1.1 exercis esmsenux2.redwoods.edu/intalgtext/chapter1/chapter1solutions.pdfsection 1.1 number...

Section 1.1 Number Systems 11

Version: Fall 2007

1.1 Exercises

In Exercises 1-8, find the prime factor-ization of the given natural number.

1. 80

2. 108

3. 180

4. 160

5. 128

6. 192

7. 32

8. 72

In Exercises 9-16, convert the given dec-imal to a fraction.

9. 0.648

10. 0.62

11. 0.240

12. 0.90

13. 0.14

14. 0.760

15. 0.888

16. 0.104

In Exercises 17-24, convert the givenrepeating decimal to a fraction.

17. 0.27

Copyrighted material. See: http://msenux.redwoods.edu/IntAlgText/1

18. 0.171

19. 0.24

20. 0.882

21. 0.84

22. 0.384

23. 0.63

24. 0.60

25. Prove that!

3 is irrational.

26. Prove that!

5 is irrational.

In Exercises 27-30, copy the given ta-ble onto your homework paper. In eachrow, place a check mark in each columnthat is appropriate. That is, if the num-ber at the start of the row is rational,place a check mark in the rational col-umn. Note: Most (but not all) rows willhave more than one check mark.

27.

N W Z Q R0"2"2/30.150.2!

5

12 Chapter 1 Preliminaries

Version: Fall 2007

28.

N W Z Q R10/2!

"60.9!

20.37

29.

N W Z Q R"4/3

120!

111.36/2

30.

N W Z Q R"3/5!

101.62510/20/511

In Exercises 31-42, consider the givenstatement and determine whether it istrue or false. Write a sentence explainingyour answer. In particular, if the state-ment is false, try to give an example thatcontradicts the statement.

31. All natural numbers are whole num-bers.

32. All whole numbers are rational num-bers.

33. All rational numbers are integers.

34. All rational numbers are whole num-bers.

35. Some natural numbers are irrational.

36. Some whole numbers are irrational.

37. Some real numbers are irrational.

38. All integers are real numbers.

39. All integers are rational numbers.

40. No rational numbers are natural num-bers.

41. No real numbers are integers.

42. All whole numbers are natural num-bers.

Section 1.1 Number Systems

Version: Fall 2007

1.1 Solutions

1. 80 = 2 · 2 · 2 · 2 · 5

3. 180 = 2 · 2 · 3 · 3 · 5

5. 128 = 2 · 2 · 2 · 2 · 2 · 2 · 2

7. 32 = 2 · 2 · 2 · 2 · 2

9. There are three decimal places, so 0.648 = 6481000 = 81

125 .


25 .

13. There are two decimal places, so 0.14 = 14100 = 7

50 .


125 .

17. 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 = 3

11 .


100x = 24.24x = 0.24


33 .


100x = 84.84x = 0.84


33 .


100x = 63.63x = 0.63

Chapter 1 Preliminaries

Version: Fall 2007


11 .

25. Suppose that!

3 is rational. Then it can be expressed as the ratio of two integersp and q as follows:

!3 = pq

Square both sides,

3 = p2

q2,

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

p2 = 3q2 (1)

Now p and q each have their own unique prime factorizations. Both p2 and q2 have aneven 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.

27.

N W Z Q R0 x x x x"2 x x x"2/3 x x0.15 x x0.2 x x!

5 x

29.

N W Z Q R"4/3 x x

12 x x x x x0 x x x x!

11 x1.3 x x6/2 x x x x x

Section 1.1 Number Systems

Version: Fall 2007

31. True. The only di!erence between the two sets is that the set of whole numberscontains the number 0.

33. False. For example, 12 is not an integer.

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

37. True. For example, ! and!

2 are real numbers which are irrational.

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

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

Section 1.2 Solving Equations 27

Version: Fall 2007

1.2 Exercises

In Exercises 1-12, solve each of the givenequations for x.

1. 45x+ 12 = 0

2. 76x! 55 = 0

3. x! 7 = !6x+ 4

4. !26x+ 84 = 48

5. 37x+ 39 = 0

6. !48x+ 95 = 0

7. 74x! 6 = 91

8. !7x+ 4 = !6

9. !88x+ 13 = !21

10. !14x! 81 = 0

11. 19x+ 35 = 10

12. !2x+ 3 = !5x! 2

In Exercises 13-24, solve each of thegiven equations for x.

