Download Exercises with Solutions

Section 1.4
Compound Inequalities
63
1.4 Exercises
In Exercises 1-12, solve the inequality.
Express your answer in both interval and
set notations, and shade the solution on
a number line.
1.
2.
8x + 5 ≤ −1 and 4x − 2 > −1
18.
−x − 1 < 7 and − 6x − 9 ≥ 8
19.
−3x + 8 ≤ −5 or − 2x − 4 ≥ −3
20.
−6x − 7 < −3 and − 8x ≥ 3
21.
9x − 9 ≤ 9 and 5x > −1
22.
−7x + 3 < −3 or − 8x ≥ 2
23.
3x − 5 < 4 and − x + 9 > 3
24.
−8x − 6 < 5 or 4x − 1 ≥ 3
25.
9x + 3 ≤ −5 or − 2x − 4 ≥ 9
26.
−7x + 6 < −4 or − 7x − 5 > 7
27.
4x − 2 ≤ 2 or 3x − 9 ≥ 3
28.
−5x + 5 < −4 or − 5x − 5 ≥ −5
29.
5x + 1 < −6 and 3x + 9 > −4
30.
7x + 2 < −5 or 6x − 9 ≥ −7
31.
−7x − 7 < −2 and 3x ≥ 3
32.
4x + 1 < 0 or 8x + 6 > 9
33.
7x + 8 < −3 and 8x + 3 ≥ −9
34.
3x < 2 and − 7x − 8 ≥ 3
35.
−5x + 2 ≤ −2 and − 6x + 2 ≥ 3
36.
4x − 1 ≤ 8 or 3x − 9 > 0
37.
2x − 5 ≤ 1 and 4x + 7 > 7
38.
3x + 1 < 0 or 5x + 5 > −8
−8x − 3 ≤ −16x − 1
6x − 6 > 3x + 3
3.
−12x + 5 ≤ −3x − 4
4.
7x + 3 ≤ −2x − 8
5.
−11x − 9 < −3x + 1
6.
4x − 8 ≥ −4x − 5
7.
4x − 5 > 5x − 7
8.
−14x + 4 > −6x + 8
9.
2x − 1 > 7x + 2
10.
−3x − 2 > −4x − 9
11.
−3x + 3 < −11x − 3
12.
6x + 3 < 8x + 8
In Exercises 13-50, solve the compound
inequality. Express your answer in both
interval and set notations, and shade the
solution on a number line.
1
17.
13.
2x − 1 < 4 or 7x + 1 ≥ −4
14.
−8x + 9 < −3 and − 7x + 1 > 3
15.
−6x − 4 < −4 and − 3x + 7 ≥ −5
16.
−3x + 3 ≤ 8 and − 3x − 6 > −6
Copyrighted material. See: http://msenux.redwoods.edu/IntAlgText/
Version: Fall 2007
64
Chapter 1
Preliminaries
39.
−8x + 7 ≤ 9 or − 5x + 6 > −2
40.
x − 6 ≤ −5 and 6x − 2 > −3
41.
−4x − 8 < 4 or − 4x + 2 > 3
42.
9x − 5 < 2 or − 8x − 5 ≥ −6
43.
−9x − 5 ≤ −3 or x + 1 > 3
44.
−5x − 3 ≤ 6 and 2x − 1 ≥ 6
45.
−1 ≤ −7x − 3 ≤ 2
46.
0 < 5x − 5 < 9
47.
5 < 9x − 3 ≤ 6
48.
−6 < 7x + 3 ≤ 2
49.
−2 < −7x + 6 < 6
50.
−9 < −2x + 5 ≤ 1
In Exercises 51-62, solve the given inequality for x. Graph the solution set on
a number line, then use interval and setbuilder notation to describe the solution
set.
51.
1
x 1
1
− < + <
3
2 4
3
52.
x 1
1
1
− < − <
5
2 4
5
53.
1
1 x
1
− < − <
2
3 2
2
54.
2
1 x
2
− ≤ − ≤
3
2 5
3
55.
−1 < x −
x+1
<2
5
56.
−2 < x −
2x − 1
<4
3
57.
−2 <
x+1 x+1
−
≤2
2
3
Version: Fall 2007
x − 1 2x − 1
−
≤2
3
5
58.
−3 <
59.
x<4−x<5
60.
−x < 2x + 3 ≤ 7
61.
