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1.5: Chapter 1 Exercises with Solutions

( \newcommand{\kernel}{\mathrm{null}\,}\)

In Exercises 1.5.1 - 1.5.8, find the prime factorization of the given natural number.

Exercise 1.5.1

80

Answer

80=22225

Exercise 1.5.2

108

Exercise 1.5.3

180

Answer

180=22335

Exercise 1.5.4

160

Exercise 1.5.5

128

Answer

128=2222222

Exercise 1.5.6

192

Exercise 1.5.7

32

Answer

32=22222

Exercise 1.5.8

72

In Exercises 1.5.9-1.5.16, convert the given decimal to a fraction.

Exercise 1.5.9

0.648

Answer

There are three decimal places, so 0.648=6481000=81125

Exercise 1.5.10

0.62

Exercise 1.5.11

0.240

Answer

There are three decimal places, so 0.240=2401000=625

Exercise 1.5.12

0.90

Exercise 1.5.13

0.14

Answer

There are two decimal places, so 0.14=14100=750

Exercise 1.5.14

0.760

Exercise 1.5.15

0.888

Answer

There are three decimal places, so 0.888=8881000=111125

Exercise 1.5.16

0.104

In Exercises 1.5.17-1.5.24, convert the given repeating decimal to a fraction.

Exercise 1.5.17

0.¯27

Answer

Let x=0.¯27. Then 100x=27.¯27. Subtracting on both sides of these equations.

100x=27.¯27x=0.¯27

yields 99x=27. Finally, solve for x by dividing by 99:x=2799=311.

Exercise 1.5.18

0.¯171

Exercise 1.5.19

0.¯24

Answer

Let x=0.¯24. Then 100x=24.¯24. Subtracting on both sides of these equations 100x=24.¯24x=0.¯24

yields 99x=24. Finally, solve for x by dividing by 99:x=2499=833

Exercise 1.5.20

0.¯882

Exercise 1.5.21

0.¯84

Answer

Let x=0.¯84. Then 100x=84.¯84. Subtracting on both sides of these equations

100x=84.¯.84x=0.¯84

yields 99x=84. Finally, solve for x by dividing by 99:x=8499=2833

Exercise 1.5.22

0.¯384

Exercise 1.5.23

0.¯63

Answer

Let x=0.¯63. Then 100x=63.¯63. Subtracting on both sides of these equations

100x=63.¯63x=0.¯63

yields 99x=63. Finally, solve for x by dividing by 99:x=6399=711

Exercise 1.5.24

0.¯60

Exercise 1.5.25

Prove that 3 is irrational.

Answer

Suppose that 3 is rational. Then it can be expressed as the ratio of two integers p and q as follows:

3=pq

Square both sides, 3=p2q2

then clear the equation of fractions by multiplying both sides by q2:

p2=3q2

Now p and q each have their own unique prime factorizations. Both p2 and q2 have an even number of factors in their prime factorizations. But this contradicts equation (1), because the left side would have an even number of factors in its prime factorization, while the right side would have an odd number of factors in its prime factorization (there’s one extra 3 on the right side).

Therefore, our assumption that 3 was rational is false. Thus, 3 is irrational.

Exercise 1.5.26

Prove that 5 is irrational.

In Exercises 1.5.27-1.5.30, copy the given table onto your homework paper. In each row, place a check mark in each column that is appropriate. That is, if the number at the start of the row is rational, place a check mark in the rational column. Note: Most (but not all) rows will have more than one check mark.

Exercise 1.5.27
  N W Z Q R
0          
-2          
-2/3          
0.15          
0.¯2          
5          
Answer
  N W Z Q R
0   x x x x
-2     x x x
-2/3       x x
0.15       x x
0.¯2       x x
5         x
Exercise 1.5.28
  N W Z Q R
10/2          
π          
-6          
0.¯9          
2          
0.37          
Exercise 1.5.29
  N W Z Q R
-4/3          
12          
0          
11          
1.¯3          
6/2          
Answer
  N W Z Q R
-4/3       x x
12 x x x x x
0   x x x x
11         x
1.¯3       x x
6/2 x x x x x
Exercise 1.5.30
  N W Z Q R
-3/5          
10          
1.625          
10/2          
0/5          
11          

In Exercises 1.5.31-1.5.42, consider the given statement and determine whether it is true or false. Write a sentence explaining your answer. In particular, if the statement is false, try to give an example that contradicts the statement.

Exercise 1.5.31

All natural numbers are whole numbers.

Answer

True. The only difference between the two sets is that the set of whole numbers contains the number 0.