13. 6! 3(x+ 1) = !4(x+ 6) + 2

14. (8x+ 3)! (2x+ 6) = !5x+ 8

15. !7 ! (5x! 3) = 4(7x+ 2)

16. !3! 4(x+ 1) = 2(x+ 4) + 8

17. 9! (6x! 8) = !8(6x! 8)

18. !9! (7x! 9) = !2(!3x+ 1)

19. (3x! 1)! (7x! 9) = !2x! 6


20. !8! 8(x! 3) = 5(x+ 9) + 7

21. (7x! 9)! (9x+ 4) = !3x+ 2

22. (!4x! 6) + (!9x+ 5) = 0

23. !5! (9x+ 4) = 8(!7x! 7)

24. (8x! 3) + (!3x+ 9) = !4x! 7

In Exercises 25-36, solve each of thegiven equations for x. Check your solu-tions using your calculator.

25. !3.7x! 1 = 8.2x! 5

26. 8.48x! 2.6 = !7.17x! 7.1

27. !23x+ 8 = 4

5x+ 4

28. !8.4x = !4.8x+ 2

29. !32x+ 9 = 1

4x+ 7

30. 2.9x! 4 = 0.3x! 8

31. 5.45x+ 4.4 = 1.12x+ 1.6

32. !14x+ 5 = !4

5x! 4

33. !32x! 8 = 2

5x! 2

34. !43x! 8 = !1

4x+ 5

35. !4.34x! 5.3 = 5.45x! 8.1

36. 23x! 3 = !1

4x! 1


Version: Fall 2007

In Exercises 37-50, solve each of thegiven equations for the indicated vari-able.

37. P = IRT for R

38. d = vt for t

39. v = v0 + at for a

40. x = v0 + vt for v

41. Ax+By = C for y

42. y = mx+ b for x

43. A = !r2 for !

44. S = 2!r2 + 2!rh for h

45. F = kqq0r2

for k

46. C = Q

mTfor T

47. Vt

= k for t

48. " = h

mvfor v

49. P1V1n1T1

= P2V2n2T2

for V2

50. ! = nRTVi for n

51. Tie a ball to a string and whirl itaround in a circle with constant speed.It is known that the acceleration of theball is directly toward the center of thecircle and given by the formula

a = v2

r, (1)

where a is acceleration, v is the speed ofthe 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, findthe radius of the circle of motion.

52. A particle moves along a line withconstant acceleration. It is known thevelocity of the particle, as a function ofthe amount of time that has passed, isgiven by the equation

v = v0 + at, (2)

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

i. Solve formula (2) for t.ii. You know that the current velocity

of the particle is 120 m/s. You alsoknow that the initial velocity was 40 m/sand the acceleration has been a con-stant a = 2 m/s2. How long did ittake the particle to reach its currentvelocity?

53. Like Newton’s Universal Law of Grav-itation, the force of attraction (repulsion)between two unlike (like) charged parti-cles is proportional to the product of thecharges and inversely proportional to thedistance between them.

F = kCq1q2r2

(3)

In this formula, kC " 8.988#109 Nm2/C2

and is called the electrostatic constant.The variables q1 and q2 represent the charges(in Coulombs) on the particles (whichcould either be positive or negative num-bers) and r represents the distance (inmeters) between the charges. Finally, Frepresents the force of the charge, mea-sured in Newtons.

Section 1.2 Solving Equations 29

Version: Fall 2007

i. Solve formula (3) for r.ii. Given a force F = 2.0 # 1012 N, two

equal charges q1 = q2 = 1 C, find theapproximate distance between the twocharged particles.


Version: Fall 2007

1.2 Solutions

1.

45x+ 12 = 0=$ 45x = !12

=$ x = !1245 = ! 4

15

3.

x! 7 = !6x+ 4=$ x+ 6x = 4 + 7=$ 7x = 11

=$ x = 117

5.

37x+ 39 = 0=$ 37x = !39

=$ x = !3937

7.

74x! 6 = 91=$ 74x = 97

=$ x = 9774

9.

! 88x+ 13 = !21=$ ! 88x = !34

=$ x = !34!88 = 17

44

Section 1.2 Solving Equations

Version: Fall 2007

11.

19x+ 35 = 10=$ 19x = !25

=$ x = !2519

13.

