Solution manual for college algebra 12th edition by lial

The intersection of sets A and B is made up of all the elements belonging to both set A and set B.. The union of sets A and B is made up of all the elements belonging to set A or set

2 Set A is a subset of set B if every element of

set A is also an element of set B

3 The set of all elements of the universal set U

that do not belong to set A is the complement

of set A

4 The intersection of sets A and B is made up of

all the elements belonging to both set A and

set B

5 The union of sets A and B is made up of all the

elements belonging to set A or set B (or both)

6 The set  1 1 1 

3 9 27

1, , , , is infinite

7 Using set notation, the set {x|x is a natural

number less than 6} is {1, 2, 3, 4, 5}

11 The set {4, 5, 6, …, 15} has a limited number

of elements, so it is a finite set Yes, 10 is an

element of the set

12 The set {1, 2, 3, 4, 5, …, 75} has a limited

number of elements, so it is a finite set Yes,

10 is an element of the set

13 The set  1 1 1 

2 4 8

1, , , , has an unlimited

number of elements, so it is an infinite set No,

10 is not an element of the set

14 The set {4, 5, 6, …} has an unlimited number

of elements, so it is an infinite set Yes, 10 is

an element of the set

15 The set {x | x is a natural number larger than

11}, which can also be written as {11, 12, 13,

14, …}, has an unlimited number of elements,

so it is an infinite set No, 10 is not an element

of the set

16 The set {x | x is a natural number greater than

or equal to 10}, which can also be written as {10, 11, 12, 13, …}, has an unlimited number

of elements, so it is an infinite set Yes, 10 is

an element of the set

17 There are infinitely many fractions between 1

and 2, so {x | x is a fraction between 1 and 2}

is an infinite set No, 10 is not an element of the set

18 The set {x | x is an even natural number} has

no largest element Because it has an unlimited number of elements, this is an infinite set Yes, 10 is an element of the set

19 The elements of the set {12, 13, 14, …, 20}

are all the natural numbers from 12 to 20 inclusive There are 9 elements in the set, {12,

13, 14, 15, 16, 17, 18, 19, 20}

20 The elements of the set {8, 9, 10, …, 17} are

all the natural numbers from 8 to 17 inclusive There are 10 elements in the set, {8, 9, 10, 11,

12, 13, 14, 15, 16, 17}

21 Each element of the set  1 1 1

1, , ,, after the first is found by multiplying the preceding number by 1

2 There are 6 elements in the set,

 1 1 1 1 1

2 4 8 16 32

1, , , , ,

22 Each element of the set {3, 9, 27, …, 729}

after the first is found by multiplying the preceding number by 3 There are 6 elements

in the set, {3, 9, 27, 81, 243, 729}

23 To find the elements of the set

{17, 22, 27, …, 47}, start with 17 and add 5 to find the next number There are 7 elements in the set, {17, 22, 27, 32, 37, 42, 47}

24 To find the elements of the set

{74, 68, 62, …, 38}, start with 74 and subtract

6 (or add –6) to find the next number There are 7 elements in the set, {74, 68, 62, 56, 50,

44, 38}

25 When you list all elements in the set {all

natural numbers greater than 8 and less than 15}, you obtain {9, 10, 11, 12, 13, 14}