−x < x + 5 ≤ 11
62.
−2x < 3 − x ≤ 8
63. Aeron has arranged for a demonstration of “How to make a Comet” by
Professor O’Commel. The wise professor has asked Aeron to make sure the
auditorium stays between 15 and 20 degrees Celsius (C). Aeron knows the thermostat is in Fahrenheit (F) and he also
knows that the conversion formula between the two temperature scales is C =
(5/9)(F − 32).
a) Setting up the compound inequality
for the requested temperature range
in Celsius, we get 15 ≤ C ≤ 20. Using the conversion formula above, set
up the corresponding compound inequality in Fahrenheit.
b) Solve the compound inequality in part
(a) for F. Write your answer in set
notation.
c) What are the possible temperatures
(integers only) that Aeron can set the
thermostat to in Fahrenheit?
Section 1.4
Compound Inequalities
1.4 Solutions
1.
− 8x − 3 ≤ −16x − 1
=⇒
− 8x + 16x ≤ −1 + 3
=⇒
8x ≤ 2
=⇒
x≤
1
4
Thus, the solution interval is (−∞, 14 ] = {x|x ≤ 14 }.
1
4
3.
− 12x + 5 ≤ −3x − 4
=⇒
− 12x + 3x ≤ −4 − 5
=⇒
− 9x ≤ −9
=⇒
x≥1
Thus, the solution interval is [1, ∞) = {x|x ≥ 1}.
1
5.
− 11x − 9 < −3x + 1
=⇒
− 11x + 3x < 1 + 9
=⇒
− 8x < 10
=⇒
x>−
5
4
Thus, the solution interval is (− 45 , ∞) = {x|x > − 54 }.
− 54
Version: Fall 2007
Chapter 1
Preliminaries
7.
4x − 5 > 5x − 7
=⇒
4x − 5x > −7 + 5
=⇒
− x > −2
=⇒
x<2
Thus, the solution interval is (−∞, 2) = {x|x < 2}.
2
9.
2x − 1 > 7x + 2
=⇒
2x − 7x > 2 + 1
=⇒
− 5x > 3
=⇒
x<−
3
5
Thus, the solution interval is (−∞, − 35 ) = {x|x < − 53 }.
− 35
11.
− 3x + 3 < −11x − 3
=⇒
− 3x + 11x < −3 − 3
=⇒
8x < −6
=⇒
x<−
3
4
Thus, the solution interval is (−∞, − 34 ) = {x|x < − 43 }.
− 34
13.
2x − 1 < 4 or 7x + 1 ≥ −4
Version: Fall 2007
=⇒
2x < 5
or
7x ≥ −5
=⇒
x<
5
2
or
x≥−
5
7
Section 1.4
Compound Inequalities
5/2
−5/7
For the union, shade anything shaded in either graph. The solution is the set of all real
numbers (−∞, ∞).
15.
− 6x − 4 < −4 and − 3x + 7 ≥ −5
=⇒
− 6x < 0
and
− 3x ≥ −12
=⇒
x>0
and
x≤4
=⇒
0<x≤4
0
4
The intersection is all points shaded in both graphs, so the solution is (0, 4] = {x|0 <
x ≤ 4}.
0
4
17.
8x + 5 ≤ −1 and 4x − 2 > −1
=⇒
8x ≤ −6
and
4x > 1
=⇒
x≤−
3
4
and
x>
1
4
−3/4
1/4
Shade all numbers that are shaded in both graphs. There are no such numbers, so the
solution set is empty. No solution.
Version: Fall 2007
Chapter 1
Preliminaries
19.
− 3x + 8 ≤ −5 or − 2x − 4 ≥ −3
− 3x ≤ −13 or
=⇒
x≥
=⇒
13
3
or
− 2x ≥ 1
x≤−
1
2
13/3
−1/2
For the union, shade all points that are shaded in either graph:
1
1 [ 13
13
−∞, −
, ∞ = {x|x ≤ − or x ≥ }
2
3
2
3
− 21
13
3
21.
9x − 9 ≤ 9 and 5x > −1
=⇒
=⇒
9x ≤ 18 and 5x > −1
x≤2
and x > −
1
5
2
−1/5
For the intersection, shade any points that are shaded in both graphs. The solution set
is (− 51 , 2] = {x| − 15 < x ≤ 2}.
− 15
2
23.
