F G

Chapter 3,
continued
F G
3. C;
175
270
BA 5 [ 0.03 0.05 0.08] 370
225
200
255
5 [0.03(175) 1 0.05(370) 1 0.08(200)
0.03(270) 1 0.05(225) 1 0.08(255)]
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 defined when the number
of columns in the first matrix is the same as the number
of rows in the second matrix.
4.
y
2
5 [39.75 39.75]
(5, 1)
x
22
Mary and Mark each made a total of $39.75 in
commission.
4. H;
H 1 N 1 3O 5 63
The lines appear to intersect at (5, 1).
2N 1 O 5 44
F G
Check: 2(5) 2 1 0 9
2H 1 O 5 18
1 1 3
A5
0
2
1
2
0
1
1
0
2
1
2
0
det A 5
{
5 1 3(1) 0 8
9 5 9 5.
{
3
1
1
1
0
2
1
2
0
2
858
y
x
21
(24, 22)
5 (2 1 2 1 0) 2 (12 1 0 1 0) 5 28
{
1 63 3
0 44 1
2 18 1
1 63
0 44
2 18
{
The lines appear to intersect at (24, 22).
Check: 2(24) 2 3(22) 0 22
22 5 22 N 5 }}
28
(44 1 126 1 0) 2 (264 1 18 1 0)
5 14
5 }}}
28
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
24 1 (22) 0 26
6.
26 5 26 y
(0, 6)
The atomic weight of nitrogen is 14.
5. Let c 5 bushels of corn.
1
Let s 5 bushels of soybeans.
x
22
Let w 5 bushels of wheat.
2.35c 1 5.40s 1 3.60w 5 4837
The lines appear to intersect at (0, 6).
c 1 s 1 w 5 1700
Check: 3(0) 1 6 0 6
F
GF G F G
G
c 2 3.25s 2 3.25w 5 0
2.35
5.40
3.60
c
4837
1
1
s
w
5 1700
0
0.7647
0.2353
0.5556 21.4690
0.1634
1
1
A21 5
F
23.25 23.25
0
F G
20.5556
1.7042 20.3987
1300
X 5 A21B 5
190
210
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.
20 1 2(6) 0 12
6 5 6 7.
12 5 12 3x 1 2y 5 5
32
6x 1 4y 5 10
22x 1 3y 5 27
33
26x 1 9y 5 81
13y 5 91
y57
3x 1 2(7) 5 5 l x 5 23
The solution is (23, 7).
8. 3x 1 5y 5 5
2x 2 3y 5 16
33
9x 1 15y 5 15
35
10x 2 15y 5 80
19x
5 95
x55
3(5) 1 5y 5 5 l y 5 22
The solution is (5, 22).
Algebra 2
Worked-Out Solution Key
163
Chapter 3,
2x 1 3y 5 9
2x 1 3y 5
23x 1 y 5 25
15. 6x 2 y 1 4z 5 6
9
9x 2 3y 5 275
3 (23)
11x
6x 2 y 1 4z 5
2x 1 2y 2 5z 5 242
26x 2 6y 1 15z 5 126
3 (23)
27y 1 19z 5 132
5 266
x
2x 2 3y 1 z 5 31
5 26
2(26) 1 3y 5 9 l y 5 7
22x 2 6y 1 2z 5
32
24y 2 3z 5
10. r 5 price of regular gasoline.
27y 1 19z 5 132
p 5 price of premium gasoline.
2r 1 p 5 0.30
24y 2 3z 5 20
3 14
28y 1 21z 5 2140
3 (27)
97z 5
214r 1 14p 5 4.2
z54
6x 2 (28) 1 4(4) 5 6 l x 5 23
p 5 2.12
2r 1 2.12 5 0.30 l r 5 1.82
The solution is (23, 28, 4).
Regular gas costs $1.82 per gallon and premium gas
costs $2.12 per gallon.
16. 5x 1 y 2 z 5 40
2x 1 3y 1 z 5 16
12. 2x 1 3y > 6
2x 1 2ya5
25x 1 15y 1 5z 5 80
35
2x 1 3y 1 z 5 16
10y 1 5z 5 60
1
1
y 2 z 5 40
x 1 7y 1 4z 5 44
y
y
5x 1
16y 1 4z 5 120
2x 2 ya8
x
388
27y 1 19(4) 5 132 l y 5 28
24p 5 50.88
11. 4x 1 y < 1
20
228y 1 76z 5 528
34
14r 1 10p 5 46.68
21
62
2x 1 2y 2 5z 5 242
2x 1 2y 2 5z 5 242
The solution is (26, 7).