26 When you list all elements in the set

{all natural numbers not greater than 4} you

37 0 is not an element of  because the empty ,

set contains no elements Thus, 0

38  is a subset of , not an element of 

The empty set contains no elements Thus we

44. True 3 is not an element of {7, 6, 5, 4}

45. True Both sets contain exactly the same four

48. False These two sets are not equal because {3, 7, 12, 14, 0} contains the element 0, which

of U, the statement is true

52. True This statement says “C is a subset of U.” Because every element of C is also an element

of U, the statement is true

53. True Because both elements of D, 2 and 10, are also elements of B, D is a subset of B

54. True Because both elements of D, 2, and 10, are also elements of A, D is a subset of A

55. False Set A contains two elements, 6 and 12, that are not elements of B Thus, A is not a subset of B

56. False Set B contains two elements, 2 and 8, that are not elements of C Thus, B is not a subset of C

57. True The empty set is a subset of every set

58. True The empty set is a subset of every set

59. True Because 4, 8, and 10 are all elements of

B, {4, 8, 10} is a subset of B

60. False Because 0 is not an element of D, {0, 2}

is not a subset of D

61. False Because B contains two elements, 4 and

8, that are not elements of D, B is not a subset

of D

62. False There are three elements of A (2, 6, and 8) that are not elements of C, so A is not a subset of C

63. Every element of {2, 4, 6} is also an element

of {2, 3, 4, 5, 6}, so {2, 4, 6} is a subset of {2, 3, 4, 5, 6}

We write {2, 4, 6} {2, 3, 4, 5, 6}

64. Every element of {1, 5} is also an element of

{0, 1 2, 3, 5}, so {1, 5} is a subset of the set

74. False The sets {6, 8, 9} and {9, 8, 6} are

equal because they contain exactly the same

three elements Their union contains the same

elements, namely 8, 9, and 6

75. {3, 5, 9, 10} {3, 5, 9, 10}

In order to belong to the intersection of two

sets, an element must belong to both sets

Because the empty set contains no elements,

{3, 5, 9, 10}  , so the statement is

false

76. True For any set A, A A

77. True Because the two sets are equal, their

union contains the same elements, namely 1,

82. MU Because MU, the intersection of M and U will contain the same elements as M

MR = {0, 1, 2, 3, 4, 6, 8}

85. MN There are no elements which belong to both M and N, so MN . M and N are disjoint

sets

86. UN Because NU, the elements belonging to U and N are all the elements belonging to N,

UN = N or {1, 3, 5, 7, 9, 11, 13}

87. NR = {0, 1, 2, 3, 4, 5, 7, 9, 11, 13}

88. MQ Because MQ, the elements belonging to M

or Q are all the elements belonging to Q

MQ = Q or {0, 2, 4, 6, 8, 10, 12}

89. N

The set N is the complement of set N, which

means the set of all elements in the universal

set U that do not belong to N

N Q or {0, 2, 4, 6, 8, 10, 12}

90. Q

The set Q is the complement of set Q, which

means the set of all elements in the universal

set U that do not belong to Q

Q N or {1, 3, 5, 7, 9, 11, 13}

91. M Q

First form M, the complement of M M

contains all elements of U that are not

First form R, the complement of R R

contains all elements of U the are not elements

of R Thus, R {5, 6, 7, 8, 9, 10, 11, 12,

13} Now form the intersection of Q and R

Thus, we have Q R {6, 8, 10, 12}

93.  R

Because the empty set contains no elements,

there are no elements belonging to both and

R Thus,  and R are disjoint sets, and

R

  

94.  Q

Because the empty set contains no elements,

there are no elements belonging to both 

and Q Thus,  and Q are disjoint sets, and

Q

  

95. N 

Because  contains no elements, the only

elements belonging to N or  are the

elements of N Thus,  and N are disjoint

sets, and N  N or {1, 3, 5, 7, 9, 11, 13}

96. R

Because  contains no elements, the only

elements belonging to R or  are the

elements of R Thus,  and R are disjoint

sets, and R R or {0, 1, 2, 3, 4}

97. (MN) R

First, form the intersection of M and N

Because M and N have no common elements

(they are disjoint), MN   Thus,

(MN) R   R. Now, because 

contains no elements, the only elements

belonging to R or  are the elements of R

Thus,  and R are disjoint sets, and

100. (RN) M

First form the union of R and N We have

RN = {0, 1, 2, 3, 4, 5, 7, 9, 11, 13} Now find the complement of M We have

M = {1, 3, 5, 7, 9, 10, 11, 12, 13} Now, find the intersection of these two sets We have

(RN)MN or {1, 3, 5, 7, 9, 11, 13}

101. ( M  Q) R First, find M, the complement of M We have M{1, 3, 5, 7, 9, 10, 11, 12, 13} Next,

form the union of M and Q We have

M  Q = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,

12, 13}U Thus, we have

( M  Q)RURR or {0, 1, 2, 3, 4}

102. Q (MN) First, form the union of M and N We have

MN = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 13} Now form the intersection of Q with this set

105. x x| U x, M

This means all elements of U except those in set M This gives the set {1, 3, 5, 7, 9, 10, 11,