3x − 5 < 4 and − x + 9 > 3
=⇒
3x < 9
and
− x > −6
=⇒
x<3
and
x<6
3
Version: Fall 2007
Section 1.4
Compound Inequalities
6
For the intersection, shade all points shaded in both graphs. The solution set is
(−∞, 3) = {x|x < 3}.
3
25.
9x + 3 ≤ −5 or − 2x − 4 ≥ 9
=⇒
9x ≤ −8
or
− 2x ≥ 13
=⇒
x≤−
8
9
or
x≤−
13
2
−8/9
−13/2
Note that − 89 > − 13
2 . For the union, shade any points that are shaded in either graph.
The solution set is (−∞, − 89 ] = {x|x ≤ − 89 }.
− 89
27.
4x − 2 ≤ 2 or 3x − 9 ≥ 3
=⇒
4x ≤ 4
or 3x ≥ 12
=⇒
x≤1
or
x≥4
1
4
For the union, shade any points that are shaded in either graph:
[
(−∞, 1] [4, ∞) = {x|x ≤ 1 or x ≥ 4}
1
4
Version: Fall 2007
Chapter 1
Preliminaries
29.
5x + 1 < −6 and 3x + 9 > −4
=⇒
5x < −7
=⇒
x<−
and 3x > −13
7
5
and
x>−
13
3
−7/5
−13/3
For the intersection, shade any points that are shaded in both graphs. The solution set
7
13
7
is (− 13
3 , − 5 ) = {x| − 3 < x < − 5 }.
− 13
3
− 75
31.
− 7x − 7 < −2 and 3x ≥ 3
=⇒
− 7x < 5
=⇒
x>−
and 3x ≥ 3
5
7
and x ≥ 1
−5/7
1
For the intersection, shade any points that are shaded in both graphs. The solution set
is [1, ∞) = {x|x ≥ 1}.
1
33.
7x + 8 < −3 and 8x + 3 ≥ −9
=⇒
7x < −11 and 8x ≥ −12
=⇒
x<−
−11/7
Version: Fall 2007
11
7
and
x≥−
3
2
Section 1.4
Compound Inequalities
−3/2
For the intersection, shade all points that are shaded in both graphs. There are no
such points, so there is no intersection.
35.
− 5x + 2 ≤ −2 and − 6x + 2 ≥ 3
=⇒
=⇒
− 5x ≤ −4
x≥
4
5
and
− 6x ≥ 1
and
x≤−
1
6
4/5
−1/6
For the intersection, shade all points that are shaded in both graphs. There are no
such points, so there is no solution.
37.
2x − 5 ≤ 1 and 4x + 7 > 7
=⇒
2x ≤ 6
and
4x > 0
=⇒
x≤3
and
x>0
3
0
For the intersection, shade all points that are shaded in both graphs. Thus, the solution
set is (0, 3] = {x|0 < x ≤ 3}.
0
3
Version: Fall 2007
Chapter 1
Preliminaries
39.
− 8x + 7 ≤ 9 or − 5x + 6 > −2
=⇒
− 8x ≤ 2
=⇒
x≥−
1
4
− 5x > −8
or
x<
or
8
5
−1/4
8/5
For the union, shade all points that are shaded in either graph. Every number is shaded
in one graph or the other, so the solution is the set of all real numbers (−∞, ∞).
41.
− 4x − 8 < 4 or − 4x + 2 > 3
=⇒
=⇒
− 4x < 12 or
x > −3
or
− 4x > 1
x<−
1
4
−3
−1/4
For the union, shade all numbers that are shaded in either graph. Every number is
shaded in one of the graphs, so the solution is the set of all real numbers (−∞, ∞).
43.
− 9x − 5 ≤ −3 or x + 1 > 3
=⇒
− 9x ≤ 2
=⇒
x≥−
2
9
or
x>2
or
x>2
−2/9
2
Version: Fall 2007
Section 1.4
Compound Inequalities
For the union, shade all numbers that shaded in either graph. The solution interval is
[− 29 , ∞) = {x|x ≥ − 29 }.
− 29
45.
− 1 ≤ −7x − 3 ≤ 2
2 ≤ −7x ≤ 5
=⇒
2
5
≥x≥−
7
7
5
2
− ≤x≤−
7
7
−
=⇒
=⇒
Thus, the solution interval is [− 75 , − 72 ] = {x| −
− 75
5
7
≤ x ≤ − 72 }.
