 19 G
X 5 A21B 5 F 3 } 19
2 1 } 19
1 } 19
6 } 19 GF 22
225 G 5 F 1 28G
The solution is (1, 28).
32. F 2 1 2 5GF x
yG 5 F 22
38G
A21 5 1 } 8 F 5 21
22 2G 5 F 5 } 8 2 1 } 8
2 1 } 4
1 } 4 G
X 5 A21B 5 F 5 } 8 2 1 } 8
2 1 } 4
1 } 4 GF 22
38G 5 F 26
10G The solution is (26, 10).
33. F 5 7 3 5GF x
yG 5 F 20
16G
A21 5 1 } 4 F 5 27
23 5G 5 F 5 } 4 2 7 } 4
2 3 } 4
5 } 4 G
X 5 A21B 5 F 5 } 4 2 7 } 4
2 3 } 4
5 } 4 GF 20
16G 5 F 23
5G The solution is (23, 5).
34. C;F 3 25 21 2GF x
yG 5 F 226
10G
A21 5 1F 2 5 1 3G 5 F 2 5
1 3G X 5 A21B 5 F 2 5
1 3GF 226
10G 5 F 22
4G The solution is (22, 4).
35. F 1 21 23
5 2 1
23 21 0GF x
y
zG 5 F 2
217
8G
A21 5 F 1 3 5
23 29 216
1 4 7G
X 5 A21B 5 F 1 3 5
23 29 216
1 4 7GF 2
217
8G 5 F 29
19
210G
The solution is (29, 19, 210).
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176Algebra 2Worked-Out Solution Key
36. F 23 1 28
1 22 1
2 22 5GF x
y
zG 5 F 18
211
217G
A21 5 F 21.6 2.2 23
20.6 0.2 21
0.4 20.8 1G
X 5 A21B 5 F 21.6 2.2 23
20.6 0.2 21
0.4 20.8 1GF 18
211
217G 5 F 22
4
21G
The solution is (22, 4, 21).
37. F 2 4 5
1 2 3
5 24 22GF x
y
zG 5 F 5
4
23G
A21 ø F 0.5714 20.8571 0.1429
1.2143 22.0714 20.0714
21 2 0G
X 5 A21B 5 F 21
22
3G
The solution is (21, 22, 3).
38. F 4 21 21
6 0 21
21 4 5GF x
y
zG 5 F 220
227
23G
A21 ø F 0.1905 0.0476 0.0476
21.3810 0.9048 20.0952
1.1429 20.7143 0.2857G
X 5 A21B 5 F 24
1
3G
The solution is (24, 1, 3).
39. F 3 2 21
21 25 4
4 1 1GF x
y
zG 5 F 14
248
2G
A21 5 F 0.75 0.25 20.25
21.41 } 6 20.58 } 3 0.916
21.58 } 3 21.41 } 6 1.08 } 3 G
X 5 A21B 5 F 0.75 0.25 20.25
21.41 } 6 20.58 } 3 0.91 } 6 21.58 } 3 21.41 } 6 1.08 } 3
GF 14
248
2G
5 F 22
10
0G
The solution is (22, 10, 0).
40. F 6 1 2 1 21 1
21 4 21GF x
y
zG 5 F 11
25
14G
A21 5 F 0.25 20.75 20.25
0 0. } 3 0. } 3 0.25 2.08 } 3 0.58 } 3
GX 5 A21B 5 F 0.25 20.75 20.25
0 0. } 3 0. } 3 0.25 2.08 } 3 0.58 } 3
GF 11
25
14G 5 F 3
3
25G
The solution is (3, 3, 25).
41. Sample answer: F 4 10 2 5G
42. F 2 5 24 6
0 2 1 27
4 8 27 14
3 6 25 10
G F w
x
y
z
G 5 F 0
52
225
216
G A21 ø F 210 4 27 229
5 22 216 18
4 22 217 20
2 21 27 8
G
X 5 A21B 5 F 23
8
1
25
G The solution is (23, 8, 1, 25).
Problem Solving
43. s 5 hours in a single-engine plane
t 5 hours in a twin-engine plane
s 1 t 5 200
60s 1 240t 5 21,000
F 1 1 60 240GF s
t G 5 F 200 21,000G
A21 5 1 } 180 F 240 21
260 1G 5 F 4 }
3 2 1
} 180
2 1 } 3 1 }
180 G
X 5 A21B 5 F 4 }
3 2 1
} 180
2 1 } 3 1 }
180 GF 200
21,000G 5 F 150 50G
The pilot spent 150 hours fl ying a single-engine plane and 50 hours fl ying a twin-engine plane.
Chapter 3, continued
n2ws-03-b.indd 176 6/27/06 9:57:11 AM
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177Algebra 2
Worked-Out Solution Key
Chapter 3, continued
44. x 5 three-point fi eld goals
y 5 two-point fi eld goals
z 5 free throws
x 1 y 1 z 5 976
3x 1 2y 1 z 5 1680
y 2 z 5 135
F 1 1 1
3 2 1
0 1 21G
F x
y
zG 5 F 976
1680
135G
A21 5 F 21 2 } 3 2 1 } 3
1 2 1 } 3 2 } 3
1 2 1 } 3 2 1 } 3 G
X 5 A21B 5 F 21 2 } 3 2 1 } 3
1 2 1 } 3 2 } 3
1 2 1 } 3 2 1 } 3 GF 976
1680
135G 5 F 99
506
371G
Dirk made 99 three-point fi eld goals, 506 two-point fi eld goals, and 371 free throws.
45. a. m 5 number of batches of muffi ns
r 5 number of batches of rolls
m 1 2r 5 8 (cups of buttermilk)
m 1 3r 5 11 (number of eggs)
b. F 1 2 1 3GF m
r G 5 F 8 11G
c. A21 5 1 } 1 F 3 22
21 1G 5 F 3 22 21 1G
X 5 A21B 5 F 3 22 21 1GF 8
11G 5 F 2 3G
The class should make two batches of muffi ns and three batches of rolls.
46 a. c 5 cost of cheese
m 5 cost of meat
2c 1 3m 5 18
3c 1 5m 5 28
F 2 3 3 5GF c
mG 5 F 18 28G
A21 5 1F 5 23 23 2
G 5 F 5 23 23 2
G X 5 A21B 5 F 5 23
23 2GF 18
28G 5 F 6
2G Each cheese costs $6.00 and each meat costs $2.00.
b. 3c 1 5m 5 28
7c 1 10m 5 60
F 3 5 7 10GF c
mG 5 F 28 60G
A21 5 2 1 } 5 F 10 25
27 3G 5 F 22 1
7 } 5 2
3 } 5 G
X 5 A21B 5 F 22 1
7 } 5 2
3 } 5 GF 28
60G 5 F 4 3.2G
Each cheese costs $4.00 and each meat costs $3.20.
c. Sample answer: The results of parts (a) and (b) are not the same. The larger volume of cheese reduces the unit price. The amount of meat is larger and raises the unit price.