12, 13}

106. x x| U x, R

This means all elements of U except those in

set R This gives the set {5, 6, 7, 8, 9, 10, 11,

12, 13}

107. x x| M and xQ

This means all elements that are members of

both set M and set Q, or MQ Note that set

M is a subset of set Q because all members of

set M are included in set Q Thus, the answer

is set M or {0, 2, 4, 6, 8}

108. x x| Q and xR

This means all elements that are members of

both set Q and set R, or QR Thus, the

answer is {0, 2, 4}

109. x x| M or xQ

This means all elements that are either

members of set M or set Q, or M Q Note

that set M is a subset of set Q because all

members of set M are included in set Q Thus,

the answer is set Q or {0, 2, 4, 6, 8, 10, 12}

110. x x| Qor xR

This means all elements that are either

members of set Q or set R, or QR Thus,

4. If the real number a is to the left of the real

number b on a number line, then a < (is less

than) b

5. The distance on a number line from a number

to 0 is the absolute value of that number

6 (a) 0 is a whole number Therefore, it is also

an integer, a rational number, and a real

number 0 belongs to B, C, D, F

(b) 34 is a natural number Therefore, it is

also a whole number, an integer, a

rational number, and a real number 34

belongs to A, B, C, D, F

(c) 94 is a rational number and a real

number 94 belongs to D, F

(d) 36 is a natural number Therefore, 6

it is also a whole number, an integer, a rational number, and a real number

a real number 2.16 belongs to D, F

11 1 and 3 are natural numbers

12 0, 1, and 3 are whole numbers

13 –6,  (or –3), 0, 1, and 3 are integers 124

14  6, 124 (or 3), 58, 0, 14, 1, and 3 are rational numbers

15  3, 2 and 12 are irrational numbers

16 All are real numbers

61 No; in general a  b b a Examples will

vary, i.e if a = 15 and b = 0, then

76 The process in your head should be the

82. This statement is true by the algebraic

definition of absolute value stated in the text

The P d value for a woman whose actual

systolic blood pressure is 116 and whose

normal value should be 120 is 4

90 Consider the relation, P d  P120 17

Because 103 and 137 both differ from 120 by

17, these are the two possible values for the

patient’s systolic blood pressure

For Exercises 91−98, x = −4 and y = 2

99 True because 2525 and 2525

100 True because the absolute value of any number

is always greater than or equal to 0

105 d P Q ,    1  4    1 4 3 3 or d P Q ,          4  1 4 1 3 3

106. d P R ,   8  4   8 4 12 12 or d P R ,     4 8 12 12

107. d Q R ,      8  1 8 1 9  or 9 d Q R ,      1 8 9 9

108 d Q S ,  12   1 13 13 or d Q S ,   1 12  13 13

109. xy0 if x and y have the same sign

110. x y2 0 if y is positive, because x2 is positive for any nonzero x

x  if x is negative, because y2 is positive for any nonzero y

113 Because x3 has the same sign as x,

  if x and y have different signs

115. Because     18 ( 1) 19 19 and 1 ( 18)  19 19, the number of strokes between their scores is 19

For exercises 121−132, we use the formula

250 1000 12.5 6.25 1250

,3

126 First, find the value of the numerator

132 First, find the value of the numerator

136 We use the maximum values for C, T, and Y,

A A A but the minimum value for I

A (because we want the fewest

38. 4 y5 is a polynomial of degree 5 and is a monomial

39. 6x3x4 is a polynomial It has degree 4 because the term 3x4 has the greatest degree

It is a binomial because there are two terms

40.  9y 5y3 is a polynomial It has degree 3

because the term 5 y3 has the greatest degree

It is a binomial because there are two terms

41. 7z52z31 is a polynomial It has degree 5

because the term 7z5 has the greatest

degree It is a trinomial because there are three

terms

42. 9t48t37 is a polynomial It has degree 4

because the term 9t4 has the greatest degree

It is a trinomial because there are three terms

43. 15a b2 312a b3 813b512b6 is a

polynomial It has degree 11 because the term

3 8

12a b has the greatest degree It is not a

monomial, binomial, or trinomial

44. 16x y5 712x y3 84xy918x10 is a

polynomial It has degree 12 because the term

5 7

16x y

 has the greatest degree It is not a

monomial, binomial, or trinomial

2

9

8x  x  is not a polynomial because one

term has a variable in the denominator

46. 2 6 35

1

3t t  is not a polynomial because one

term has a variable in the denominator

2 32

Note that (3y − 5)(3y + 5) is a special product,

the difference of squares

x x x x x x x

81212

20200

104 Both polynomials have missing terms Insert

each missing term with a 0 coefficient

2

2 2

x y The area of each rectangle is

xy Therefore, the area of the largest

square can be written as x22xyy2

(c) Answers will vary Sample answer: The

expressions are equivalent because they represent the same area

(d) It reinforces the special product for

squaring a binomial:

(xy) x 2xyy

106 Answers will vary Sample answer:

The total area of the largest rectangle is

60, 501, 000 ft

V  h a abb



(b) The shape becomes a rectangular box

with a square base Its volume is given by length × width × height or b h2

The pyramid is slightly smaller than the

Superdome in New Orleans

(c) The base area of the Great Pyramid is

5 5

4.2 4.2

2 322.1

4 4

15 15

3 815

Section R.4 Factoring Polynomials

1 The process of finding polynomials whose

product equals a given polynomial is called

factoring

2 A polynomial is factored completely when it is

written as a product of prime polynomials

3. Factoring is the opposite of multiplying

4. When a polynomial has more than three terms,

it can sometimes be factored using factoring

by grouping

5. There is no factoring pattern for a sum of squares in the real number system In particular, x2y2 does not factor as

can be factored as (x1)(x1) The other choices are not correct factorizations of 4