Exercise 1.5.32

All whole numbers are rational numbers.

Exercise 1.5.33

All rational numbers are integers.

Answer

False. For example, 12 is not an integer.

Exercise 1.5.34

All rational numbers are whole numbers.

Exercise 1.5.35

Some natural numbers are irrational.

Answer

False. All natural numbers are rational, and therefore not irrational.

Exercise 1.5.36

Some whole numbers are irrational.

Exercise 1.5.37

Some real numbers are irrational.

Answer

True. For example, π and √2 are real numbers which are irrational.

Exercise 1.5.38

All integers are real numbers.

Exercise 1.5.39

All integers are rational numbers.

Answer

True. Every integer b can be written as a fraction b/1.

Exercise 1.5.40

No rational numbers are natural numbers.

Exercise 1.5.41

No real numbers are integers.

Answer

False. For example, 2 is a real number that is also an integer.

Exercise 1.5.42

All whole numbers are natural numbers.

In Exercises 1.5.43-1.5.54, solve each of the given equations for x.

Exercise 1.5.43

45x + 12 = 0

Answer

45x+12=045x=12x=1245=415

Exercise 1.5.44

76x − 55 = 0

Exercise 1.5.45

x − 7 = −6x + 4

Answer

x7=6x+4x+6x=4+77x=11x=117

Exercise 1.5.46

−26x + 84 = 48

Exercise 1.5.47

37x + 39 = 0

Answer

37x+39=037x=39x=3937

Exercise 1.5.48

−48x + 95 = 0

Exercise 1.5.49

74x − 6 = 91

Answer

74x6=9174x=97x=9774

Exercise 1.5.50

−7x + 4 = −6

Exercise 1.5.51

−88x + 13 = −21

Answer

88x+13=2188x=34x=3488=1744

Exercise 1.5.52

−14x − 81 = 0

Exercise 1.5.53

19x + 35 = 10

Answer

19x+35=1019x=25x=2519

Exercise 1.5.54

−2x + 3 = −5x − 2

In Exercises 1.5.55-1.5.66, solve each of the given equations for x.

Exercise 1.5.55

6 − 3(x + 1) = −4(x + 6) + 2

Answer

63(x+1)=4(x+6)+263x3=4x24+23x+3=4x223x+4x=223x=25

Exercise 1.5.56

(8x + 3) − (2x + 6) = −5x + 8

Exercise 1.5.57

−7 − (5x − 3) = 4(7x + 2)

7(5x3)=4(7x+2)75x+3=28x+85x4=28x+85x28x=8+433x=12x=1233=411

Exercise 1.5.58

−3 − 4(x + 1) = 2(x + 4) + 8

Exercise 1.5.59

9 − (6x − 8) = −8(6x − 8)

Answer

9(6x8)=8(6x8)96x+8=48x+646x+17=48x+646x+48x=641742x=47x=4742

Exercise 1.5.60

−9 − (7x − 9) = −2(−3x + 1)

Exercise 1.5.61

(3x − 1) − (7x − 9) = −2x − 6

Answer

(3x1)(7x9)=2x63x17x+9=2x64x+8=2x64x+2x=682x=14x=7

Exercise 1.5.62

−8 − 8(x − 3) = 5(x + 9) + 7

Exercise 1.5.63

(7x − 9) − (9x + 4) = −3x + 2

Answer

(7x9)(9x+4)=3x+27x99x4=3x+22x13=3x+22x+3x=2+13x=15

Exercise 1.5.64

(−4x − 6) + (−9x + 5) = 0

Exercise 1.5.65

−5 − (9x + 4) = 8(−7x − 7)

Answer

5(9x+4)=8(7x7)59x4=56x569x9=56x569x+56x=56+947x=47x=1

Exercise 1.5.66

(8x − 3) + (−3x + 9) = −4x − 7

In Exercises 1.5.67-1.5.78, solve each of the given equations for x. Check your solutions using your calculator.

Exercise 1.5.67

−3.7x − 1 = 8.2x − 5

Answer

First clear decimals by multiplying by 10.

3.7x1=8.2x537x10=82x5037x82x=50+10119x=40x=40119

Here is a check of the solutions on the graphing calculator. The left-hand side of the equation is evaluated at the solution in (a), the right-hand side of the equation is evaluated at the solution in (b). Note that they match.

Screen Shot 2019-07-26 at 3.54.32 PM.png

Exercise 1.5.68

8.48x − 2.6 = −7.17x − 7.1

Exercise 1.5.69

23x+8=45x+4

Answer

First clear fractions by multiplying by 15.