14r 1 10p 5 46.68
6
x
21
16y 1 4z 5 120
35
10y 1 5z 5 60
34
280y 2 20z 5 2600
40y 1 20z 5
240y
13. x 1 3yq5
59
y
2x 1 2y < 4
240
5 2360
16(9) 1 4z 5 120 l z 5 26
y
5x 1 9 2 (26) 5 40 l x 5 5
The solution is (5, 9, 26).
17. w 5 number of wind instruments.
1
x
s 5 number of string instruments.
21
14.
p 5 number of percussion instruments.
x 2 y 1 z 5 10
Equation 1: w 1 s 1 p 5 15
23x 1 5y 2 z 5 218
22x 1 4y
Equation 2: w 5 2(s 1 p)
5 28
Equation 3: s 5 3
4x 1
4x 1 y 2 2z 5 15
23x 1 5y 2 z 5 218
3 (22)
y 2 2z 5 15
10x 2 9y
22x 1 4y 5 28
35
10x 2 9y 5 51
w 5 2(3 1 p) l w 5 6 1 2p
(6 1 2p) 1 3 1 p 5 15 l p 5 2
6x 2 10y 1 2z 5 36
w 5 6 1 2(2) l w 5 10
5 51
210x 1 20y 5 240
10x 2 9y 5
51
11y 5
11
y51
Ten students play wind instruments, three play string
instruments, and two play percussion.
18.
22x 1 4(1) 5 28 l x 5 6
F GF GF
F G
F GF GF
F G
4 25
2
1
3
21
3
27
4
5
5
6 2 1 1 z 5 10 l z 5 5
The solution is (6, 1, 5).
19.
21
8
2 23
1
7 24
6
21
5
5
164
Algebra 2
Worked-Out Solution Key
G
4 1 (21)
25 1 3
2 1 (27)
314
3 22
25
7
21 1 7
8 1 (24)
216
23 1 (21)
6
4
8 24
G
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
9.
continued
Chapter 3,
20.
continued
F GF GF
F
F
GF
F
F G
F GF
F
GF
F
F G
10 24
5
0
2
1
9
2
10 2 0
5
7
522
10 213
5
21.
22
3
5
21
6 22
5
5
22
3
6
8
23. 8
5
28 0 29
620
22 2 (29)
2 24
0
7
7
6
23(22)
23(3)
23(6)
8
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
25.
8(5)
8(21)
8(6)
8(22)
32
5
5
27.
7]
0 25
1 25
4 23
28
48 216
7
4
3
6
12
4(5) 1 (21)(3)
1(5) 1 7(3)
17 220
10
26
46
82
33.
G
6 23
5
2
1
0
G
G
22(5) 1 5(21)
0(6) 1 3(2)
0(23) 1 3(0)
0(5) 1 3(21)
215
6
0
23
G
G
{
2 25
y
G
11
5
9
25 24
1
5}
213 22
1
5
}
5
13
}
2
13
2}
13
4
13
1
}
}
5
13
}
11
2
}
13
1
2}
13
9
4
13
5
7
1
The solution is (7, 1).
F GF G F G
F GF
3
A
22(23) 1 5(0)
6
x
1
x
21 2
y
21
22(6) 1 5(2)
22
4
X 5 A21B 5
34.
5
1
A
1(4) 1 7(6)
3
1 0
0
1 0 50
1 70 20
F GF G F G
F GF G
F GF G F G
21
1(22) 1 7(12)
0
{
0
0
0 50
70 20
1 ft2
4(4) 1 (21)(6)
22
G
(1750 in.2) 1 }
ø 12.2 ft2
144 in.2 2
4(22) 1 (21)(12)
G
G
2
8
1
32. Area 5 6}
2
10
5 22
G
319.99
549.99
You will need approximately 12.2 square feet
of material.
1(25) 1 (25)(23)
1
5
31.
21(2) 1 (21)(29)]
1(0) 1 (25)(4)
21
6,389,810
GF
5 1750 square inches
11(25) 1 7(23)
220
6,869,810
1
G
G
5000
5 6}2 [(0 1 0 1 0) 2 (3500 1 0 1 0)]
11(0) 1 7(4)
4
5
G
30.