47. b 5 ounces of Bran Crunchies
t 5 ounces of Toasted Oats
w 5 ounces of Whole Wheat Flakes
78b 1 104t 1 198w 5 500
b 1 0.6w 5 5
22b 1 25.5t 1 23.8w 5 100
F 78 104 198
1 0 0.6
22 25.5 23.8GF b
t
wG 5 F 500
5 100
G A21 5 F 20.0056 0.9348 0.0223
20.0039 20.9079 0.0549
0.0093 0.1086 20.0378G
X 5 A21B ø F2.3
0.8
1.2G
About 2.3 ounces of Bran Crunchies, 0.8 ounces of Toasted Oats, and 1.2 ounces of Whole Wheat Flakes should be combined.
48. a. Equation 1: Area per
sheet p ( Sheets
of red 1 Sheets of
yellow 1 Sheets
of blue )
5 Total area
Equation 2: Area of red
5 Area of yellow
1 Area of blue
Equation 3:
Cost p Sheets of red
1 Cost p Sheets of yellow
1 Cost p Sheets
of blue 5
Total cost
Equation 1: 0.75r 1 0.75y 1 0.75b 5 9
Equation 2: r 2 y 2 b 5 0
Equation 3: 6.5r 1 4.5y1 8.5b 5 80
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178Algebra 2Worked-Out Solution Key
b. F 0.75 0.75 0.75
1 21 21
6.5 4.5 8.5GF r
y
bG 5 F 9
0
80G
c. X 5 A21B 5 F 6
2.5
3.5G
You should buy 6 sheets of red, 2.5 sheets of yellow, and 3.5 sheets of blue tiles.
49. a. AT 5 F 0 1 21 0GF 1 3 5
1 4 2G 5 F 0 1 1 0 1 4 0 1 2
21 1 0 23 1 0 25 1 0G 5 F 1 4 2
21 23 25G AAT 5 A(AT ) 5 F 0 1
21 0GF 1 4 2
21 23 25G 5 F 0 2 1 0 2 3 0 2 5
21 1 0 24 1 0 22 1 0G 5 F 21 23 25
21 24 22G Matrix A rotates the triangle 908 clockwise about
the origin.
x
y
1
4(1, 21)(21, 21)
(23, 24)
(25, 22)
(1, 1)
(5, 2)
(3, 4)
(4, 23)
(2, 25)
ATAAT
A
b. To get back to the original triangle, you must fi nd AAAAT, or A2(AAT). This means the triangle will be rotated 908 clockwise four times from T, or rotated 908 clockwise twice from AAT, so it will be back to its original position.
50. A 5 F a b
c dG, B 5 F d }
ad 2 cb 2b
} ad 2 cb
2c }
ad 2 cb
a }
ad 2 cb G
AB 5 F ad 2 bc }
ad 2 cb 2ab 1 ba
} ad 2 cb
cd 2 dc
} ad 2 cb
2cb 1 da
} ad 2 cb
G 5 F 1 0
0 1G 5 I
BA 5 F ad 2 bc }
ad 2 bc db 2 bd
} ad 2 bc
2ca 1 ac
} ad 2 bc
2bc 1 ad
} ad 2 bc
G 5 F 1 0
0 1G 5 I
Because AB 5 BA 5 I, B is the inverse of A.
Mixed Review
51.
1
x
y
21
52.
1
x
y
21
53.
1
x
y
21
54.
2
x
y
21
55.
1
x
y
21
56. 1
x
y
1
57. The points have approximately no correlation because they show no linear pattern.
58. The points have a positive correlation because y tends to increase as x increases.
59. The points have a negative correlation because y tends to decrease as x increases.
60. F 5 4x 18 6
G 5 F 5 220 3y 6G
4x 5 220 3y 5 18
x 5 25 y 5 6
61. F 23x 29 13 25G 1 F 4 12
25y 16G 5 F 220 3 18 11G
F 23x 1 4 3 13 2 5y 11G 5 F 220 3
18 11G 23x 1 4 5 220 13 2 5y 5 18
23x 5 224 25y 5 5
x 5 8 y 5 21
Quiz 3.6–3.8 (p. 217)
1. 2AB 5 2F 1 24 5 2
GF 2 23 0 1
G 5 F 2 28 10 4
GF 2 23 0 1
G 5 F 4 1 0 26 2 8
20 1 0 230 1 4G 5 F 4 214
20 226G
Chapter 3, continued
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179Algebra 2
Worked-Out Solution Key
2. AB 1 AC 5 (F 1 24 5 2
GF 2 23 0 1
G) 1 (F 1 24
5 2GF 26 21
2 4G)
5 F 2 1 0 23 2 4 10 1 0 215 1 2
G 1 F 26 2 8 21 2 16
230 1 4 25 1 8G
5 F 2 2 14 27 2 17 10 2 26 213 1 3
G 5 F 212 224 216 210
G 3. A(B 1 C) 5 F 1 24
5 2G(F 2 23
0 1G 1 F 26 21
2 4G)
5 F 1 24 5 2
GF 24 24 2 5
G 5 F 24 2 8 24 2 20
220 1 4 220 1 10G 5 F 212 224
216 210G
4. (B 2 A)C 5 (F 2 23 0 1
G 2 F 1 24 5 2
G)F 26 21 2 4
G 5 F 1 1
25 21GF 26 21
2 4G
5 F 26 1 2 21 1 4 30 2 2 5 2 4
G 5 F 24 3 28 1
G 5. 5 4
22 23 5 5(23) 2 (22)(4) 5 27
6. 1 0 22 23 1 4
2 3 21 1 0
23 1
2 3
5 (21 1 0 1 18) 2 (24 1 12 1 0) 5 9
7. 2 21 5 23 6 9
22 3 1 2 21
23 6
22 3
5 (12 1 18 2 45) 2 (260 1 54 1 3) 5 212
8. F 1 3
2 7GF x
y G 5 F 22
26G A21 5
1 } 1 F 7 23
22 1G 5 F 7 23
22 1G X 5 A21B 5 F 7 23
22 1GF 22
26G 5 F 4
22G The solution is (4, 22).
9. F 3 24
2 23GF x
y G 5 F 5
3G A21 5 21F 23 4
22 3G 5 F 3 24
2 23G X 5 A21B 5 F 3 24
2 23GF 5
3G 5 F 3
1G The solution is (3, 1).
10. F 23 2
6 25GF x
y G 5 F 213
24G A21 5
1 } 3 F 25 22
26 23G 5 F 2 5 } 3 2 2 } 3
22 21G
X 5 A21B 5 F 2 5 } 3 2 2 } 3
22 21GF 213
24G 5 F 17 }
3
2G
The solution is 1 17 }
3 , 2 2 .
11. F 3 21
2 22GF x
y G 5 F 24
28G A21 5 2
1 } 4 F 22 1
22 3G 5 F 1 } 2 2 1 } 4
1 } 2 2 3 } 4
G X 5 A21B 5 F 1 }
2 2 1 } 4
1 } 2 2 3 } 4
GF 24
28G 5 F 0
4G The solution is (0, 4).