23x+8=45x+410x+120=12x+6010x12x=6012022x=60x=6022=3011

Here is a check of the solutions on the graphing calculator. The left-hand side of the equation is evaluated at the solution in (a), the right-hand side of the equation is evaluated at the solution in (b). Note that they match.

Screen Shot 2019-07-26 at 4.00.59 PM.png

Exercise 1.5.70

−8.4x = −4.8x + 2

Exercise 1.5.71

32x+9=14x+7

Answer

First clear fractions by multiplying by 4.

32x+9=14x+76x+36=x+286xx=28367x=8x=87

Here is a check of the solutions on the graphing calculator. The left-hand side of the equation is evaluated at the solution in (a), the right-hand side of the equation is evaluated at the solution in (b). Note that they match.

Screen Shot 2019-07-26 at 4.02.24 PM.png

Exercise 1.5.72

2.9x − 4 = 0.3x − 8

Exercise 1.5.73

5.45x + 4.4 = 1.12x + 1.6

Answer

First clear decimals by multiplying by 100.

5.45x+4.4=1.12x+1.6545x+440=112x+160545x112x=160440433x=280x=280433

Here is a check of the solutions on the graphing calculator. The left-hand side of the equation is evaluated at the solution in (a), the right-hand side of the equation is evaluated at the solution in (b). Note that they match.

Screen Shot 2019-07-26 at 4.03.55 PM.png

Exercise 1.5.74

14x+5=45x4

Exercise 1.5.75

32x8=25x2

Answer

First clear fractions by multiplying by 10. 32x8=25x215x80=4x2015x4x=20+8019x=60x=6019

Here is a check of the solutions on the graphing calculator. The left-hand side of the equation is evaluated at the solution in (a), the right-hand side of the equation is evaluated at the solution in (b). Note that they match.

Screen Shot 2019-07-26 at 4.06.54 PM.png

Exercise 1.5.76

43x8=14x+5

Exercise 1.5.77

−4.34x − 5.3 = 5.45x − 8.1

Answer

First clear decimals by multiplying by 100.

4.34x5.3=5.45x8.1434x530=545x810434x545x=810+530979x=280x=280979

Here is a check of the solutions on the graphing calculator. The left-hand side of the equation is evaluated at the solution in (a), the right-hand side of the equation is evaluated at the solution in (b). Note that they match.

Screen Shot 2019-07-26 at 4.11.18 PM.png

Exercise 1.5.78

23x3=14x1

In Exercises 1.5.79-50, solve each of the given equations for the indicated variable.

Exercise 1.5.79

P = IRT for R

Answer

P=IRTP=(IT)RPIT=(IT)RITPIT=R

Exercise 1.5.80

d = vt for t

Exercise 1.5.81

v=v0+at for a

Answer

v=v0+atvv0=atvv0t=a

Exercise 1.5.82

x=v0+vt for v

Exercise 1.5.83

Ax + By = C for y

Answer

Ax+By=CBy=CAxy=CAxB

Exercise 1.5.84

y = mx + b for x

Exercise 1.5.85

A=πr2 for π

Answer

A=πr2Ar2=π

Exercise 1.5.86

S=2πr2+2πrh for h

Exercise 1.5.87

F=kqq0r2 for k

Answer

F=kqq0r2Fr2=kqq0Fr2qq0=k

Exercise 1.5.88

C=QmT for T

Exercise 1.5.89

Vt=k for t

Answer

Vt=kV=ktVk=t

Exercise 1.5.90

λ=hmv for v

Exercise 1.5.91

P1V1n1T1=P2V2n2T2 for V2

Answer

Cross multiply, then divide by the coefficient of V2.

P1V1n1T1=P2V2n2T2n2P1V1T2=n1P2V2T1n2P1V1T2n1P2T1=V2

Exercise 1.5.92

π=nRTVi for n

Exercise 1.5.93

Tie a ball to a string and whirl it around in a circle with constant speed. It is known that the acceleration of the ball is directly toward the center of the circle and given by the formula a=v2r where a is acceleration, v is the speed of the ball, and r is the radius of the circle of motion.

i. Solve formula (1) for r.

ii. Given that the acceleration of the ball is 12 m/s2 and the speed is 8 m/s, find the radius of the circle of motion.

Answer

Cross multiply, then divide by the coefficient of r.

a=v2rar=v2r=v2a

To find the radius, substitute the acceleration a=12m/s2 and speed v = 8 m/s.

r=v2a=(8)212=6412=163

Hence, the radius is r=16/3m, or 513 meters.