28 276
5
26.
40
G
G
4000 10,000
109.99
{ { 5 24(8) 2 5(2) 5 242
3 25
{ 2 6 { 5 3(6) 2 2(25) 5 28
3 0
{ 1 6 { 5 3(6) 2 1(0) 5 18
24
5
2
7
5
29 218
8(4)
64
6
8000
F
26 29
F GF
F
F
F GF
F
F
F GF
F
F
11
215
5
8(8)
5 [21(8) 1 (21)(26)
5 [22
GF
6000
Warehouse 2 6,389,810
29.
23(5)
F
F
5000
Total Value ($)
Warehouse 1 6,869,810
G
21 2 (28)
5
24. [21 21]
G
5
525
5
6 22
21
5
G
28. AB 5
327
5
4
26
127
G
22 2 (24)
5
22. 23
3
24 7
2
24 2 9
5
1 2 21
5 }7
1
3
X5A B5
12
1
}
2
7
2}7
1
7
}
5
3
7
}
G
F GF G F G
1
2
7
2}7
21
1
7
}
3
7
12
}
21
21
}
5
22
5
The solution is (22, 5).
Algebra 2
Worked-Out Solution Key
165
Chapter 3,
F GF G F G
F GF G
F GF G F G
3
2
x
4 23
y
1
A21 5 2}
17
5
211
8
23 22
3
17
}
21
X5A B5
5
3
24
3
}
17
2
}
17
4
17
3
17
4.
y
2
17
211
8
3
4
} 2}
17
17
5
21
24
(4, 0)
2
The lines appear to intersect at (4, 0).
Check: 3(4) 2 0 0 12
6. x 2 3yq9
1
}x 2 ya
3
y22
3x 2 y 5 2
1
y
1
x
21
(1, 1)
1
1
24 5 24 5. 2x 1 y < 6
Chapter 3 Test (p. 227)
y
24 1 8(0) 0 24
12 5 12 The solution is (21, 24).
1.
x
3x 2 y 5 12
} 2}
}
2x 1 8y 5 24
2
x
The solution is a line.
7. x 2 2ya214
Check: 3(1) 2 1 0 2
y22{x{ 1 5
4(1) 1 1 0 5
252
y
8. 23x 1 4y > 212
yq{x{
The lines appear to intersect at (1, 1).
1
y
y
555
26x 2 2y 5 214
1
1
x
2
(4, 25)
9. 3x 1 y 5 29
x
32
26 5 26 y
1
7x
26(4) 2 2(25) 0 214
214 5 214 3
x 2 2 y 5 23
x
5 228
x 5 24
3(24) 1 y 5 29 l y 5 3
The solution is (24, 3).
10. 2x 1 3y 5 22 3 (22)
1
6x 1 2y 5 218
x 2 2y 5 210
x 2 2y 5 210
The lines appear to intersect at (4, 25).
Check: 4 1 2(25) 0 26
x
21
22
x 1 2y 5 26
3.
x
22
4x 1 y 5 5
2.
y
24x 2 6y 5
4
4x 1 7y 5 26
4x 1 7y 5 26
y 5 22
2x 2 3y 5 15
The lines do not intersect, so there is no solution.
2x 1 3(22) 5 22 l x 5 2
The solution is (2, 22).
11. x 1 4y 5 226 l 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 l x 5 6
The solution is (6, 28).
166
Algebra 2
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
35.
continued
Chapter 3,
continued
12. x 2 y 1 z 5 23 3 2
2x 2 2y 1 2z 5 26
4x 1 2y 2 z 5
4x 1 2y 2 z 5 2
6x
2x 2 y 1 5z 5 4
1 z 5 24
4x 2 2y 1 10z 5 8
32
4x 1 2y 2
4x 1 2y 2 z 5 2
8x
6x 1 z 5 24
16. C 2 3B 5
2
5
z5 2
5 46
246x
5
x 5 21
6(21) 1 z 5 24 l z 5 2
The solution is (21, 4, 2).