12. F 7 4
5 3GF x
y G 5 F 6
225G A21 5 1F 3 24
25 7G 5 F 3 24
25 7G X 5 A21B 5 F 3 24
25 7GF 6
225G 5 F 118
2205G The solution is (118, 2205).
13. F 4 1
26 1GF x
y G 5 F 22
18G A21 5
1 } 10 F 1 21
6 4G 5 F 1 } 10
2 1 } 10
3 } 5 2 } 5 G X 5 A21B 5 F 1 }
10 2 1 } 10
3 } 5 2 } 5 GF 22
18G 5 F 22
6G The solution is (22, 6).
Chapter 3 continued
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180Algebra 2Worked-Out Solution Key
14. Area 5 6 1 } 2
0 2 1 12 2 1
12 26 1 0 2 12 2
12 26
5 6 1 } 2 [(0 1 24 1 312) 2 (24 1 0 1 24)] 5 144
The area of the sail is 144 square feet.
Problem Solving Workshop 3.8 (p. 219)
1. 2m 1 p 5 17.75
2m 1 2p 1 d 5 34.50
4m 1 3p 1 2d 5 67.25 F 2 1 0 � 17.75
2 2 1 � 34.50
4 3 2 � 67.25G
(22)R1 1 R3 F 2 1 0 � 17.75
2 2 1 � 34.50
0 1 2 � 31.75G
(21)R1 1 R2
F 2 1 0 � 17.75
0 1 1 � 16.75
0 1 2 � 31.75G
(21)R2 1 R3 F 2 1 0 � 17.75
0 1 1 � 16.75
0 0 1 � 15G
From the third row, d 5 15. From the second row, p 1 d 5 16.75, so p 1 15 5 16.75, or p 5 1.75. From the fi rst row, 2m 1 p 5 17.75, so 2m 1 1.75 5 17.75, or m 5 8.
A movie pass costs $8, a package of popcorn costs $1.75, and a DVD costs $15.
2. s 5 amount in stocks; b 5 amount in bonds
m 5 amount in money market funds
s 1 b 1 m 5 18000
0.1s 1 0.07b 1 0.05m 5 1440
s 2 b 2 m 5 0
F 1 1 1 � 18000
0.1 0.07 0.05 � 1440
1 21 21 � 0G
100R2
R1 1 (21)R3 F 1 1 1 � 18,000
10 7 5 � 144,000
0 2 2 � 18,000G
(210)R1 1 R2
F 1 1 1 � 18,000
0 23 25 � 236,000
0 2 2 � 18,000G
2R2 1 3R3 F 1 1 1 � 18,000
0 23 25 � 236,000
0 0 24 � 218,000G
(21)R2
1 2 1 } 4 2 R3
F 1 1 1 � 18,000
0 3 5 � 36,000
0 0 1 � 4500G
From row 3, m 5 4500. From row 2, 3b 1 5m 5 36,000, so 3b 1 5(4500) 5 36,000, or b 5 4500. From row 1, s 1 b 1 m 5 18,000, so s 1 4500 1 4500 5 18,000, or s 5 9000. You should invest $9000 in stocks, $4500 in bonds, and $4500 in money market funds.
3. s 5 pounds of sunfl ower seed
t 5 pounds of thistle seed
s 1 t 5 20
0.34s 1 0.79t 5 10.85
F 1 1 � 20 0.34 0.79 � 10.85G
(20.34)R1 1 R2
F 1 1 � 20 0 0.45 � 4.05G
R2
} 0.45
F 1 1 � 20 0 1 � 9G
From row 2, t 5 9. From row 1, s 1 t 5 20. So, s 1 9 5 20, or s 5 11.
The mixture contains 11 pounds of sunfl ower seed and 9 pounds of thistle seed.
4. x 2 2y 1 4z 5 210
5x 1 y 2 z 5 24
3x 2 6y 1 12z 5 230 F 1 22 4 � 210
5 1 21 � 24
3 26 12 � 230G
(23)R1 1 R3 F 1 22 4 � 210
5 1 21 � 24
0 0 0 � 0G
Because row 3 produces the equation 0 5 0, the system has infi nitely many solutions.
Mixed Review of Problem Solving (p. 220)
1. a. A 5 F 4.5 6 2.5 5.5 8 2.5G B 5 F 4 6.5 3.25
5 8.5 3.25G b. B 2 A 5 F 4 6.5 3.25
5 8.5 3.25G 2 F 4.5 6 2.5 5.5 8 2.5G
5 F 20.5 0.5 0.75 20.5 0.5 0.75G
Each element in each row of B 2 A represents the difference in cost (in thousands of dollars) of a daytime, primetime, and late-night commercial on a cable TV network between the two cities.
c. A 5 1.10F 4.5 6 2.5 5.5 8 2.5G 5 F 4.95 6.6 2.75
6.05 8.8 2.75G B 5 1.10F 4 6.5 3.25
5 8.5 3.25G 5 F 4.4 7.15 3.575 5.5 9.35 3.575G
2. a. n 5 number of nickels
d 5 number of dimes
q 5 number of quarters
n 1 d 1 q 5 85
0.05n 1 0.1d 1 0.25q 5 13.25
22d 1 q 5 0
b. F 1 1 1
0.05 0.1 0.25
0 22 1GF n
d
qG 5 F 85
13.25
0G
Chapter 3, continued
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181Algebra 2
Worked-Out Solution Key
c. A21 5 F 1. } 3 26. } 6 0. } 3 20. } 1 2. } 2 20. } 4 20. } 2 4. } 4 0. } 1
G X 5 A21B 5 F 1. } 3 26. } 6 0. } 3
20. } 1 2. } 2 20. } 4 20. } 2 4. } 4 0. } 1
GF 85
13.25
0G
5 F 25
20
40G
The person has 25 nickels, 20 dimes, and40 quarters.
3. a. Sample answer:
A 5 F 3 2 8 4G C 5 F 2 5
6 13G det A 5 det C 5 24
b. Sample answer:
B 5 F 7 0 5
1 4 3
2 4 6G
D 5 F 1 2 4
5 0 4
1 6 4G
det B 5 det D 5 64
4. Matrix BA is defi ned because B has three columns andA has three rows.
BA 5 [ 0.03 0.05 0.08]F 175 270
370 225
200 255G
5 [0.03(175) 1 0.05(370) 1 0.08(200)
0.03(270) 1 0.05(225) 1 0.08(255)]
5 [39.75 39.75]
This matrix represents the total commission made by each salesperson. Mary and Mark each made a total of $39.75 in commission.
5. a. H 1 N 1 3O 5 63
2N 1 O 5 44
2H 1 O 5 18
b. A 5 F 1 1 3
0 2 1
2 0 1G
det A 5 1 1 3 0 2 1
2 0 1 1 1
0 2
2 0
5 (2 1 2 1 0) 2 (12 1 0 1 0) 5 28
c. You can use Cramer’s rule to solve the system.