Exercise 1.5.94

A particle moves along a line with constant acceleration. It is known the velocity of the particle, as a function of the amount of time that has passed, is given by the equation

v=v0+at where v is the velocity at time t, v0 is the initial velocity of the particle (at time t = 0), and a is the acceleration of the particle.

i. Solve formula (2) for t.

ii. You know that the current velocity of the particle is 120 m/s. You also know that the initial velocity was 40 m/s and the acceleration has been a constant a=2m/s2. How long did it take the particle to reach its current velocity?

Exercise 1.5.95

Like Newton’s Universal Law of Gravitation, the force of attraction (repulsion) between two unlike (like) charged particles is proportional to the product of the charges and inversely proportional to the distance between them. F=kCq1q2r2 In this formula, kC8.988×109Nm2/C2 and is called the electrostatic constant. The variables q1 and q2 represent the charges (in Coulombs) on the particles (which could either be positive or negative numbers) and r represents the distance (in meters) between the charges. Finally, F represents the force of the charge, measured in Newtons.

i. Solve formula (3) for r.

ii. Given a force F=2.0×1012N, two equal charges q1=q2=1C, find the approximate distance between the two charged particles.

Answer

Cross multiply, then divide by the coefficient of r.

F=kCq1q2r2Fr2=kCq1q2r2=kCq1q2F

Finally, to find r, take the square root.

r=kCq1q2F

To find the distance between the charged particles, substitute kC=8.988×109Nm2/C2,
q1=q2=1C, and F=2.0×1012N.

r=(8.988×109)(1)(1)2.0×1012

A calculator produces an approximation, r0.067 meters.

Screen Shot 2019-07-26 at 4.24.57 PM.png

Perform each of the following tasks in Exercises 1.5.96-1.5.99.

i. Write out in words the meaning of the symbols which are written in set-builder notation.

ii. Write some of the elements of this set.

iii. Draw a real line and plot some of the points that are in this set.

Exercise 1.5.96

A={xN:x>10}

Answer

i. A is the set of all x in the natural numbers such that x is greater than 10.

ii. A={11,12,13,14,}

iii.

Screen Shot 2019-08-05 at 10.42.27 AM.png

Exercise 1.5.97

B={xN:x10}

Exercise 1.5.98

C={xZ:x2}

Answer

i. C is the set of all x in the integers such that x is less than or equal to 2.

ii. C={,4,3,2,1,0,1,2}

iii.

Screen Shot 2019-08-05 at 10.43.52 AM.png

Exercise 1.5.99

D={xZ:x>3}

In Exercises 1.5.100-1.5.103, use the sets A, B, C, and D that were defined in Exercises 1.5.96-1.5.99. Describe the following sets using set notation, and draw the corresponding Venn Diagram.

Exercise 1.5.100

AB

Answer

AB={xN:x>10}={11,12,13,}

Screen Shot 2019-08-05 at 10.44.45 AM.png

Exercise 1.5.101

AB

Exercise 1.5.102

AC

Answer

AC={xZ:x2 or x>10}={,3,21,0,1,2,11,12,13}

Screen Shot 2019-08-05 at 10.45.52 AM.png

Exercise 1.5.103

CD

In Exercises 1.5.104-1.5.111, use both interval and set notation to describe the interval shown on the graph.

Exercise 1.5.104

Screen Shot 2019-07-29 at 10.11.48 PM.png

Answer

The filled circle at the endpoint 3 indicates this point is included in the set. Thus, the set in interval notation is [3,), and in set notation {x:x3}.

Exercise 1.5.105

Screen Shot 2019-07-29 at 10.12.57 PM.png

Exercise 1.5.106

Screen Shot 2019-07-29 at 10.13.31 PM.png

Answer

The empty circle at the endpoint −7 indicates this point is not included in the set. Thus, the set in interval notation is (,7), and in set notation is {x:x<7}.

Exercise 1.5.107

Screen Shot 2019-07-29 at 10.15.43 PM.png

Exercise 1.5.108

Screen Shot 2019-07-29 at 10.16.40 PM.png

Answer

The empty circle at the endpoint 0 indicates this point is not included in the set. Thus, the set in interval notation is (0,), and in set notation is {x:x>0}.

Exercise 1.5.109

Screen Shot 2019-07-29 at 10.18.30 PM.png

Exercise 1.5.110

Screen Shot 2019-07-29 at 10.19.29 PM.png

Answer

The empty circle at the endpoint −8 indicates this point is not included in the set. Thus, the set in interval notation is (8,), and in set notation is {x:x>8}.