6! 3(x+ 1) = !4(x+ 6) + 2=$ 6! 3x! 3 = !4x! 24 + 2=$ ! 3x+ 3 = !4x! 22=$ ! 3x+ 4x = !22! 3=$ x = !25

15.

! 7 ! (5x! 3) = 4(7x+ 2)=$ ! 7 ! 5x+ 3 = 28x+ 8=$ ! 5x! 4 = 28x+ 8=$ ! 5x! 28x = 8 + 4=$ ! 33x = 12

=$ x = !1233 = ! 4

11

17.

9! (6x! 8) = !8(6x! 8)=$ 9! 6x+ 8 = !48x+ 64=$ ! 6x+ 17 = !48x+ 64=$ ! 6x+ 48x = 64! 17=$ 42x = 47

=$ x = 4742


Version: Fall 2007

19.

(3x! 1)! (7x! 9) = !2x! 6=$ 3x! 1! 7x+ 9 = !2x! 6=$ ! 4x+ 8 = !2x! 6=$ ! 4x+ 2x = !6! 8=$ ! 2x = !14=$ x = 7

21.

(7x! 9)! (9x+ 4) = !3x+ 2=$ 7x! 9! 9x! 4 = !3x+ 2=$ ! 2x! 13 = !3x+ 2=$ ! 2x+ 3x = 2 + 13=$ x = 15

23.

! 5! (9x+ 4) = 8(!7x! 7)=$ ! 5! 9x! 4 = !56x! 56=$ ! 9x! 9 = !56x! 56=$ ! 9x+ 56x = !56 + 9=$ 47x = !47=$ x = !1

25. First clear decimals by multiplying by 10.

! 3.7x! 1 = 8.2x! 5=$ ! 37x! 10 = 82x! 50=$ ! 37x! 82x = !50 + 10=$ ! 119x = !40

=$ x = 40119

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


Version: Fall 2007

(a) (b)

27. First clear fractions by multiplying by 15.

! 23x+ 8 = 4

5x+ 4

=$ ! 10x+ 120 = 12x+ 60=$ ! 10x! 12x = 60! 120=$ ! 22x = !60

=$ x = !60!22 = 30

11Here is a check of the solutions on the graphing calculator. The left-hand side of theequation is evaluated at the solution in (a), the right-hand side of the equation isevaluated at the solution in (b). Note that they match.

(a) (b)


! 32x+ 9 = 1

4x+ 7

=$ ! 6x+ 36 = x+ 28=$ ! 6x! x = 28! 36=$ ! 7x = !8

=$ x = 87



Version: Fall 2007

(a) (b)


5.45x+ 4.4 = 1.12x+ 1.6=$ 545x+ 440 = 112x+ 160=$ 545x! 112x = 160! 440=$ 433x = !280

=$ x = !280433


(a) (b)


! 32x! 8 = 2

5x! 2

=$ ! 15x! 80 = 4x! 20=$ ! 15x! 4x = !20 + 80=$ ! 19x = 60

=$ x = !6019



Version: Fall 2007

(a) (b)


! 4.34x! 5.3 = 5.45x! 8.1=$ ! 434x! 530 = 545x! 810=$ ! 434x! 545x = !810 + 530=$ ! 979x = !280

=$ x = 280979


(a) (b)

37.

P = IRT=$ P = (IT )R

=$ P

IT= (IT )RIT

=$ P

IT= R


Version: Fall 2007

39.

v = v0 + at=$ v ! v0 = at

=$ v ! v0t

= a

41.

Ax+By = C=$ By = C !Ax

=$ y = C !AxB

43.

A = !r2

=$ A

r2= !

45.

F = kqq0r2

=$ Fr2 = kqq0

=$ Fr2

qq0= k

47.V

t= k

=$ V = kt

=$ V

k= t

49. Cross multiply, then divide by the coe!cient of V2.P1V1n1T1

= P2V2n2T2

=$ n2P1V1T2 = n1P2V2T1

=$ n2P1V1T2n1P2T1

= V2


Version: Fall 2007

51. Cross multiply, then divide by the coe!cient of r.

a = v2

rar = v2

r = v2

a

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

r = v2

a= (8)2

12 = 6412 = 16

3

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

53. Cross multiply, then divide by the coe!cient of r.

F = kCq1q2r2

Fr2 = kCq1q2

r2 = kCq1q2F

Finally, to find r, take the square root.

r =!kCq1q2F

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

r =!

