Chapter 1, continued

Chapter 1,
continued
8. 11y 1 2xy 5 9
5. 4x 1 9 < 25
y (11 1 2x) 5 9
4x < 16
x<4
9
y5}
11 1 2x
2
4
6
4
6
8
24
22
0
6. 1 2 3x q 214
9
y5}
11 1 2(25)
23x q 215
x a5
9
y5}
11 2 10
0
9
y 5 }1
2
7. 5x 2 7 a6x
2x 2 7 a0
2x a7
x q 27
y59
9. 0
13 26 39
113 113 113
25 22 1
4
13 1313
y 5 13x
y 5 3x 2 5
28
11.
26
8. 3 2 x > x 2 9
Fee for
5 1.5 + later
lessons
5 1.5 +
x
Total
amount
315
Number
Fee for
1 of later + later
lessons
lessons
1
9
+
x
The first lesson costs 1.5(30) 5 $45 and each additional
lesson costs $30.
Cost
per
5
calla
lily
Total
Cost
52
Number
Cost
+ of calla 1 per
lilies
peony
Number
+ of
peonies
+
+ (12 2 x)
3.50
5
x
1
5.50
52 5 3.5x 1 5.5(12 2 x)
28
75x
1. x > 25
8
0
22
0
212 210 28 26 24 22
3
2
4
or x 2 3 q 7
x q 10
or
5
7
12. 3x 2 1 < 21
9
or
11
2x 1 5 q 11
3x < 0
or
2x q 6
x<0
or
xq3
0
21
1
2
3
13. 15 aC a30
2
15 a}5(F 2 32) a30
2. x a3
9
0
2
4
27 aF 2 32 a54
6
59 aF a86
3. 23 ax < 1
23
21
4
211 ax a3
1.6 Guided Practice (pp. 41– 44)
4. x < 1
6
23 a2x a11
Lesson 1.6
22
4
10. 28 a2x 2 5 a6
x a5
You bought 7 calla lilies and 12 2 7 5 5 peonies.
0
24
11. x 1 4 a9
52 5 3.5x 1 66 2 5.5x
24
2
28 < 2x < 12
24 < x < 6
214 5 22x
26
0
9. 21 < 2x 1 7 < 19
30 5 x
12.
3 2 2x > 29
22x > 212
x<6
22
315 5 10.5x
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
0
22
When x 5 25:
22
0
21
1
Skill Practice
or x q 2
0
1
1.6 Exercises (pp. 44–47)
2
3
1. The set of all points on a number line that represent
solutions of an inequality is called the graph of
the inequality.
Algebra 2
Worked-Out Solution Key
25
Chapter 1,
continued
2. Subtraction
18. 5ax < 10
If a, b, and c are real numbers and a > b, then
a 2 c > b 2 c.
3
If a, b, and c are real numbers, c > 0, and a > b, then
ac > bc.
Division (positive)
If a, b, and c are real numbers, c > 0, and a > b, then
a b
} > }.
c c
1
2
4
6
20. xa21 or x > 1
22
0
21
1
2
21. x > 22 or x < 25
25
If a, b, and c are real numbers, c < 0, and a > b, then
ac < bc.
24
23
22
21
22. x 1 4 > 10
x>6
Division (negative)
If a, b, and c are real numbers, c < 0, and a > b, then
a b
}
c<}
c.
0
2
4
6
8
23. x 2 3a25
xa22
3. x > 4
0
22
2
4
23
6
22
0
21
1
24. 4x 2 8q24
4. x < 21
4xq4
26
24
22
2
0
2
24
22
0
2
xq1
5. xa25
1
0
21
26
2
3
25. 15 2 3x > 3
6. xq3
23x > 212
2
4
6
x<4
8
7. 6qx
0
22
0
2
4
6
2
4
6
26. 11 1 8xq7
8
8xq24
8. 22 < x
1
24
0
22
2
xq2}2
4
9. xq23.5
24
22
23
22
21
0
1
2
3
4
10. x < 2.5
0
12. xa0 or x > 2
13. x < 22 or x > 4
14. 23 < xa9
1
xa6
0
2
3
4
16. 2axa5
2
2
4
6
8
28. 2x 2 6 > 3 2 x
3x 2 6 > 3
3x > 9
4
6
x>3
8
17. 23 < x < 4
22
5
2
3
27. 4 1 }xa13
2
3
}xa9
2
2
xa9 }3
15. C; x < 21 or xq3
0
1
1 2
11. 23axa1
23 22 21
0
21
0
0
2
Algebra 2
Worked-Out Solution Key
4
2
4
6
8
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
0
26
9
0
22
Multiplication (negative)
24
7
19. x < 0 or x > 2
Multiplication (positive)
0
5
Chapter 1,
continued
39.