H 5
63 1 3 44 2 1
18 0 1 63 1
44 2
18 0 }} 28
5 (126 1 18 1 0) 2 (108 1 0 1 44)
}}} 28 5 1
N 5
1 63 3 0 44 1
2 18 1 1 63
0 44
2 18 }} 28
5 (44 1 126 1 0) 2 (264 1 18 1 0)
}}} 28 5 14
O 5
1 1 63 0 2 44
2 0 18 1 1
0 2
2 0 }} 28
5 (36 1 88 1 0) 2 (252 1 0 1 0)
}}} 28
5 16
The atomic weights of hydrogen, nitrogen, and oxygen are 1, 14, and 16, respectively.
6. c 5 bushels of corn
s 5 bushels of soybeans
w 5 bushels of wheat
2.35c 1 5.40s 1 3.60w 5 4837
c 1 s 1 w 5 1700
c 2 3.25s 2 3.25w 5 0
F 2.35 5.40 3.60
1 1 1
1 23.25 23.25GF c
s
wG 5 F 4837
1700
0G
A21 5 F 0 0.7647 0.2353
0.5556 21.4690 0.1634
20.5556 1.7042 20.3987G
X 5 A21B 5 F1300
190
210G
The farmer harvested 210 bushels of wheat.
Chapter 3 Review (pp. 222–226)
1. A system of linear equations with at least one solution is consistent, while a system with no solution is inconsistent.
2. A solution (x, y, z) of a system of linear equations in three variables is called an ordered triple.
3. The product of two matrices is defi ned when the number of columns in the fi rst matrix is the same as the number of rows in the second matrix.
4.
2x
y
22
(5, 1)
The lines appear to intersect at (5, 1).
Check: 2(5) 2 1 0 9 5 1 3(1) 0 8
9 5 9 ✓ 8 5 8 ✓
Chapter 3, continued
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182Algebra 2Worked-Out Solution Key
5.2
x
y
21
(24, 22)
The lines appear to intersect at (24, 22).
Check: 2(24) 2 3(22) 0 22 24 1 (22) 0 26
22 5 22 ✓ 26 5 26 ✓
6.
1
x
y
22
(0, 6)
The lines appear to intersect at (0, 6).
Check: 3(0) 1 6 0 6 20 1 2(6) 0 12
6 5 6 ✓ 12 5 12 ✓
7. 3x 1 2y 5 5 3 2 6x 1 4y 5 10
22x 1 3y 5 27 3 3 26x 1 9y 5 81
13y 5 91
y 5 7
3x 1 2(7) 5 5 → x 5 23
The solution is (23, 7).
8. 3x 1 5y 5 5 3 3 9x 1 15y 5 15
2x 2 3y 5 16 3 5 10x 2 15y 5 80
19x 5 95
x 5 5
3(5) 1 5y 5 5 → y 5 22
The solution is (5, 22).
9. 2x 1 3y 5 9 2x 1 3y 5 9
23x 1 y 5 25 3 (23) 9x 2 3y 5 275
11x 5 266
x 5 26
2(26) 1 3y 5 9 → y 5 7
The solution is (26, 7).
10. r 5 price of regular gasoline
p 5 price of premium gasoline
14r 1 10p 5 46.68 14r 1 10p 5 46.68
2r 1 p 5 0.30 3 14 214r 1 14p 5 4.2
24p 5 50.88
p 5 2.12
2r 1 2.12 5 0.30 → r 5 1.82
Regular gas costs $1.82 per gallon and premium gas costs $2.12 per gallon.
11. 4x 1 y < 1 12. 2x 1 3y > 6
2x 1 2y ≤ 5 2x 2 y ≤ 8
1
x
y
21
1
x
y
21
13. x 1 3y ≥ 5 2x 1 2y < 4
1x
y
21
14. x 2 y 1 z 5 10
23x 1 5y 2 z 5 218
22x 1 4y 5 28
4x 1 y 2 2z 5 15 4x 1 y 2 2z 5 15
23x 1 5y 2 z 5 218 3 (22) 6x 2 10y 1 2z 5 36
10x 2 9y 5 51
22x 1 4y 5 28 3 5 210x 1 20y 5 240
10x 2 9y 5 51 10x 2 9y 5 51
11y 5 11
y 5 1
22x 1 4(1) 5 28 → x 5 6
6 2 1 1 z 5 10 → z 5 5
The solution is (6, 1, 5).
15. 6x 2 y 1 4z 5 6 6x 2 y 1 4z 5 6
2x 1 2y 2 5z 5 242 3 (23) 26x 2 6y 1 15z 5 126
27y 1 19z 5 132
2x 2 3y 1 z 5 31 3 2 22x 2 6y 1 2z 5 62
2x 1 2y 2 5z 5 242 2x 1 2y 2 5z 5 242
24y 2 3z 5 20
27y 1 19z 5 132 3 4 228y 1 76z 5 528
24y 2 3z 5 20 3 (27) 28y 1 21z 5 2140
97z 5 388
z 5 4
27y 1 19(4) 5 132 → y 5 28
6x 2 (28) 1 4(4) 5 6 → x 5 23
The solution is (23, 28, 4).
Chapter 3, continued
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183Algebra 2
Worked-Out Solution Key
16. 5x 1 y 2 z 5 40 5x 1 y 2 z 5 40
2x 1 3y 1 z 5 16 3 5 25x 1 15y 1 5z 5 80
16y 1 4z 5 120
x 1 7y 1 4z 5 44
2x 1 3y 1 z 5 16
10y 1 5z 5 60
16y 1 4z 5 120 3 5 280y 2 20z 5 2600
10y 1 5z 5 60 3 4 40y 1 20z 5 240
240y 5 2360
y 5 9
16(9) 1 4z 5 120 → z 5 26
5x 1 9 2 (26) 5 40 → x 5 5
The solution is (5, 9, 26).
17. w 5 number of wind instruments
s 5 number of string instruments
p 5 number of percussion instruments
Equation 1: w 1 s 1 p 5 15
Equation 2: w 5 2(s 1 p)
Equation 3: s 5 3
w 5 2(3 1 p) → w 5 6 1 2p
(6 1 2p) 1 3 1 p 5 15 → p 5 2
w 5 6 1 2(2) → w 5 10
Ten students play wind instruments, three play string instruments, and two play percussion.