(8.988# 109)(1)(1)2.0# 1012

A calculator produces an approximation, r " 0.067 meters.

Section 1.3 Logic 47

Version: Fall 2007

1.3 Exercises

Perform each of the following tasks inExercises 1-4.

i. Write out in words the meaning ofthe 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.

1. A = {x ! N : x > 10}

2. B = {x ! N : x " 10}

3. C = {x ! Z : x # 2}

4. D = {x ! Z : x > $3}

In Exercises 5-8, use the sets A, B, C,andD that were defined in Exercises 1-4. Describe the following sets using setnotation, and draw the corresponding VennDiagram.

5. A %B

6. A &B

7. A & C.

8. C %D.

In Exercises 9-16, use both interval andset notation to describe the interval shownon the graph.

9.

3


10.

0

11.

$7

12.

1

13.

0

14.

1

15.

$8

16.

9

In Exercises 17-24, sketch the graph ofthe given interval.

17. [2, 5)

18. ($3, 1]

19. [1,')


Version: Fall 2007

20. ($', 2)

21. {x : $4 < x < 1}

22. {x : 1 # x # 5}

23. {x : x < $2}

24. {x : x " $1}

In Exercises 25-32, use both intervaland set notation to describe the inter-section of the two intervals shown on thegraph. Also, sketch the graph of the in-tersection on the real number line.

25.

1$3

26.

$6$3

27.

2

$4

28.

118

29.

$62

30.

15

31.

95

32.

$14

$6

In Exercises 33-40, use both intervaland set notation to describe the unionof the two intervals shown on the graph.Also, sketch the graph of the union onthe real number line.

33.

$10$8

34.

$3$2

35.

159

Section 1.3 Logic 49

Version: Fall 2007

36.

514

37.

$53

38.

119

39.

109

40.

7$2

In Exercises 41-56, use interval nota-tion to describe the given set. Also, sketchthe graph of the set on the real numberline.

41. {x : x " $6 and x > $5}

42. {x : x # 6 and x " 4}

43. {x : x " $1 or x < 3}

44. {x : x > $7 and x > $4}

45. {x : x " $1 or x > 6}

46. {x : x " 7 or x < $2}

47. {x : x " 6 or x > $3}

48. {x : x # 1 or x > 0}

49. {x : x < 2 and x < $7}

50. {x : x # $3 and x < $5}

51. {x : x # $3 or x " 4}

52. {x : x < 11 or x # 8}

53. {x : x " 5 and x # 1}

54. {x : x < 5 or x < 10}

55. {x : x # 5 and x " $1}

56. {x : x > $3 and x < $6}


Version: Fall 2007

1.3 Solutions

1.

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.

11 17

3.

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.

$4 2

5. A %B = {x ! N : x > 10} = {11, 12, 13, . . .}

A

B

7. A & C = {x ! Z : x # 2 or x > 10} = {. . . ,$3,$2$ 1, 0, 1, 2, 11, 12, 13 . . .}

A C

9. 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 : x " 3}.

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

13. 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}.

Section 1.3 Logic

Version: Fall 2007

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

17.

2 5

19.

1

21.

$4 1

23.

$2

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

1

[1,') = {x : x " 1}

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

no intersection

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

$6 2

[$6, 2] = {x : $6 # x # 2}


Version: Fall 2007

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

9

[9,') = {x : x " 9}

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

$8

($',$8] = {x : x # $8}

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

159

($', 9] & (15,')={x : x # 9 or x > 15}

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

3

($', 3) = {x : x < 3}

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

9

[9,') = {x : x " 9}

41. This set is the same as {x : x > $5}, which is ($5,') in interval notation.Graph of the set:

$5

Section 1.3 Logic

Version: Fall 2007

43. Every real number is in one or the other of the two intervals. Therefore, the setis the set of all real numbers ($','). Graph of the set:

45. This set is the same as {x : x " $1}, which is [$1,') in interval notation.Graph of the set:

$1

47. This set is the same as {x : x > $3}, which is ($3,') in interval notation.Graph of the set:

$3

49. This set is the same as {x : x < $7}, which is ($',$7) in interval notation.Graph of the set:

$7

51. This set is the union of two intervals, ($',$3] & [4,'). Graph of the set:

4$3

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

55. This set is the same as {x : $1 # x # 5}, which is [$1, 5] in interval notation.Graph of the set:

$1 5

Section 1.4 Compound Inequalities 63

Version: Fall 2007

1.4 Exercises

In Exercises 1-12, solve the inequality.Express your answer in both interval andset notations, and shade the solution ona number line.