29. 4x 1 14 < 3x 1 6
23 < 4 2 xa3
23 2 4 < 4 2 x 2 4a3 2 4
x 1 14 < 6
27 < 2xa21
x < 28
28
26
24
7 > xq1
0
22
0
30. 5 2 8xa19 2 10x
5 1 2xa19
40.
2xa14
4
6
2 1 1 < 3x 2 1 1 1a6 1 1
3 < 3xa7
2
4
6
1 < xa}7
8
3
31. 21x 1 7 < 3x 1 16
1
23
18x 1 7 < 16
0
18x < 9
41.
1
x < }2
21
1
8
2 < 3x 2 1a6
xa7
0
2
1
2
3
4
24a2 1 4x < 0
24 2 2a2 1 4x 2 2 < 0 2 2
0
1
2
26 a4x < 22
3
3
1
2}2ax < 2}2
32. 18 1 2xa9x 1 4
18a7x 1 4
23
14a7x
2ax
2
4
4
23a}3xa1
2x 2 8 > 4x 1 6
4
28 > 2x 1 6
1 2
1 2
4
4
23 }3 axa1 }3
214 > 2x
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
1
0 2 3a}3x 1 3 2 3a4 2 3
3
33. 2(x 2 4) > 4x 1 6
24axa}4
27 > x
28
0
21
0a}3x 1 3a4
42.
1
0
21
22
3
26
24
1
0
22
13
34. When divided by a negative number, the inequality must
be reversed.
24
0
22
43. x 1 1 < 23
2x 1 8a6x 2 4
xq3
24
35. The inequality should not reverse with subtraction.
or
x>2
0
22
44. x 2 4a26
10 1 3x > 5x
4
or x 2 2 > 0
x < 24
24xa212
2
2
4
or x 1 2 > 5
or
xa22
x>3
10 > 2x
24
5>x
36. Sample answer: 2x 1 3 > 13 and x 1 5 > 10
37.
45. 2x 2 3a24
22
38.
4
3x 1 1q4
4
8
3xq3
or
xa2}2
26 < x < 3
0
or
1
25 2 1 < x 1 1 2 1 < 4 2 1
24
2
2xa21
25 < x 1 1 < 4
28
0
22
0
21
46. 2 1 3x < 213
2ax 2 3a6
xq1
1
or
2
4 1 2x > 7
3x < 215
2 1 3ax 2 3 1 3a6 1 3
x < 25
5axa9
2x > 3
3
or
x > }2
1
3
5
7
9
12
11
26
24
22
0
2
Algebra 2
Worked-Out Solution Key
27
Chapter 1,
47. 0.3x 2 0.5 < 21.7
continued
or
55. a. Lowland elevations: 0ae < 0.4xq2.4
b. Alpine elevations: 2000ae < 2429
0.3x < 21.2
x < 24
0
24
48. 2x 2 4q1
4
or
Subalpine elevations: 1400ae < 2000
xq6
Alpine and subalpine elevations: 1400ae < 2429
8
c. Elevations not in mountain zone:
2 2 5xa28
2xq5
0 a e < 500 or 2429 > eq1400
25xa210
xa25
56. D;
xq2
25
28
0
24
4
49. 2(x 2 4) > 2x 1 1
8
18
50. 4x 2 5a4(x 1 2)
2x 2 8 > 2x 1 1
4x 2 5a4x 1 8
28 ò 1
all real numbers
51. 2(3x 2 1) > 3(2x 1 3)
30
An inequality is 18 1 3(t 2 2)a30.
57. 50aFa95
5
18a}9 Ca63
5
Cost per
1 class
5
is m < 45 or m > 70.
Number
Amount you
+ of
a
can spend
classes
+
x
a
100
5xa50
xa10
c. Let k 5 1.61m.
45ama70
1.61(45)a1.61ma1.61(70)
72.45aka112.7
An inequality for the legal speeds (in kilometers per
hour) is 72.45aka112.7.
m < 45
You can attend up to ten classes.
Amount
Camera
Number
Cost of
+ of days a you can
1 cost
videotapes
spend
per day
35
1
45
+
x
a
250
45xa215
43
xa}
9
7
xa4}9
You can rent the video camera for 4 complete days
or fewer.
Temperature
Gear
Inequality
608F to 658F
Full wetsuit
60aT < 65
658F to 728F
Full leg wetsuit
65aT < 72
728F to 808F
Wetsuit trunks
72aT < 80
808F or warmer
No special gear
80aT
or
1.61m < 1.61 (45)
or
k < 72.45
or
m > 70
1.61m > 1.61(70)
k > 112.7
An inequality for the illegal speeds (in kilometers per
hour) is k < 72.45 or k > 112.7.