19. AC 5
3
x1 y1 z5
2x 1 3y 1 2z 5 28
5
4y 1 3z 5 25
5y 1 z 5 2
3 (23)
215y 2 3z 5 26
5
4y 1 3z 5 25
4y 1 3z 5 25
211y 5 211
x 1 1 1 (23) 5 3 l x 5 5
F
F
F
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
x 1 y 1 3z 5 19
4x 2 10y 2 2z 5
24y 2 z 5
14
4x 1 4y 1 12z 5
76
24x 1 6y 1
24x 1 6y 1 z 5 220
z 5 14
56
252y 2 13z 5 182
3 13
5
F
1
2
5
22. A(C 2 B) 5
5
5
5
4
3
,} .
The solution is 1 2}3 , 2}
3 32
15. 2A 1 B 5 2
5
17 26
23.
F GF G
F
GF G
1 22
4 23
1
3
5
21 0
2(1) 1 3
2(22) 1 5
2(4) 1 (21)
2(23) 1 0
3
6 22
1
G
4(3) 1 (21)
4(22) 1 3
4(2) 1 6
4(0) 1 (22)
4(21) 1 1
0
G
11 25
14 22 23
8
26
10 15
G
5
5
24.
1
7 26
25.
(F G F G)F
1 22
4 23
1
F GF
5
21
3 22
21 0
2
0 21
3
3
21
3 22
3 23
2
0 21
4
G
4(3) 1 3(0)
G
G
4(22) 1 3(21)
3(21) 1 (23)(2) 3(3) 1 (23)(0) 3(22) 1 (23)(21)
2x 2 5 2} 2 } 5 19 l x 5 2}
4
4 21
4(21) 1 4
4(21) 1 3(2)
26
26
3
G
254 213
241 2}
2 z 5 14 l z 5 }
32
3
17
3
15
226 222
17
y 5 2}
3
17
1
4(8) 1 (23)(15)
5 238
242y
GF
4(26) 1 (23)(10)
10y 1 13z 5 56
10y 1 13z 5 56
13
1(8) 1 (22)(15)
z 5 220
10y 1 13z 5
24y 2
215 27
5
1(26) 1 (22)(10)
5
34
24x 1 6y 1 z 5 220
34
2 0 21
4 23
The solution is (5, 1, 23).
24x 1 6y 1 z 5 220
21 3 22
1 22
21. (A 1 B)D 5
32
8 2 3(5)
15 2 3(0)
the number of rows in E.
5(1) 1 z 5 2 l z 5 23
2x 2 5y 2 z 5 17
26 2 3(3)
10 2 3(21)
20. Not possible; the number of columns in D does not equal
y51
14.
5
21 0
G
F GF G
F
F G
5
21 2 y 1 2 5 23 l y 5 4
13.
3
23
10 15
18. 4D 1 E 5 4
8x 1 9z 5 10
8x 1 9z 5 10
8
26
17. Not possible; A and D do not have the same dimensions.
1 9z 5 10
254x 2 9z 5 36
3 (29)
F G F G
F
GF
F
G
F G(F G F G)
F GF G
F
G
F G
2 12 211
29
9
1 22
23
26
4 23
1 22
4 23
8
10 15
29
2
3
5
21 0
3
11 15
1(29) 1 (22)(11)
1(3) 1 (22)(15)
4(29) 1 (23)(11)
4(3) 1 (23)(15)
231
227
269
233
{ { 5 3(1) 2 (4)(22) 5 11
24
5
{ 2 21 { 5 (24)(21) 2 (5)(2) 5 26
3 22
4
1
21 3
1 21 3
0 2 23 0 2 5 (4 2 45 1 0) 2 (10 1 3 1 0)
5 1 22 5 1
5 254
{
{
Algebra 2
Worked-Out Solution Key
167
Chapter 3,
2
0 21 2
0
5 23
2 5 23
4
1
4
6 1
{ {
F GF G F G
F GF G
F GF G F G
1
5 (236 1 0 2 20) 2 (3 1 16 1 0) 5 275
27.
3
4
x
4
5
y
6
5
3
24
6
X 5 A21B 5
4 23
7
x
1 23
5
y
7
1 23
A21 5 }1
21
X5A B5
c 5 2a
216
23
7
21
2
23 7
236
21 2
216
24
5
F
5
4
GF G F G
F GF G
F GF G F G
3
x
29 26
y
12
2
1 26 23
A21 5 2}3
9
5
5
2
1
5
23 2}3
1
X 5 A21B 5
25
5
5
12
23 2}3
2
25
The solution is (2, 25).