18. F 4 25 2 3G 1 F 21 3
27 4G 5 F 4 1 (21) 25 1 3 2 1 (27) 3 1 4G
5 F 3 22 25 7G
19. F 21 8 2 23G 1 F 7 24
6 21G 5 F 21 1 7 8 1 (24) 2 1 6 23 1 (21)G
5 F 6 4 8 24G
20. F 10 24 5 1G 2 F 0 9
2 7G 5 F 10 2 0 24 2 9 5 2 2 1 2 7G
5 F 10 213 3 26G
21. F 22 3 5
21 6 22G 2 F 24 7 5
28 0 29G 5 F 22 2 (24) 3 2 7 5 2 5
21 2 (28) 6 2 0 22 2 (29)G 5 F 2 24 0
7 6 7G
22. 23F 5 22 3 6G 5 F 23(5) 23(22)
23(3) 23(6)G 5 F 215 6 29 218G
23. 8F 8 4 5
21 6 22G 5 F 8(8) 8(4) 8(5)
8(21) 8(6) 8(22)G 5 F 64 32 40
28 48 216G 24. [21 21]F 8 2
26 29G 5 [21(8) 1 (21)(26) 21(2) 1 (21)(29)]
5 [22 7]
25. F 11 7 1 25GF 0 25
4 23G 5 F 11(0) 1 7(4) 11(25) 1 7(23)
1(0) 1 (25)(4) 1(25) 1 (25)(23)G 5 F 28 276
220 10G 26. F 4 21
1 7GF 5 22 4 3 12 6G
5 F 4(5) 1 (21)(3) 4(22) 1 (21)(12) 4(4) 1 (21)(6) 1(5) 1 7(3) 1(22) 1 7(12) 1(4) 1 7(6)G
5 F 17 220 10 26 82 46G
27. F 22 5 0 3GF 6 23 5
2 0 1G 5 F 22(6) 1 5(2) 22(23) 1 5(0) 22(5) 1 5(21)
0(6) 1 3(2) 0(23) 1 3(0) 0(5) 1 3(21)G 5 F 22 6 215
6 0 23G 28. C 5 BI 5 F 109.99 0 0
0 319.99 0
0 0 549.99G
AC 5 F 5000 6000 8000 4000 10,000 5000G
F 109.99 0 0
0 319.99 0
0 0 549.99G
5 F 549,950 1,919,940 4,399,920 439,960 3,199,900 2,749,950G
Chapter 3, continued
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184Algebra 2Worked-Out Solution Key
The labels for the matrix are shown below. Total Value (Dollars)
19 in. 27 in. 32 in.
Warehouse 1
Warehouse 2 F 549,950 1,919,940 4,399,920 439,960 3,199,900 2,749,950G
In warehouse 1, the total value of the 19 in. TVs was $549,950, the total value of the 27 in. TVs was $1,919,940, and the total value of the 32 in. TVs was $4,399,920. In warehouse 2, the total value of the 19 in. TVs was $439,960, the total value of the 27 in. TVs was $3,199,900, and the total value of the 32 in. TVs was $2,749,950.
29. 24 2 5 8 5 24(8) 2 5(2) 5 242
30. 3 25 2 6 5 3(6) 2 2(25) 5 28
31. 3 0 1 6 5 3(6) 2 1(0) 5 18
32. Area 5 6 1 } 2
0 0 1 0 50 1
70 20 1 0 0
0 50
70 20
5 6 1 } 2 [(0 1 0 1 0) 2 (3500 1 0 1 0)]
5 1750 square inches
(1750 in.2) 1 1 ft2 }
144 in.2 2 ø 12.2 ft2
You will need approximately 12.2 square feet of material.
33. F 1 4 2 25GF x
yG 5 F 11 9G
A21 5 1 }
213 F 25 24
22 1G 5 F 5 }
13 4 }
13
2 } 13
2 1 } 13 G X 5 A21B 5 F
5 }
13 4 }
13
2 } 13
2 1 } 13 GF 11 9G 5 F 7
1G The solution is (7, 1).
34. F 3 1
21 2GF x yG 5 F 21
12G A21 5
1 } 7 F 2 21
1 3G 5 F 2 } 7 2 1 } 7
1 } 7 3 } 7 G X 5 A21B 5 F 2 } 7 2 1 } 7
1 } 7 3 } 7 GF 21 12G 5 F 22
5G The solution is (22, 5).
35. F 3 2
4 23GF x yG 5 F 211
8G A21 5 2
1 } 17 F 23 22
24 3G 5 F 3 } 17
2 } 17
4 } 17
2 3 } 17 G X 5 A21B 5 F 3 }
17 2 }
17
4 } 17
2 3 } 17 GF 211 8G 5 F 21
24G The solution is (21, 24).
Chapter 3 Test (p. 227)
1.
x
y
1
1
4x 1 y 5 5
3x 2 y 5 2
(1, 1)
The lines appear to intersect at (1, 1).
Check: 3(1) 2 1 0 2 4(1) 1 1 0 5
2 5 2 ✓ 5 5 5 ✓
2.
x
y1
1
x 1 2y 5 26
26x 2 2y 5 214
(4, 25)
The lines appear to intersect at (4, 25).
Check: 4 1 2(25) 0 26 26(4) 2 2(25) 0 214
26 5 26 ✓ 214 5 214 ✓
3.
x
y
1
1
x 2 y 5 2332
2x 2 3y 5 15
The lines do not intersect, so there is no solution.
Chapter 3, continued
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185Algebra 2
Worked-Out Solution Key
4.
x
y
2
2
2x 1 8y 5 24
3x 2 y 5 12
(4, 0)
The lines appear to intersect at (4, 0).
Check: 3(4) 2 0 0 12 24 1 8(0) 0 24
12 5 12 ✓ 24 5 24 ✓
5. 2x 1 y < 6 6. x 2 3y ≥ 9 y > 22
1 }
3 x 2 y ≤ 3
1
x
y
21
1
x
y
22
The solution is a line.
7. x 2 2y ≤ 214 8. 23x 1 4y > 212
y ≥ x y < 22x 1 5
2
x
y
22
1
x
y
21
9. 3x 1 y 5 29 3 2 6x 1 2y 5 218
x 2 2y 5 210 x 2 2y 5 210
7x 5 228
x 5 24
3(24) 1 y 5 29 → y 5 3
The solution is (24, 3).
10. 2x 1 3y 5 22 3 (22) 24x 2 6y 5 4
4x 1 7y 5 26 4x 1 7y 5 26
y 5 22
2x 1 3(22) 5 22 → x 5 2
The solution is (2, 22).
11. x 1 4y 5 226 → x 5 226 2 4y
25x 2 2y 5 214
25(226 2 4y) 2 2y 5 214
18y 5 2144
y 5 28
x 1 4(28) 5 226 → x 5 6
The solution is (6, 28).
12. x 2 y 1 z 5 23 3 2 2x 2 2y 1 2z 5 26
4x 1 2y 2 z 5 2 4x 1 2y 2 z 5 2
6x 1 z 5 24
2x 2 y 1 5z 5 4 3 2 4x 2 2y 1 10z 5 8
4x 1 2y 2 z 5 2 4x 1 2y 2 z 5 2
8x 1 9z 5 10
6x 1 z 5 24 3 (29) 254x 2 9z 5 36
8x 1 9z 5 10 8x 1 9z 5 10
246x 5 46
x 5 21
6(21) 1 z 5 24 → z 5 2
21 2 y 1 2 5 23 → y 5 4
The solution is (21, 4, 2).
13. x 1 y 1 z 5 3
2x 1 3y 1 2z 5 28
4y 1 3z 5 25
5y 1 z 5 2 3 (23) 215y 2 3z 5 26
4y 1 3z 5 25 4y 1 3z 5 25
211y 5 211
y 5 1
5(1) 1 z 5 2 → z 5 23
x 1 1 1 (23) 5 3 → x 5 5
The solution is (5, 1, 23).