59. a. Amy: (0.65)84 1 (0.15)80 1 (0.20)wq85
Brian: (0.65)80 1 (0.15)100 1 (0.20)xq85
Clara: (0.65)75 1 (0.15)95 1 (0.20)yq85
Dan: (0.65)80 1 (0.15)90 1 (0.20)zq85
b. Amy:
(0.65)84 1 (0.15)80 1 (0.20)wq85
54.6 1 12 1 (0.2)wq85
(0.2)wq18.4
wq92
Brian:
(0.65)80 1 (0.15)100 1 (0.20)xq85
52 1 15 1 (0.2)xq85
(0.2)xq18
xq90
28
a
is 45ama70.
52.
54.
(t 2 2)
b. An inequality for the illegal speeds (in miles per hour)
Problem Solving
53.
+
3
Amount
a you can
spend
58. a. An inequality for the legal speeds (in miles per hour)
22 ò 9
no solution
1
1
10aCa35
6x 2 2 > 6x 1 9
50
Hours
Cost
st
+ after 1
1 per
two hours
hour
50a}9 C 1 32a95
25a8
no solution
Fee
to
join
Cost for
1st two
hours
Algebra 2
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
28
or
Chapter 1,
continued
Problem Solving Workshop 1.6 (pp. 48–49)
Clara:
(0.65)75 1 (0.15)95 1 (0.20)yq85
1. An equation is y 5 200 2 35x.
48.75 1 14.25 1 (0.2)yq85
y < 2500
(0.2)yq22
200 2 35x < 2500
yq110
235x < 2700
Dan:
x > 20
(0.65)80 1 (0.15)90 1 (0.20)z q
2. Using a table:
1 1 zq85
Total
cost
(0.2)zq19.5
zq97.5
y
c. A grade of 85 or better is possible for Amy, Brian,
and Dan. Clara needs at least a 110 score on her final
exam and this is impossible because the maximum test
score is 100 points.
X
13
14
15
16
17
18
19
X=18
Number Cost
Minimum Cost
Number Maximum
per
of
per
+
you can a
1
+ of flash a you can
daylight daylight flash
spend
camera cameras camera cameras spend
a 2.75 +
1 4.25 + (20 2 x)a
x
1
1
Cost per
chocolate
0.75
+
Number of
chocolates
+
x
Entering y 5 3 1 0.75x into a graphing calculator and
scrolling through the table of values shows that y 5 16.5
when x 5 18. So, you can buy 18 chocolates or fewer.
60.
65
Cost
5 of
card
5
3
Y1
12.75
13.5
14.25
15
15.75
16.5
17.25
75
Using a graph:
65a2.75x 1 4.25(20 2 x)a75
Entering y 5 (3 1 0.75xa16.5) into a graphing
calculator and using the trace feature suggests that
the inequality is true when xa18.
65a2.75x 1 85 2 4.25xa75
65a85 2 1.5xa75
220a21.5xa210
Y1=(3+.75x 16.5)
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
1
2
13}3 qxq6}3
The possible numbers of each camera that you can buy
are shown in the table.
X=18.085106 Y=0
Number of daylight
cameras, x
7
8
9
10
11
12
13
3. Using a table:
Number of flash
cameras, 20 2 x
13
12
11
10
9
8
7
Weekly
earnings
y
Mixed Review for TAKS
61. C; s 5 Steve’s cards
k 5 Kevin’s cards
Weekly
5 salary
5
1550
1 Commission
rate
1
0.05
+
Weekly
sales
+
x
Entering y 5 1550 1 0.05x into a graphing calculator
and scrolling through the table of values shows that
y 5 1900 when x 5 7000. So, the sales person can sell
$7000 or more to earn at least $1900 per week.
t 5 Thomas’ cards
s5k26
t 5 2s l t 5 2(k 2 6)
s 1 k 1 t 5 22
Substitute: (k 2 6) 1 k 1 2(k 2 6) 5 22
X
5000
5500
6000
6500
7000
7500
8000
X=7000
Y1
1800
1825
1850
1875
1900
1925
1950
62. G; SA(original can) 5 2:rh 1 2:r2
SA(new can) 5 2:(2r)(2h) 1 2:(2r)2
5 4(2:rh) 1 4(2:r2)
Y1=(1550+.05x 1900)
5 4(2:rh 1 2:r2)
5 4 SA(original can)
The new surface area is four times the original
surface area.
X=6978.7234 Y=0
Using a graph:
Entering
y 5 (1550 1 0.05xq1900)
into a graphing calculator
and using the trace feature
suggests that the inequality is
true when xq7000.
Algebra 2
Worked-Out Solution Key
29