Exercise 1.5.111

Screen Shot 2019-07-29 at 10.20.24 PM.png

In Exercises 1.5.112-1.5.119, sketch the graph of the given interval.

Exercise 1.5.112

[2,5)

Answer

Screen Shot 2019-08-05 at 10.49.22 AM.png

Exercise \PageIndex{113}

(-3,1]

Exercise \PageIndex{114}

[1, \infty)

Answer

Screen Shot 2019-08-05 at 10.50.00 AM.png

Exercise \PageIndex{115}

(-\infty, 2)

Exercise \PageIndex{116}

\{x :-4<x<1\}

Answer

Screen Shot 2019-08-05 at 10.50.50 AM.png

Exercise \PageIndex{117}

\{x : 1 \leq x \leq 5\}

Exercise \PageIndex{118}

\{x : x<-2\}

Answer

Screen Shot 2019-08-05 at 10.51.30 AM.png

Exercise \PageIndex{119}

\{x : x \geq-1\}

In Exercises \PageIndex{120}-\PageIndex{127}, use both interval and set notation to describe the intersection of the two intervals shown on the graph. Also, sketch the graph of the intersection on the real number line.

Exercise \PageIndex{120}

Screen Shot 2019-07-29 at 10.25.01 PM.png

Answer

The intersection is the set of points that are in both intervals (shaded on both graphs). Graph of the intersection:

Screen Shot 2019-08-05 at 10.52.32 AM.png

[1, \infty)=\{x : x \geq 1\}

Exercise \PageIndex{121}

Screen Shot 2019-07-29 at 10.26.55 PM.png

Exercise \PageIndex{122}

Screen Shot 2019-07-29 at 10.27.31 PM.png

Answer

There are no points that are in both intervals (shaded in both), so there is no intersection. Graph of the intersection:

Screen Shot 2019-08-05 at 10.53.41 AM.png

no intersection

Exercise \PageIndex{123}

Screen Shot 2019-07-29 at 10.28.18 PM.png

Exercise \PageIndex{124}

Screen Shot 2019-07-29 at 10.29.22 PM.png

Answer

The intersection is the set of points that are in both intervals (shaded in both). Graph of the intersection:

Screen Shot 2019-08-05 at 10.54.24 AM.png

[-6,2]=\{x :-6 \leq x \leq 2\}

Exercise \PageIndex{125}

Screen Shot 2019-07-29 at 10.30.25 PM.png

Exercise \PageIndex{126}

Screen Shot 2019-07-29 at 10.31.23 PM.png

Answer

The intersection is the set of points that are in both intervals (shaded in both). Graph of the intersection:

Screen Shot 2019-08-05 at 10.55.19 AM.png

[9, \infty)=\{x : x \geq 9\}

Exercise \PageIndex{127}

Screen Shot 2019-07-29 at 10.33.57 PM.png

In Exercises \PageIndex{128}-\PageIndex{135}, use both interval and set notation to describe the union of the two intervals shown on the graph. Also, sketch the graph of the union on the real number line.

Exercise \PageIndex{128}

Screen Shot 2019-07-29 at 10.37.45 PM.png

Answer

The union is the set of all points that are in one interval or the other (shaded in either graph). Graph of the union:

Screen Shot 2019-08-05 at 10.58.47 AM.png

(-\infty,-8]=\{x : x \leq-8\}

Exercise \PageIndex{129}

Screen Shot 2019-07-29 at 10.39.08 PM.png

Exercise \PageIndex{130}

Screen Shot 2019-07-29 at 10.39.38 PM.png

Answer

The union is the set of all points that are in one interval or the other (shaded in either graph). Graph of the union:

Screen Shot 2019-08-05 at 11.01.44 AM.png

(-\infty, 9] \cup(15, \infty)
=\{x : x \leq 9 \text { or } x>15\}

Exercise \PageIndex{131}

Screen Shot 2019-07-29 at 10.40.55 PM.png

Exercise \PageIndex{132}

Screen Shot 2019-07-29 at 10.42.41 PM.png

Answer

The union is the set of all points that are in one interval or the other (shaded in either). Graph of the union:

Screen Shot 2019-08-05 at 11.03.02 AM.png

(-\infty, 3)=\{x : x<3\}

Exercise \PageIndex{133}

Screen Shot 2019-07-29 at 10.44.52 PM.png

Exercise \PageIndex{134}

Screen Shot 2019-07-29 at 10.45.57 PM.png

Answer

The union is the set of all points that are in one interval or the other (shaded in either). Graph of the union:

Screen Shot 2019-08-05 at 11.04.38 AM.png

[9, \infty)=\{x : x \geq 9\}

Exercise \PageIndex{135}

Screen Shot 2019-07-29 at 10.46.54 PM.png

In Exercises \PageIndex{136}-56, use interval notation to describe the given set. Also, sketch the graph of the set on the real number line.