1

x 

10 C: x3 8 (x2)(x22x4) Use the pattern for factoring the sum of cubes,

16 The greatest common factor is hj

5( 3) 2( 3) ( 3)( 3)[5( 3) 2 ( 3)]

(8 3)(2 5) 16 40 6 15 and(3 8 )(5 2 ) 15 6 40 16

combinations of these factors until the correct one is found

2

6a 11a 4 2a1 3a 4

36 The positive factors of 8 could be 2 and 4, or 1

and 8 As factors of –21, we could have –3

and 7, 3 and –7, 1 and –21, or –1 and 21 Try

different combinations of these factors until

the correct one is found

2

8h 2h21 4h7 2h 3

37 The positive factors of 3 are 1 and 3 Because

the middle term is positive, we know the

factors of 8 must both be positive As factors

of 8, we could have 1 and 8, or 2 and 4 Try

different combinations of these factors until

the correct one is found

2

3m 14m 8 3m2 m 4

38. The positive factors of 9 are 3 and 3, or 1 and

9 Because the middle term is negative, we

know the factors of 8 must both be negative

As factors of 8, we could have –1 and –8, or

–2 and –4 Try different combinations of these

factors until the correct one is found

2

9y 18y 8 3y4 3y 2

39 The positive factors of 15 are 1 and 15, or 3

and 5 Because the middle term is positive, we

know the factors of 8 must both be positive

As factors of 8, we could have 1 and 8, or 2

and 4 Trying different combinations of these

factors we find that 15p224p is prime 8

40 The positive factors of 9 are 1 and 9, or 3 and

3 As factors of −2, we could have −1 and 2,

or 1 and −2 Trying different combinations of

these factors we find that 9x24x is 2

6k 5kp6p  2k3p 3k2p

44. The positive factors of 14 could be 2 and 7, or

1 and 14 As factors of –15, we could have 3 and –5, –3 and 5, 1 and –15, or –1 and 15 Try different combinations of these factors until the correct one is found

14m 11mr15r  7m5r 2m3r

45 The positive factors of 5 can only be 1 and 5

As factors of –6, we could have –1 and 6, –6 and 1, –2 and 3, or –3 and 2 Try different combinations of these factors until the correct one is found

5a 7ab6b  5a3b a2b

46. The positive factors of 12 could be 4 and 3,

2 and 6, or 1 and 12 As factors of –5 we could have –1 and 5 or –5 and 1 Try different combinations of these factors until the correct one is found

12s 11st5t  4s5t 3s t

47 The positive factors of 12 could be 4 and 3,

2 and 6, or 1 and 12 The factors of −1 are 1 and −1 Try different combination of these factors until the correct one is found

12x xyy (4xy)(3xy)

48 The positive factors of 30 could be 5 and 6, 3

and 10, 2 and 15, or 1 and 30 The only factors

of –1 (the coefficient of the last term) are 1 and –1 Try different combinations of these factors until the correct one is found

50 First, factor out the greatest common factor,

2 2

(21 7 2)(3 1 4)(21 9)(3 3) 3(7 3)(3)( 1)9(7 3)( 1)

(8 6 3)(12 9 1)(8 3)(12 10) 2(8 3)(6 5)

86. Let u = 5x + 7 This substitution gives

96. Let x = 2y – 1 This substitution gives

2(2 1) 4(2 1) 4 [(2 1) 2]

Try to factor this trinomial by trial and error

Trying all the possible combinations of factors

will show that none of them are correct

2 2

2 2 2 2

x x

(3 5) 18(3 5) 81 [(3 5) 9]

(3 5 9)(3 4)

a a

115 Answers will vary Sample answer: In general,

a sum of squares is not factorable over the real

number system If there is a greatest common

factor, as in 4x216, it may be factored out,

as here, to obtain 4x24 

116 Answers will vary Sample answer: The

smaller figures are pieced together to form the

large square with side length x + y Thus, the

sum of the areas of the smaller figures,

x  xyy will be equivalent to the area

of the large square,  2

127. From Exercise 125, we have

commutative property of addition

129 The answer in Exercise 127 and the final line

in Exercise 128 are the same

We use the result from Exercise 128

Section R.5 Rational Expressions

1 The quotient of two polynomials in which the

denominator is not equal to zero is a rational

expression

2 The domain of a rational expression consists

of all real numbers except those that make the

denominator equal to 0

3. In the rational expression 1,

5

x x



 the domain cannot include the number 5

4. A rational expression is in lowest terms when

the greatest common factor of its numerator

and its denominator is 1

(4x + 2)(x – 1) = 0 is excluded from the

domain

(4x2)(x  1) 0

1 2

(2x + 3)(x – 5) = 0 is excluded from the

domain

(2x + 3)(x – 5) = 0

3 2