1. !8x! 3 " !16x! 1

2. 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 compoundinequality. Express your answer in bothinterval and set notations, and shade thesolution on a number line.

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


17. 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


Version: Fall 2007

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 in-equality for x. Graph the solution set ona number line, then use interval and set-builder notation to describe the solutionset.

51. !13 <x

2 + 14 <

13

52. !15 <x

2 !14 <

15

53. !12 <

13 !x

2 <12

54. !23 "

12 !x

5 "23

55. !1 < x! x+ 15 < 2

56. !2 < x! 2x! 13 < 4

57. !2 < x+ 12 ! x+ 1

3 " 2

58. !3 < x! 13 ! 2x! 1

5 " 2

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 demon-stration of “How to make a Comet” byProfessor O’Commel. The wise profes-sor has asked Aeron to make sure theauditorium stays between 15 and 20 de-grees Celsius (C). Aeron knows the ther-mostat is in Fahrenheit (F) and he alsoknows that the conversion formula be-tween the two temperature scales is C =(5/9)(F ! 32).

a) Setting up the compound inequalityfor the requested temperature rangein Celsius, we get 15 " C " 20. Us-ing the conversion formula above, setup the corresponding compound in-equality in Fahrenheit.

b) Solve the compound inequality in part(a) for F. Write your answer in setnotation.

c) What are the possible temperatures(integers only) that Aeron can set thethermostat to in Fahrenheit?

Section 1.4 Compound Inequalities

Version: Fall 2007

1.4 Solutions

1.

! 8x! 3 " !16x! 1=$ ! 8x+ 16x " !1 + 3=$ 8x " 2

=$ x " 14

Thus, the solution interval is (!%, 14 ] = {x|x " 1

4}.

14

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 > !54

Thus, the solution interval is (!54 ,%) = {x|x > !5

4}.

!54


Version: Fall 2007

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 < !35

Thus, the solution interval is (!%,!35) = {x|x < !3

5}.

!35

11.

! 3x+ 3 < !11x! 3=$ ! 3x+ 11x < !3! 3=$ 8x < !6

=$ x < !34

Thus, the solution interval is (!%,!34) = {x|x < !3

4}.

!34

13.

2x! 1 < 4 or 7x+ 1 # !4=$ 2x < 5 or 7x # !5

=$ x <52 or x # !5

7


Version: Fall 2007

5/2

!5/7

For the union, shade anything shaded in either graph. The solution is the set of all realnumbers (!%,%).

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 " !34 and x >

14

!3/4

1/4

Shade all numbers that are shaded in both graphs. There are no such numbers, so thesolution set is empty. No solution.


Version: Fall 2007

19.

! 3x+ 8 " !5 or ! 2x! 4 # !3=$ ! 3x " !13 or ! 2x # 1

=$ x # 133 or x " !1

2

13/3

!1/2

For the union, shade all points that are shaded in either graph:!!%,!1

2

"#$133 ,%

%= {x|x " !1

2 or x # 133 }

!12

133

21.

9x! 9 " 9 and 5x > !1=$ 9x " 18 and 5x > !1

=$ x " 2 and x > !15

2

!1/5

For the intersection, shade any points that are shaded in both graphs. The solution setis (!1

5 , 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

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 " !89 or x " !13

2

!8/9

!13/2

Note that !89 > !

132 . For the union, shade any points that are shaded in either graph.

The solution set is (!%,!89 ] = {x|x " !8

9}.

!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

29.

5x+ 1 < !6 and 3x+ 9 > !4=$ 5x < !7 and 3x > !13

=$ x < !75 and x > !13

3

!7/5

!13/3

For the intersection, shade any points that are shaded in both graphs. The solution setis (!13

3 ,!75) = {x|! 13

3 < x < !75}.

!133 !7

5

31.

! 7x! 7 < !2 and 3x # 3=$ ! 7x < 5 and 3x # 3

=$ x > !57 and x # 1

!5/7

1

For the intersection, shade any points that are shaded in both graphs. The solution setis [1,%) = {x|x # 1}.

1

33.