30.
F GF G F G
F GF G
F GF G F G
3 2
x
21 4
y
21
A
1
5}
14
5
4 22
1
21
X5A B5
3
15
5
1
1
2
7
2}7
1
}
14
3
}
14
}
2
7
2}7
15
1
}
14
3
}
14
233
5
9
26
The solution is (9, 26).
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
0.05 0.07
168
GF G F G
1
15,000
f
s
5
Algebra 2
Worked-Out Solution Key
880
6500
3a 1 s 5 800
14a 1 5s 5 3775
3a 1 s 5 800
3 (25)
14a 1 5s 5 3775
215a 2 5s 5 24000
14a 1 5s 5
2a
3775
5 2225
a 5 225
c 5 2(225) 5 450
450 1 225 1 s 5 800 l 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
233
}
8500
3(2a) 1 8a 1 5s 5 3775
25
5
5
3c 1 8a 1 5s 5 3775
2a 1 a 1 s 5 800
The solution is (24, 4).
29.
880
236
5
21 2
15,000
50
c 1 a 1 s 5 800
F GF G F G
F GF G
F GF G F G
2 27
3.5 250
22.5
50
22.5
s 5 number of senior citizen tickets.
3
The solution is (22, 3).
28.
F
1
a 5 number of adult tickets.
22
5
20.05
3.5 250
5
32. c 5 number of children’s tickets.
4 23
4
25
4
25
5
X 5 A21B 5
GF G
GF G F G
0.07 21
The investor should invest $8500 in 5% interest bonds
and $6500 in 7% interest bonds.
7
5 24
1
A21 5 2}1
F
A21 5 }
0.02
31 1 c 5 34 l c 5 3
The boat travels 31 miles per hour in still water. The
speed of the current is 3 miles per hour.
TAKS Practice (pp. 230 –231)
1. B;
Choice B represents the front view of the pyramid.
2. G;
(x, y) l (4x, 4y)
R(21, 4) l R9(24, 16)
The coordinates of vertex R9 are (24, 16).
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
26.
continued
Chapter 3,
continued
3. B;
9. A;
The front view shows that the solid has four levels
of cubes.
Use the Linear Pair Postulate and the fact that the interior
angles in a pentagon add up to 5408.
The top view shows that the first level has 11 cubes.
(180 2 5x) 1 (180 2 3x) 1 (180 2 5x) 1
The front and side views show that the second level has
3 cubes.
(180 2 9x) 1 (180 2 2x)
The front and side views show that the third and fourth
levels each have 1 cube.
224x 5 2360
So, the total number of cubes needed to construct the
solid is 11 1 3 1 1 1 1 5 16 cubes.
The value of x is 15.
4. J;
SA 5 SA (top) 1 SA (bottom) 1 SA (front) 1
SA (back) 1 2SA (side)
5 540
x 5 15
10. H;
1
V 5 }3Bh
1
5 3(40)(12) 1 36(40) 1 3(8)(40) 1 24(40) 1
Original pyramid: 36 5 }3 Bh
New pyramid: V 5 }3 B(2h) 5 21 }3Bh 2 5 2(36) 5 72
1
2[8(36) 1 8(24) 1 8(12)]
5 5952
The solid has a surface area of 5952 square inches.
The volume of the new pyramid is 72 cubic centimeters.
11. C;
5. C;
3x 1 2y 5 8
x 2 4y 5 216
3(0) 1 2(4) 0 8
0 2 4(4) 0 216
858
3 4
2 7
1
2
9
4
} }x 2 1 5 }
3
2
4
7
}x 2 1 5 }
216 5 216 6. H;
5
2
4
7
}x 5 }
Let t 5 gross of top-selling DVD.
35
x5}
8
Let s 5 gross of second top-selling DVD.
35
t 1 s 5 600.9
t 5 s 1 39.9 l s 5 t 2 39.9
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
1
t 1 (t 2 39.9) 5 600.9
2t 5 640.8
t 5 320.4
The top-selling DVD grossed $320.4 million.
7. C;
y 2 y1 5 m(x 2 x1)
2
y 2 1 5 }5 (x 2 (24))
2
8
2
13
The solution of the equation is x 5 }
.