Exercise \PageIndex{136}

\{x : x \geq-6 \text { and } x>-5\}

Answer

This set is the same as \{x : x>-5\}, which is (-5, \infty) in interval notation. Graph of the set:

Screen Shot 2019-08-05 at 11.06.12 AM.png

Exercise \PageIndex{137}

\{x : x \leq 6 \text { and } x \geq 4\}

Exercise \PageIndex{138}

\{x : x \geq-1 \text { or } x<3\}

Answer

Every real number is in one or the other of the two intervals. Therefore, the set is the set of all real numbers (-\infty, \infty). Graph of the set:

Screen Shot 2019-08-05 at 11.07.04 AM.png

Exercise \PageIndex{139}

\{x : x>-7 \text { and } x>-4\}

Exercise \PageIndex{140}

\{x : x \geq -1 \text { or } x>6\}

Answer

This set is the same as \{x : x \geq-1\}, which is [-1, \infty) in interval notation. Graph of the set:

Screen Shot 2019-08-05 at 11.10.10 AM.png

Exercise \PageIndex{141}

\{x : x \geq 7 \text { or } x<-2\}

Exercise \PageIndex{142}

\{x : x \geq 6 \text { or } x>-3\}

Answer

This set is the same as \{x : x>-3\}, which is (-3, \infty) in interval notation. Graph of the set:

Screen Shot 2019-08-05 at 11.12.01 AM.png

Exercise \PageIndex{143}

\{x : x \leq 1 \text { or } x>0\}

Exercise \PageIndex{144}

\{x : x<2 \text { and } x<-7\}

Answer

This set is the same as \{x : x<-7\}, which is (-\infty,-7) in interval notation. Graph of the set:

Screen Shot 2019-08-05 at 11.13.07 AM.png

Exercise \PageIndex{145}

\{x : x \leq-3 \text { and } x<-5\}

Exercise \PageIndex{146}

\{x : x \leq-3 \text { or } x \geq 4\}

Answer

This set is the union of two intervals, (-\infty,-3] \cup[4, \infty). Graph of the set:

Screen Shot 2019-08-05 at 11.14.04 AM.png

Exercise \PageIndex{147}

\{x : x<11 \text { or } x \leq 8\}

Exercise \PageIndex{148}

\{x : x \geq 5 \text { and } x \leq 1\}

Answer

There are no numbers that satisfy both inequalities. Thus, there is no intersection. Graph of the set:

Screen Shot 2019-08-05 at 11.14.56 AM.png

Exercise \PageIndex{149}

\{x : x<5 \text { or } x<10\}

Exercise \PageIndex{150}

\{x : x \leq 5 \text { and } x \geq-1\}

Answer

This set is the same as \{x :-1 \leq x \leq 5\}, which is [−1, 5] in interval notation. Graph of the set

Screen Shot 2019-08-05 at 11.17.54 AM.png

Exercise \PageIndex{151}

\{x : x>-3 \text { and } x<-6\}

In Exercises \PageIndex{152}-\PageIndex{163}, solve the inequality. Express your answer in both interval and set notations, and shade the solution on a number line.

Exercise \PageIndex{152}

-8 x-3 \leq-16 x-1

Answer

\begin{aligned} & -8 x-3 \leq-16 x-1 \\ \Longrightarrow \quad & − 8x + 16x \leq −1 + 3 \\ \Longrightarrow \quad& 8x \leq 2 \\ \Longrightarrow \quad & x \leq \frac{1}{4}\end{aligned} \nonumber

Thus, the solution interval is (−\infty, \frac{1}{4}] = \{x|x \leq \frac{1}{4}\}.

Screen Shot 2019-08-07 at 10.21.02 PM.png

Exercise \PageIndex{153}

6 x-6>3 x+3

Exercise \PageIndex{154}

-12 x+5 \leq-3 x-4

Answer

\begin{aligned} & -12 x+5 \leq-3 x-4 \\ \Longrightarrow \quad & -12x + 3x \leq −4 − 5 \\ \Longrightarrow \quad& -9x \leq -9 \\ \Longrightarrow \quad & x \geq 1\end{aligned} \nonumber

Thus, the solution interval is [1,\infty) = \{x|x \geq 1\}.