7x+ 8 < !3 and 8x+ 3 # !9=$ 7x < !11 and 8x # !12

=$ x < !117 and x # !3

2

!11/7


Version: Fall 2007

!3/2

For the intersection, shade all points that are shaded in both graphs. There are nosuch points, so there is no intersection.

35.

! 5x+ 2 " !2 and ! 6x+ 2 # 3=$ ! 5x " !4 and ! 6x # 1

=$ x # 45 and x " !1

6

4/5

!1/6

For the intersection, shade all points that are shaded in both graphs. There are nosuch 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 solutionset is (0, 3] = {x|0 < x " 3}.

0 3


Version: Fall 2007

39.

! 8x+ 7 " 9 or ! 5x+ 6 > !2=$ ! 8x " 2 or ! 5x > !8

=$ x # !14 or x <

85

!1/4

8/5

For the union, shade all points that are shaded in either graph. Every number is shadedin 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 ! 4x > 1

=$ x > !3 or x < !14

!3

!1/4

For the union, shade all numbers that are shaded in either graph. Every number isshaded 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 or x > 2

=$ x # !29 or x > 2

!2/9

2


Version: Fall 2007

For the union, shade all numbers that shaded in either graph. The solution interval is[!2

9 ,%) = {x|x # !29}.

!29

45.

! 1 " !7x! 3 " 2=$ 2 " !7x " 5

=$ ! 27 # x # !

57

=$ ! 57 " x " !

27

Thus, the solution interval is [!57 ,!

27 ] = {x|! 5

7 " x " !27}.

!57 !2

7

47.

5 < 9x! 3 " 6=$ 8 < 9x " 9

=$ 89 < x " 1

Thus, the solution interval is (89 , 1] = {x|89 < x " 1}.

89

1

49.

! 2 < !7x+ 6 < 6=$ ! 8 < !7x < 0

=$ 87 > x > 0

=$ 0 < x < 87

Thus, the solution set is (0, 87 ){x|0 < x < 8

7}.

0 87


Version: Fall 2007

51. Multiply by 12 to clear the fractions.

!13 <x

2 + 14 <

13

12!!1

3

%< 12!x

2 + 14

%< 12!1

3

%

!4 < 6x+ 3 < 4

Subtract 3 from all three members, then divide all three members of the resultinginequality by 6.

!7 < 6x < 1!7

6 < x <16

Thus, the solution interval is (!7/6, 1/6), or equivalently, {x : !7/6 < x < 1/6}.

!7/6 1/6


!12 <

13 !x

2 <12

6!!1

2

%< 6!1

3 !x

2

%< 6!1

2

%

!3 < 2! 3x < 3

Subtract 2 from all three members, then divide all three members of the resultinginequality by !3. Remember to reverse the inequality symbols.

!5 < !3x < 153 > x > !

13

It is conventional to change the order of this solution to match the order of the shadedsolution on the number line. So, equivalently,

!13 < x <

53 .

Thus, the solution interval is (!1/3, 5/3), or equivalently, {x : !1/3 < x < 5/3}.

!1/3 5/3


Version: Fall 2007


!1 < x! x+ 15 < 2

5(!1) < 5!x! x+ 1

5

%< 5(2)

!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 inequalityby 4.

!4 < 4x < 11!1 < x < 11

4Thus, the solution interval is (!1, 11/4), or equivalently, {x : !1 < x < 11/4}.

!1 11/4


!2 < x+ 12 ! x+ 1

3 " 2

6(!2) < 6!x+ 1

2 ! x+ 13

%" 6(2)

!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 11

59. 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

x < 4! x and 4! x < 52x < 4 and ! x < 1x < 2 and x > !1

Thus, the solution interval is (!1, 2), or equivalently, {x : !1 < x < 2}.

!1 2

61. 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 and x+ 5 " 11! 2x < 5 and x " 6x > !5/2

Thus, the solution interval is (!5/2, 6], or equivalently, {x : !5/2 < x " 6}.

!5/2 6

63.

a) 15 " 59(F ! 32) " 20

b)

15 " 59(F ! 32) " 20

=$ 9(15) " (9)59(F ! 32) " (9)20

=$ 135 " 5(F ! 32) " 180=$ 135 " 5F ! 160 " 180=$ 295 " 5F " 340

=$ 2955 "

5F5 "

3405

=$ 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}.

1.1 exercis esmsenux2.redwoods.edu/intalgtext/chapter1/chapter1solutions.pdfsection 1.1 number...

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