14. 2x 2 5y 2 z 5 17 3 2 4x 2 10y 2 2z 5 34
24x 1 6y 1 z 5 220 24x 1 6y 1 z 5 220
24y 2 z 5 14
x 1 y 1 3z 5 19 3 4 4x 1 4y 1 12z 5 76
24x 1 6y 1 z 5 220 24x 1 6y 1 z 5 220
10y 1 13z 5 56
24y 2 z 5 14 3 13 252y 2 13z 5 182
10y 1 13z 5 56 10y 1 13z 5 56
242y 5 238
y 5 2 17
} 3
24 1 2 17
} 3 2 2 z 5 14 → z 5 26
} 3
2x 2 5 1 2 17
} 3 2 2 26
} 3 5 19 → x 5 2 4 } 3
The solution is 1 2 4 } 3 , 2
17 } 3 ,
26 }
3 2 .
15. 2A 1 B 5 2F 1 22 4 23
G 1 F 3 5
21 0G
5 F 2(1) 1 3 2(22) 1 5
2(4) 1 (21) 2(23) 1 0G 5 F 5 1
7 26G
Chapter 3, continued
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186Algebra 2Worked-Out Solution Key
16. C 2 3B 5 F 26 8
10 15G 2 3F 3 5
21 0G
5 F 26 2 3(3) 8 2 3(5)
10 2 3(21) 15 2 3(0)G 5 F 215 27
13 15G
17. Not possible; A and D do not have the same dimensions.
18. 4D 1 E 5 4F 21 3 22
2 0 21G 1 F 4 21 3
6 22 1G
5 F 4(21) 1 4 4(3) 1 (21) 4(22) 1 3
4(2) 1 6 4(0) 1 (22) 4(21) 1 1G
5 F 0 11 25
14 22 23G
19. AC 5 F 1 22
4 23GF 26 8
10 15G
5 F 1(26) 1 (22)(10) 1(8) 1 (22)(15)
4(26) 1 (23)(10) 4(8) 1 (23)(15)G
5 F 226 222
254 213G
20. Not possible; the number of columns in D does not equal the number of rows in E.
21. (A 1 B)D 5 (F 1 22 4 23
G 1 F 3 5
21 0G)F 21 3 22
2 0 21G
5 F 4 3 3 23
GF 21 3 22
2 0 21G
5 F 4(21) 1 3(2) 4(3) 1 3(0) 4(22) 1 3(21)
3(21) 1 (23)(2) 3(3) 1 (23)(0) 3(22) 1 (23)(21)G
5 F 2 12 211
29 9 23G
22. A(C 2 B) 5 F 1 22 4 23
G(F 26 8 10 15
G 2 F 3 5
21 0G)
5 F 1 22 4 23
GF 29 3 11 15
G 5 F 1(29) 1 (22)(11) 1(3) 1 (22)(15)
4(29) 1 (23)(11) 4(3) 1 (23)(15)G 5 F 231 227
269 233G
23. 3 22 4 1 5 3(1) 2 (4)(22) 5 11
24. 24 5 2 21 5 (24)(21) 2 (5)(2) 5 26
25. 21 3 1 0 2 23
5 1 22 21 3
0 2
5 1
5 (4 2 45 1 0) 2 (10 1 3 1 0)
5 254
26. 2 0 21 5 23 2
1 4 6 2 0
5 23
1 4
5 (236 1 0 2 20) 2 (3 1 16 1 0) 5 275
27. F 3 4 4 5GF x
yG 5 F 6 7G
A21 5 2 1 } 1 F 5 24
24 3G 5 F 25 4
4 23G X 5 A21B 5 F 25 4
4 23GF 6 7G 5 F 22
3G The solution is (22, 3).
28. F 2 27 1 23GF x
yG 5 F 236
216G A21 5
1 } 1 F 23 7
21 2G 5 F 23 7
21 2G X 5 A21B 5 F 23 7
21 2GF 236
216G 5 F 24 4G
The solution is (24, 4).
29. F 5 3
29 26GF x yG 5 F 25
12G A21 5 2
1 } 3 F 26 23
9 5G 5 F 2 1
23 2 5 } 3 G X 5 A21B 5 F 2 1
23 2 5 } 3 GF 25
12G 5 F 2
25G The solution is (2, 25).
30. F 3 2
21 4GF x yG 5 F 15
233G A21 5
1 } 14 F 4 22
1 3G 5 F 2 } 7 2 1 } 7
1 } 14
3 } 14
G X 5 A21B 5 F 2 } 7 2 1 } 7
1 } 14
3 } 14
GF 15
233G 5 F 9
26G The solution is (9, 26).
Chapter 3, continued
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187Algebra 2
Worked-Out Solution Key
31. f 5 amount invested in 5% interest bond
s 5 amount invested in 7% interest bond
f 1 s 5 15,000
0.05f 1 0.07s 5 880
F 1 1 0.05 0.07GF f
sG 5 F 15,000
880G A21 5
1 } 0.02 F 0.07 21
20.05 1G 5 F 3.5 250
22.5 50G X 5 A21B 5 F 3.5 250
22.5 50GF 15,000
880G 5 F 8500
6500G The investor should invest $8500 in 5% interest bonds
and $6500 in 7% interest bonds.
32. c 5 number of children’s tickets
a 5 number of adult tickets
s 5 number of senior citizen tickets
c 1 a 1 s 5 800
3c 1 8a 1 5s 5 3775
c 5 2a
2a 1 a 1 s 5 800
3a 1 s 5 800
3(2a) 1 8a 1 5s 5 3775
14a 1 5s 5 3775
3a 1 s 5 800 3 (25) 215a 2 5s 5 24000
14a 1 5s 5 3775 14a 1 5s 5 3775
2a 5 2225
a 5 225
c 5 2(225) 5 450
450 1 225 1 s 5 800 → s 5 125
There were 450 children’s tickets, 225 adult tickets, and 125 senior citizen tickets sold.
33. b 5 speed of boat
c 5 speed of current
b 1 c 5 34
b 2 c 5 28
2b 5 62
b 5 31
31 1 c 5 34 → c 5 3
The boat travels 31 miles per hour in still water. The speed of the current is 3 miles per hour.
Standardized Test Preparation (p. 229)
1. Substitute these values into the fi rst equation:
21 1 2 1 (23) 0 2
22 Þ 2
Because 22 Þ 2, choice B can be eliminated.
2. Substitute the matrix in choice D for X in the equation:
F 3 4
2 3GF 1 0
0 1G 0 F 4 210
1 26G F 3 4
2 3G Þ F 4 210
1 26G Because the resulting matrices are not equal, choice D
can be eliminated.
Standardized Test Practice (pp. 230–231)
1. C;
d 1 (d 2 39.9) 5 600.9
2d 5 640.8
d 5 320.4
The top DVD grossed $320.4 million.