Screen Shot 2019-08-08 at 10.56.57 PM.png

Exercise \PageIndex{155}

7 x+3 \leq-2 x-8

Exercise \PageIndex{156}

-11 x-9<-3 x+1

Answer

\begin{aligned} & − 11x − 9 < −3x + 1 \\ \Longrightarrow \quad & − 11x + 3x < 1 + 9 \\ \Longrightarrow \quad& − 8x < 10 \\ \Longrightarrow \quad & x > -\frac{5}{4}\end{aligned} \nonumber

Thus, the solution interval is (−\frac{5}{4} ,\infty) = \{x|x >−\frac{5}{4} \}.

Screen Shot 2019-08-08 at 11.00.01 PM.png

Exercise \PageIndex{157}

4 x-8 \geq-4 x-5

Exercise \PageIndex{158}

4 x-5>5 x-7

Answer

\begin{aligned} & 4x − 5 > 5x − 7\\ \Longrightarrow \quad & 4x − 5x > −7 + 5 \\ \Longrightarrow \quad& − x > −2 \\ \Longrightarrow \quad &x < 2\end{aligned} \nonumber
Thus, the solution interval is (−\infty, 2) = \{x|x < 2\}.

Screen Shot 2019-08-08 at 11.02.23 PM.png

Exercise \PageIndex{159}

-14 x+4>-6 x+8

Exercise \PageIndex{160}

2 x-1>7 x+2

Answer

\begin{aligned} & 2x − 1 > 7x + 2\\ \Longrightarrow \quad & 2x − 7x > 2 + 1 \\ \Longrightarrow \quad& − 5x > 3 \\ \Longrightarrow \quad &x < −\frac{3}{5}\end{aligned} \nonumber
Thus, the solution interval is (−\infty, −\frac{3}{5}) = \{x|x < −\frac{3}{5}\}.

Screen Shot 2019-08-08 at 11.04.53 PM.png

Exercise \PageIndex{161}

-3 x-2>-4 x-9

Exercise \PageIndex{162}

-3 x+3<-11 x-3

Answer

\begin{aligned} & − 3x + 3 < −11x − 3\\ \Longrightarrow \quad & − 3x + 11x < −3 − 3 \\ \Longrightarrow \quad& 8x < −6 \\ \Longrightarrow \quad &x < -\frac{3}{4}\end{aligned} \nonumber
Thus, the solution interval is (−\infty, −\frac{3}{4}) = \{x|x < −\frac{3}{4}\}.

Screen Shot 2019-08-08 at 11.07.23 PM.png

Exercise \PageIndex{163}

6 x+3<8 x+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.

Exercise \PageIndex{164}

2 x-1<4 or 7 x+1 \geq-4

Answer

\begin{aligned} & 2x − 1 < 4 \text{ or } 7x + 1 \geq −4\\ \Longrightarrow \quad & 2x < 5\quad \text{or}\quad 7x \geq −5 \\ \Longrightarrow \quad&x<\frac{5}{2}\quad\text{or}\quad x\geq-\frac{5}{7}\end{aligned} \nonumber

Screen Shot 2019-08-08 at 11.11.56 PM.png

For the union, shade anything shaded in either graph. The solution is the set of all real numbers (−\infty,\infty).

Screen Shot 2019-08-08 at 11.13.59 PM.png

Exercise \PageIndex{165}

-8 x+9<-3 and -7 x+1>3

Exercise \PageIndex{166}

-6 x-4<-4 and -3 x+7 \geq-5

Answer

\begin{aligned} & − 6x − 4 < −4 \text{ and } − 3x + 7 \geq −5\\ \Longrightarrow \quad & -6x < 0\quad \text{and}\quad -3x \geq −12 \\ \Longrightarrow \quad&x>0\quad\text{and}\quad x\leq4 \\ \Longrightarrow \quad & 0< x \leq 4 \end{aligned} \nonumber

Screen Shot 2019-08-08 at 11.21.51 PM.png

The intersection is all points shaded in both graphs, so the solution is (0, 4] = \{x|0 < x \leq 4\}.