8
12. G;
Total charge 5 Base cost 1 Hourly cost + Hours worked
c 5 20 1 5h
13. Let r 5 regular tickets.
Let m 5 matinee tickets.
6r 1 4m 5 6000
r 5 890 1 m
6(890 1 m) 1 4m 5 6000
10m 5 660
y 5 }5 x 1 }5 1 1
y 5 }5 x 1 }
5
m 5 66
The theater sold 66 matinee tickets.
5y 5 2x 1 13
Cumulative Review, Chs. 1–3 (pp. 232–233)
22x 1 5y 5 13
1. 3x 2 2 5x 2 2 8x 1 12x 1 3x 5 22x 2 1 7x
The equation 22x 1 5y 5 13 passes through the point
2
(24, 1) and has slope }5 .
8. J;
The range could be used to determine that the
temperatures varied by 258F.
2. 15x 2 6x 1 4x 1 10y 2 3y 5 13x 1 7y
3. 3x 1 6 2 4x 2 1 3x 1 9 5 24x 2 1 6x 1 15
4. 6x 2 7 5 22x 1 9
8x 5 16
x52
Check: 6(2) 2 7 0 22(2) 1 9
555
Algebra 2
Worked-Out Solution Key
169
Chapter 3,
continued
13. {x 2 4{ < 5
5. 4(x 2 3) 5 16x 1 18
4x 2 12 5 16x 1 18
25 < x 2 4 < 5
212x 5 30
25 1 4 < x < 5 1 4
21 < x < 9
5
x 5 2}2
5
5
Check: 41 2}2 2 3 2 0 16 1 2}2 2 1 18
222 5 222 7
3
1
} x 1 3 5 2} x 2 }
6.
2
2
3
7
3
1
} x 1 } x 5 2} 2 3
2
2
3
3
6
21
2
} x 1 } x 5 2} 2 }
6
2
2
6
9
23
}x 5 2}
2
6
9 6
x 5 2}2 }
23
27
}
x 5 223
22
6
8
1 0
x 1 3a215
or
x 1 3q15
xa218
or
xq12
0
6
12
15. {6x 1 1{ < 23
223 < 6x 1 1 < 23
223 2 1 < 6x < 23 2 1
24
22
<x<}
2}
6
6
11
1
24 < x < }
3
2
7. {x 1 3{ 5 5
x1355
or
x 1 3 5 25
x52
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
or
13
x 5 2}
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
or
x 5 25
10. 6(x 2 4) > 2x 1 8
11
3
26
24
0
22
2
4
6
25 2 2
7
16. (3, 2), (21, 25), m 5 } 5 }
21 2 3
4
The line rises.
23 2 4
7
17. (27, 4), (5, 23), m 5 } 5 2}
12
5 2 (27)
The line falls.
4 2 (26)
10
18. (24, 26), (24, 4), m 5 } 5 }
0
24 2 (24)
No slope; the line is vertical.
5
323
0
2
19. 2}, 3 , }, 3 , m 5 } 5 }
50
23
5
4
3
2
}
} 2 1 2} 2
12
4
1
21
2
3
The line is horizontal.
20.
21.
y
y
1
6x 2 24 > 2x 1 8
21
4x > 32
x>8
0
11.
2
4
x
1
6
8
10
x
21
12
22.
3ax 2 2a8
y
3 1 2axa8 1 2
5axa10
1
22
21
1
3
5
7
9
x
11
12. 2x < 26
or
x12>5
x < 23
or
x>3
23.
24.
y
y
2
1
25
170
23
21
1
Algebra 2
Worked-Out Solution Key
3
5
21
x
21
x
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
2
x 5 216
4
218 212 26
7 27
3
1 27
1 3 0 2}2 2}
2 }2
Check: }3 2}
23
23
60
60
}5}
23
23
x57
2
14. {x 1 3{q15
1 2
1
0
Chapter 3,
25.
continued
26.
y
23x 2 7(4) 5 222 l x 5 22
y
1
22 2 4 1 2z 5 24 l z 5 1
x
21
The solution is (22, 4, 1).
1
34. B 2 3A 5
x
22
5
27.
28.
y
y
5
1
x
21
30.
y
y
1
1
x
The relation is a function.
31. 4x 2 3y 5 32
The relation is not
a function.
4x 2 3y 5
22x 1 y 5 214
5
x
21
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
The solution is (5, 24).