2. C;
3x 1 2y 5 8 Eq. 1 x 2 4y 5 216 Eq. 2
3(0) 1 2(4) 0 8 0 2 4(4) 0 216
8 5 8 ✓ 216 5 216 ✓
3. B;
Eliminate other possibilities.
4. A;
s 5 number of scarves
g 5 number of pairs of gloves
s 1 g 5 8 → g 5 8 2 s
6s 1 9g 5 66
6s 1 9(8 2 s) 5 66
23s 5 26
s 5 2
5. D;
A21 5 2 1 } 1 F 7 3
5 2G 5 F 27 23
25 22G X 5 A21B 5 F 27 23
25 22GF 4 6 0 21G 5 F 228 239
220 228G
Chapter 3, continued
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188Algebra 2Worked-Out Solution Key
6. B;
d 5 driving time
c 5 classroom instruction
b 5 observing
d 1 c 1 b 5 46
c 5 3d
b 5 d 2 4
d 1 3d 1 (d 2 4) 5 46
5d 5 50
d 5 10
7. C;
2x 2 3 1 2 } 3 x 1 2 2 5 9
0x 5 15
0 5 15
No solution
8. D;
Area of triangle with vertices
(1, 25), (2, 3), and (12, 2):
Area 5 6 1 } 2
1 25 1 2 3 1
12 2 1 1 25
2 3
12 2
5 6 1 } 2 [(3 2 60 1 4) 2 (36 1 2 2 10)]
5 40.5
Area of triangle with vertices
(21, 0), (5, 9), and (8, 0):
Area 5 6 1 } 2
21 0 1 5 9 1
8 0 1 21 0
5 9
8 0
5 6 1 } 2 [(29 1 0 1 0) 2 (72 1 0 1 0)]
5 40.5
9. B;
p 5 number of peony bulbs
h 5 number of phlox bulbs
l 5 number of lily bulbs
p 1 h 1 l 5 6
3.99p 1 2.61h 1 2.24l 5 19.43
h 5 2l
p 1 2l 1 l 5 6
p 1 3l 5 6
3.99p 1 2.61(2l) 1 2.24l 5 19.43
3.99p 1 7.46l 5 19.43
p 1 3l 5 6 3 (23.99) 23.99p 2 11.97l 5 223.94
3.99p 1 7.46l 5 19.43 3.99p 1 7.46l 5 19.43
24.51l 5 24.51
l 5 1
10. B; 4 6
26 29 5 4(29) 2 (26)(6) 5 0
Because the determinant is 0, the matrix has no inverse.
11. 2x 2 5y 5 210
x 1 4y 5 21 → x 5 21 2 4y
2(21 2 4y) 2 5y 5 210
213y 5 252
y 5 4
12. 21 1 2(22) 1 2z 5 7
2z 5 12
z 5 6
13. 2 1 4 3 25 0
1 0 21 2 1
3 25
1 0
5 (10 1 0 1 0) 2 (220 1 0 2 3) 5 33
14. 2F 5x 0
22 3G 2 F 3 21 4y 22G 5 F 17 1
210 8G F 10x 0
24 6G 2 F 3 21 4y 22G 5 F 17 1
210 8G F 10x 2 3 1
24 2 4y 8G 5 F 17 1
210 8G 10x 2 3 5 17 24 2 4y 5 210
10x 5 20 24y 5 26
x 5 2 y 5 3 } 2
x 1 4y 5 2 1 4 1 3 } 2 2 5 8
15. F 48 21 11 2
46 22 8 6
43 21 12 6
41 19 15 7
GF 2
0
1
1
G 5 F 109
106
104
104
G Detroit earned 109 points. Tampa Bay earned 106 points
and San José and Boston each earned 104 points.
16. m 5 number of matineé tickets
r 5 number of regular tickets
4m 1 6r 5 6000
r 5 890 1 m
4m 1 6(890 1 m) 5 6000
10m 5 660
m 5 66
r 5 890 1 66 5 956
The theater sold 66 matineé tickets and 956 regular tickets.
Chapter 3, continued
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189Algebra 2
Worked-Out Solution Key
17. a. y 5 3x 1 12.87 where x represents the number of years since 1996
b. y 5 1.86x 1 26.97 where x represents the number of years since 1996
c.
x
y
200
4 6 8 10 12 14 16 18
16
8
24
32
40
48
56
64
72
Years since 1996
Sp
end
ing
(d
olla
rs)
y 5 1.86x 1 26.97
y 5 3x 1 12.87
Sample answer: If you look at the graphs of the 2 equations, the two lines intersect at (12.5, 50.4). This means that about one-half of the way into the twelfth year, the video game spending will equal that of the box offi ce spending.
18 a. Area 5 6 1 } 2
0 0 1 20 0 1
x y 1 5 100
b. 6 1 } 2
0 0 1 20 0 1
x y 1 0 0
20 0
x y
5 100
6 1 } 2 [(0 1 0 1 20y) 2 (0 1 0 1 0)] 5 100
6 1 } 2 (20y) 5 100
20y 5 6200
y 5 610
Because (x, y) is in the fi rst quadrant, y must be positive. So, y 5 10.
c. The value of x does not matter as long as y 5 10.
Cumulative Review, Chs. 1–3 (pp. 232–233)
1. 3x2 2 5x2 2 8x 1 12x 1 3x 5 22x2 1 7x
2. 15x 2 6x 1 4x 1 10y 2 3y 5 13x 1 7y
3. 3x 1 6 2 4x2 1 3x 1 9 5 24x2 1 6x 1 15
4. 6x 2 7 5 22x 1 9
8x 5 16
x 5 2
Check: 6(2) 2 7 0 22(2) 1 9
5 5 5 ✓
5. 4(x 2 3) 5 16x 1 18
4x 2 12 5 16x 1 18
212x 5 30
x 5 2 5 } 2
Check: 4 1 2 5 } 2 2 3 2 0 16 1 2
5 } 2 2 1 18
222 5 222 ✓
6. 1 }
3 x 1 3 5 2
7 } 2 x 2
3 } 2
1 }
3 x 1
7 } 2 x 5 2
3 } 2 2 3
2 }
6 x 1
21 } 6 x 5 2
3 } 2 2
6 } 2
23
} 6 x 5 2
9 } 2
x 5 2 9 } 2 1 6 } 23
2 x 5 2
27 } 23
Check: 1 }
3 1 2
27 } 23 2 1 3 0 2
7 } 2 1 2
27 } 23 2 2
3 } 2
60
} 23
5 60
} 23 ✓
7. x 1 3 5 5
x 1 3 5 5 or x 1 3 5 25
x 5 2 or x 5 28
8. 4x 2 1 5 27
4x 2 1 5 27 or 4x 2 1 5 227
4x 5 28 or 4x 5 226
x 5 7 or x 5 2 13
} 2
9. 9 2 2x 5 41
9 2 2x 5 41 or 9 2 2x 5 241
22x 5 32 or 22x 5 250
x 5 216 or x 5 25
10. 6(x 2 4) > 2x 1 8
6x 2 24 > 2x 1 8
4x > 32
x > 8
0 2 4 6 8 10 12
11. 3 ≤ x 2 2 ≤ 8 3 1 2 ≤ x ≤ 8 1 2
5 ≤ x ≤ 10
1 3 5 7 9 1121
12. 2x < 26 or x 1 2 > 5
x < 23 or x > 3
1 3 5212325
13. x 2 4 < 5
25 < x 2 4 < 5
25 1 4 < x < 5 1 4
21 < x < 9
0 2 4 6 8 1022
14. x 1 3 ≥ 15
x 1 3 ≤ 215 or x 1 3 ≥ 15
x ≤ 218 or x ≥ 12
0 6 1226212218
Chapter 3, continued
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190Algebra 2Worked-Out Solution Key
15. 6x 1 1 < 23
223 < 6x 1 1 < 23
223 2 1 < 6x < 23 2 1
2 24
} 6 < x < 22
} 6
24 < x < 11
} 3
0 2 4 62426 22
113
16. (3, 2), (21, 25), m 5 25 2 2
} 21 2 3 5
7 } 4
The line rises.