Screen Shot 2019-08-08 at 11.23.30 PM.png

Exercise \PageIndex{167}

-3 x+3 \leq 8 and -3 x-6>-6

Exercise \PageIndex{168}

8 x+5 \leq-1 and 4 x-2>-1

Answer

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Exercise \PageIndex{169}

-x-1<7 and -6 x-9 \geq 8

Exercise \PageIndex{170}

-3 x+8 \leq-5 or -2 x-4 \geq-3

Answer

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Exercise \PageIndex{171}

-6 x-7<-3 and -8 x \geq 3

Exercise \PageIndex{172}

9 x-9 \leq 9 and 5 x>-1

Answer

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Exercise \PageIndex{1}

-7 x+3<-3 or -8 x \geq 2

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Exercise \PageIndex{1}

3 x-5<4 and -x+9>3

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Exercise \PageIndex{1}

-8 x-6<5 or 4 x-1 \geq 3

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Exercise \PageIndex{1}

9 x+3 \leq-5 or -2 x-4 \geq 9

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Exercise \PageIndex{1}

-7 x+6<-4 or -7 x-5>7

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Exercise \PageIndex{1}

4 x-2 \leq 2 or 3 x-9 \geq 3

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Exercise \PageIndex{1}

-5 x+5<-4 or -5 x-5 \geq-5

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Exercise \PageIndex{1}

5 x+1<-6 and 3 x+9>-4

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Exercise \PageIndex{1}

7 x+2<-5 or 6 x-9 \geq-7

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Exercise \PageIndex{1}

-7 x-7<-2 and 3 x \geq 3

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Exercise \PageIndex{1}

4 x+1<0 or 8 x+6>9

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Exercise \PageIndex{1}

7 x+8<-3 and 8 x+3 \geq-9

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Exercise \PageIndex{1}

3 x<2 and -7 x-8 \geq 3

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Exercise \PageIndex{1}

-5 x+2 \leq-2 and -6 x+2 \geq 3

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Exercise \PageIndex{1}

4 x-1 \leq 8 or 3 x-9>0

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Exercise \PageIndex{1}

2 x-5 \leq 1 and 4 x+7>7

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Exercise \PageIndex{1}

3 x+1<0 or 5 x+5>-8

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Exercise \PageIndex{1}

-8 x+7 \leq 9 or -5 x+6>-2

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Exercise \PageIndex{1}

x-6 \leq-5 and 6 x-2>-3

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Exercise \PageIndex{1}

-4 x-8<4 or -4 x+2>3

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Exercise \PageIndex{1}

9 x-5<2 or -8 x-5 \geq-6

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Exercise \PageIndex{1}

-9 x-5 \leq-3 or x+1>3

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Exercise \PageIndex{1}

-5 x-3 \leq 6 and 2 x-1 \geq 6

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Exercise \PageIndex{1}

-1 \leq-7 x-3 \leq 2

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Exercise \PageIndex{1}

0<5 x-5<9

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Exercise \PageIndex{1}

5<9 x-3 \leq 6

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Exercise \PageIndex{1}

-6<7 x+3 \leq 2

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Exercise \PageIndex{1}

-2<-7 x+6<6

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Exercise \PageIndex{1}

-9<-2 x+5 \leq 1

Answer

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In Exercises 51-62, solve the given inequality for x. Graph the solution set on a number line, then use interval and set-builder notation to describe the solution set.

Exercise \PageIndex{1}

-\frac{1}{3}<\frac{x}{2}+\frac{1}{4}<\frac{1}{3}

Answer

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Exercise \PageIndex{1}

-\frac{1}{5}<\frac{x}{2}-\frac{1}{4}<\frac{1}{5}

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Exercise \PageIndex{1}

-\frac{1}{2}<\frac{1}{3}-\frac{x}{2}<\frac{1}{2}

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Exercise \PageIndex{1}

-\frac{2}{3} \leq \frac{1}{2}-\frac{x}{5} \leq \frac{2}{3}

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Exercise \PageIndex{1}

-1<x-\frac{x+1}{5}<2

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Exercise \PageIndex{1}

-2<x-\frac{2 x-1}{3}<4

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Exercise \PageIndex{1}

-2<\frac{x+1}{2}-\frac{x+1}{3} \leq 2

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Exercise \PageIndex{1}

-3<\frac{x-1}{3}-\frac{2 x-1}{5} \leq 2

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Exercise \PageIndex{1}

x<4-x<5

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Exercise \PageIndex{1}

-x<2 x+3 \leq 7

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Exercise \PageIndex{1}

-x<x+5 \leq 11

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Exercise \PageIndex{1}

−2x < 3 − x \leq 8

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Exercise \PageIndex{1}

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 \leq C \leq 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?

Answer

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1.5: Chapter 1 Exercises with Solutions is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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