18x
5
36
24
5
4
x 5 }3
5
16
4
5 }3 2 2y 5 24 l y 5 }
3
4 16
The solution is }3 , }
.
3
1 2
1
33.
2x 1 3y 1 z 5 9
5
2
x 2 y 1 2z 5 24
x 2 y 1 2z 5 24
3 (22)
24x 2 6y 2 2z 5 218
23x 2 7y
3x 1
3x 1 y 2 4z 5 26
2x 1 3y 1 z 5 9
34
5 222
y 2 4z 5 26
3 11
11x 1 13y 5 30
33
9 219
2 210
3 21
1
1 4
5
2
24
8
8
24
2
27 12
F GF G
F
GF G
(F G F G)F G
F GF G
F
G
F G
1
5
6
6
2
27 12
2(1) 2 (24)
2(5) 2 8
2(6) 2 (27)
2(6) 2 12
8
24
5
6
2
19
0
6
22
2
27 12
3 21
1 4
22 2
3 21
28 8
5
5
2
2
6
24
F
(F G F
3 21
5 30
y54
8
27
12
21
7
1
0 24
22
14
22
3 21
21(1) 1 7(22)
5
F
G)F
G
1
0 24
22
3 21
21(0) 1 7(3)
G
F GF
F GF
215
21 23
230
42 26
3
1
38. A21 5 2}
1 24
21
39. A
90
238y 5 2152
24
2
G
21(24) 1 7(21)
22(1) 1 14(22) 22(0) 1 14(3) 22(24) 1 14(21)
233x 2 77y 5 2242
33x 1 39y 5
1
5
F GF
8x 1 12y 1 4z 5 36
11x 1 13y
23x 2 7y 5 222
2 2 3(4)
37. (B 1 C)D
15x 2 6y 5 212
3x 1 6y 5
4
21 2 3(6)
4
5 16
4x 2 3(24) 5 32 l x 5 5
3x 1 6y 5 36
1
22(3) 1 2(5) 22(21) 1 2(2)
5 28(3) 1 8(5) 28(21) 1 8(2)
4
y 5 24
32. 5x 2 2y 5 24 3 3
6
5 2 3(1)
5
32
2y 5
22
3 2 3(22)
36. (C 2 A)B 5
24x 1 2y 5 228
32
2
22 6
52
21
5
23
(F G F G) F G
x
52
29.
3 21
35. 2(A 1 B) 2 C
1
21
F G F G
F
G
F G
21
40. A
1 24
3
3
5}
F
24
5
29
1
1
10 24
5 2}
5
6
5
22
22
G
23
4
4
25
4
2}3 23
1
2
G
F G
1
5
G
G
1
5
2}
10
}
2
5
}
}
1
5
Algebra 2
Worked-Out Solution Key
171
Chapter 3,
G
1
41. A21 5 }
24 22
28
0
1
42. Area 5 6} 15
2
8
0
10
25
25
{
F G
1
5
2}3
1
2}
12
1
2}3
5
2}
24
1 0
0
1 15 10
1 8 25
{
46.
m
7
0
Recovered material
(millions of tons)
F
28
continued
5
6
0
6
5
5
0
0
1
5 6}2 [(0 1 0 1 375) 2 (80 1 0 1 0)]
5 147.5
R
W
43. a. } 5 }
T
R 2 1 A2
Best-fitting line approximation: m 5 2.5t 1 52
47. s > 213, j a 263, s 1 j > 472.5
The estimate shows that the Red Sox won about 98
games, which is the same as the actual number of
games won.
44. a. g 5 21 2 4t
(0, 21)
Clean and jerk weight (kg)
j
W ø 97.9
400
300
200
100
0
18
0
100 200 300 400
Snatch weight (kg)
s
12
6
0
(5.25, 0)
0
2
4
6
8 t
Time (hours)
c. Domain: 0ata5.25
Range: 0aga21
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.
Algebra 2
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Gasoline (gallons)
t
8
In 2010, about 92 million tons of material will
be recovered.
162(949)2
b. W 5 }}
(949)2 1 (768)2
172
6
m 5 92
TR2
W5}
2
R 1 A2
45.
4
Years since 1994
When t 5 16: m 5 2.5(16) 1 52
2
g
24
2
Sample answer:
The area of the playground is 147.5 square yards.
b.
0