− 72
47.
5 < 9x − 3 ≤ 6
=⇒
8 < 9x ≤ 9
=⇒
8
<x≤1
9
Thus, the solution interval is ( 89 , 1] = {x| 89 < x ≤ 1}.
1
8
9
49.
− 2 < −7x + 6 < 6
=⇒
=⇒
=⇒
− 8 < −7x < 0
8
>x>0
7
8
0<x<
7
Thus, the solution set is (0, 78 ){x|0 < x < 78 }.
0
8
7
Version: Fall 2007
Chapter 1
51.
Preliminaries
Multiply by 12 to clear the fractions.
1
x 1
1
− < + <
3 2 4 3
1
x 1
1
12 −
< 12
< 12
+
3
2 4
3
−4 < 6x + 3 < 4
Subtract 3 from all three members, then divide all three members of the resulting
inequality by 6.
−7 < 6x < 1
1
7
− <x<
6
6
Thus, the solution interval is (−7/6, 1/6), or equivalently, {x : −7/6 < x < 1/6}.
−7/6
53.
1/6
Multiply by 6 to clear the fractions.
1
1 x
1
− < − <
2
3
2
2 1
1 x
1
6 −
−
<6
<6
2
3 2
2
−3 < 2 − 3x < 3
Subtract 2 from all three members, then divide all three members of the resulting
inequality by −3. Remember to reverse the inequality symbols.
−5 < −3x < 1
1
5
>x>−
3
3
It is conventional to change the order of this solution to match the order of the shaded
solution on the number line. So, equivalently,
1
5
− <x< .
3
3
Thus, the solution interval is (−1/3, 5/3), or equivalently, {x : −1/3 < x < 5/3}.
−1/3
Version: Fall 2007
5/3
Section 1.4
55.
Compound Inequalities
Multiply by 5 to clear the fractions.
x+1
<2
−1 < x −
5 x+1
< 5(2)
5(−1) < 5 x −
5
−5 < 5x − (x + 1) < 10
−5 < 5x − x − 1 < 10
−5 < 4x − 1 < 10
Add 1 to all three members, then divide all three members of the resulting inequality
by 4.
−4 < 4x < 11
11
−1 < x <
4
Thus, the solution interval is (−1, 11/4), or equivalently, {x : −1 < x < 11/4}.
−1
57.
11/4
Multiply by 6 to clear the fractions.
x+1 x+1
−2 <
−
≤2
3 2
x+1 x+1
6(−2) < 6
−
≤ 6(2)
2
3
−12 < 3(x + 1) − 2(x + 1) ≤ 12
−12 < 3x + 3 − 2x − 2 ≤ 12
−12 < x + 1 ≤ 12
Subtract 1 from all three members.
−13 < x ≤ 11
Thus, the solution interval is (−13, 11], or equivalently, {x : −13 < x ≤ 11}.
−13
59.
11
We’ll need to split the compound inequality
x<4−x<5
and write it using “and.” Then we can solve each part independently.
Version: Fall 2007
Chapter 1
Preliminaries
x<4−x
and
4−x<5
2x < 4
and
−x<1
x<2
and
x > −1
Thus, the solution interval is (−1, 2), or equivalently, {x : −1 < x < 2}.
−1
61.
2
We’ll want to split the compound inequality
−x < x + 5 ≤ 11
and write it using “and.” Then we can solve each part independently.
−x<x+5
− 2x < 5
and
x + 5 ≤ 11
x≤6
and
x > −5/2
Thus, the solution interval is (−5/2, 6], or equivalently, {x : −5/2 < x ≤ 6}.
−5/2
6
63.
a) 15 ≤ 95 (F − 32) ≤ 20
b)
=⇒
5
15 ≤ (F − 32) ≤ 20
9
5
9(15) ≤ (9) (F − 32) ≤ (9)20
9
135 ≤ 5(F − 32) ≤ 180
=⇒
135 ≤ 5F − 160 ≤ 180
=⇒
295 ≤ 5F ≤ 340
=⇒
=⇒
=⇒
295
5F
340
≤
≤
5
5
5
59 ≤ F ≤ 68
The solution is {F |59 ≤ F ≤ 68}.
c) In roster form, the solutions are {59, 60, 61, 62, 63, 64, 65, 66, 67, 68}.
Version: Fall 2007