17. (27, 4), (5, 23), m 5 23 2 4
} 5 2 (27)
5 2 7 } 12
The line falls.
18. (24, 26), (24, 4), m 5 4 2 (26)
} 24 2 (24)
5 10
} 0
No slope; the line is vertical.
19. 1 2 5 } 4 , 3 2 , 1 2 }
3 , 3 2 , m 5
3 2 3 }
2 }
3 2 1 2
5 } 4 2 5
0 }
23
} 12
5 0
The line is horizontal.
20.
1
x
y
21
21.
1
x
y
21
22.
1
x
y
22
23.
2
x
y
21
24.
1
x
y
21
25.
1
x
y
22
26.1
x
y
21
27.
1
x
y
21
28.
1
x
y
21
29.
1
x
y
21
30.
1
x
y
21
The relation is a function. The relation is not a function.
31. 4x 2 3y 5 32 4x 2 3y 5 32
22x 1 y 5 214 3 2 24x 1 2y 5 228
2y 5 4
y 5 24
4x 2 3(24) 5 32 → x 5 5
The solution is (5, 24).
32. 5x 2 2y 5 24 3 3 15x 2 6y 5 212
3x 1 6y 5 36 3x 1 6y 5 36
18x 5 24
x 5 4 } 3
5 1 4 } 3 2 2 2y 5 24 → y 5
16 } 3
The solution is 1 4 } 3 ,
16 }
3 2 .
33. x 2 y 1 2z 5 24 x 2 y 1 2z 5 24
2x 1 3y 1 z 5 9 3 (22) 24x 2 6y 2 2z 5 218
23x 2 7y 5 222
3x 1 y 2 4z 5 26 3x 1 y 2 4z 5 26
2x 1 3y 1 z 5 9 3 4 8x 1 12y 1 4z 5 36
11x 1 13y 5 30
23x 2 7y 5 222 3 11 233x 2 77y 5 2242
11x 1 13y 5 30 3 3 33x 1 39y 5 90
238y 5 2152
y 5 4
23x 2 7(4) 5 222 → x 5 22
22 2 4 1 2z 5 24 → z 5 1
The solution is (22, 4, 1).
34. B 2 3A 5 F 3 21 5 2G 2 3F 22 6
1 4G 5 F 3 2 3(22) 21 2 3(6)
5 2 3(1) 2 2 3(4)G 5 F 9 219
2 210G
Chapter 3, continued
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191Algebra 2
Worked-Out Solution Key
35. 2(A 1 B) 2 C
5 2(F 22 6 1 4G 1 F 3 21
5 2G) 2 F 24 8
27 12G 5 2F 1 5
6 6G 2 F 24 8
27 12G 5 F 2(1) 2 (24) 2(5) 2 8
2(6) 2 (27) 2(6) 2 12G 5 F 6 2 19 0G
36. (C 2 A)B 5 (F 24 8 27 12G 2 F 22 6
1 4G)F 3 21 5 2G
5 F 22 2 28 8GF 3 21
5 2G 5 F 22(3) 1 2(5) 22(21) 1 2(2)
28(3) 1 8(5) 28(21) 1 8(2)G 5 F 4 6
16 24G37. (B 1 C)D
5 (F 3 21 5 2G 1 F 24 8
27 12G)F 1 0 24
22 3 21G 5 F 21 7
22 14GF 1 0 24
22 3 21G 5 F 21(1) 1 7(22) 21(0) 1 7(3) 21(24) 1 7(21)
22(1) 1 14(22) 22(0) 1 14(3) 22(24) 1 14(21)G 5 F 215 21 23
230 42 26G38. A21 5 2
1 } 1 F 3 24
24 5G 5 F 23 4
4 25G39. A21 5
1 } 3 F 24 29
3 6G 5 F 2 4 } 3 23
1 2G
40. A21 5 2 1 } 10 F 1 22
24 22G 5 F 2 1 } 10 1 } 5
2 } 5 1 } 5 G 41. A21 5
1 } 24 F 28 28
22 25G 5 F 2 1 } 3 2 1 } 3
2 1 } 12 2 5 } 24 G
42. Area 5 6 1 } 2
0 0 1 15 10 1
8 25 1 0 0
15 10
8 25
5 6 1 } 2 [(0 1 0 1 375) 2 (80 1 0 1 0)]
5 147.5
The area of the playground is 147.5 square yards.
43. a. W
} T 5 R2
} R2 1 A2
W 5 TR2
} R2 1 A2
b. W 5 162(949)2
}} (949)2 1 (768)2
W ø 97.9
The estimate shows that the Red Sox won about 98 games, which is about the same as the actual number of games won.
44. a. g 5 21 2 4t
b.
Time (hours)
Gas
olin
e (g
allo
ns)
0 2 4 6 8 t
g
0
6
12
18
24(0, 21)
(5.25, 0)
c. Domain: 0 ≤ t ≤ 5.25
Range: 0 ≤ g ≤ 21
45. c 5 kp
3900 5 k(78,000)
k 5 0.05
c 5 0.05p
When p 5 125,000: c 5 0.05(125,000)
c 5 6250
If a house sells for $125,000, the real estate agent’s commission will be $6250.
Chapter 3, continued
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192Algebra 2Worked-Out Solution Key
46.
Years since 1994
Rec
ove
red
mat
eria
l(m
illio
ns
of
ton
s)
0 2 4 6 8 t
m
050
55
60
65
70
Sample answer:
Best-fitting line approximation: m 5 2.5t 1 52
When t 5 16: m 5 2.5(16) 1 52
m 5 92
In 2010, about 92 million tons of material willbe recovered.
47. s > 213, j ≤ 263, s 1 j > 472.5
0 100 200 300 4000
100
200
300
400
Snatch weight (kg)
Cle
an a
nd
jerk
wei
gh
t (k
g)
s
j
Chapter 3, continued
n2ws-03-c.indd 192 6/27/06 10:00:15 AM