Solution manual for applied mathematics for the managerial life and social sciences 7th edition by tan

The number 3 is an integer, a rational number, and a real number.. The number 420 is an integer, a rational number, and a real number.. The number38is a rational real number.. The numb

1 FUNDAMENTALS OF ALGEBRA

1.1 Real Numbers

Concept Questions page 6

c. 34(answer is not unique) d. 3 (answer is not unique)

3 a The associative law of addition states that a  b  c  a  b  c.

b The distributive law states that ab  ac  a b  c.

4 a No For example, 4  2  5  4  3  7  4  2  5  2  5  3, and

4 52  4

2 5

1 The number 3 is an integer, a rational number, and a real number.

2 The number 420 is an integer, a rational number, and a real number.

3 The number38is a rational real number 4 The number 1254 is a rational real number

5 The number11 is an irrational real number 6 The number 5 is an irrational real number

7 The number2 is an irrational real number 8 The number2 is an irrational real number

9 The number 2421 is a rational real number 10 The number 271828    is an irrational real number

11 False 2 is not a whole number 12 True.

15 False No natural number is irrational 16 True.

17. 2x  y  z  z  2x  y: The Commutative Law of Addition.

1

.

32. 2x  1 x  3

2x  1 x  3

2x  1 2x  1 Property 2 of quotients.

x x  1 Property 7 of quotients and the Distributive Law.

37 False Consider a  2 and b  12 Then ab  1, but a  1 and b  1.

38 True Multiplying both sides of the equation by 1

a (which exists because a  0), we have 1

a ab  1

a 0, or b  0.

39 False Consider a  3 and b  2 Then a  b  3  2  b  a  2  3  1.

40 False Consider a  3 and b  2 Then a

18. 3x2yz3x2yz:TheAssociativeLawofAddition

19. u33u:TheCommutativeLawofMultiplication

20. a2

b2c



a2b2

c:TheAssociativeLawofMultiplication

21. u22uu:TheDistributiveLaw

22. 2u2u:TheDistributiveLaw

23. 2x3yx4y2x

3yx4y

:TheAssociativeLawofAddition

24. a2b a3ba a3b2b a3b:TheDistributiveLaw

25. a[cd]acd:Property1ofnegatives

26. 2xy

3x2y

2xy3x2y:Property3ofnegatives

27. 02a3b0:Property1involvingzero

28. Ifxyxy0,thenxyorxy.Property2involvingzero

42 False Consider a  1, b  2, and c  3 Then a

b c 

123 

Concept Questions page 12

1 a No, this is not a polynomial expression because of the term of 2x in which the power of x is not a nonnegative

integer

b Yes

c No It is a rational expression.

2 a A polynomial of degree n in x is an expression of the form a n x n  a n1 x n1      a1x  a0, where n is a nonnegative integer and a0 a1     a n are real numbers with a n  0 One polynomial of degree 4 in x is

3



2 3

 2 3

 2 3

3

 3

3 5

 3 5

 3 5

12x3

46. x  2y y  3x2xy3 x  y  1  xy3x2

2y26x y2x y3x 3y3  3x27x y2y23x 3y3.

57. Thetotalweeklyprofitisgivenbytherevenueminusthecost:



004x22000x

0000002x3

58. Thetotalrevenueisgivenbyx px 00004x1000004x2

10x.Therefore,thetotalprofitisgivenbytherevenueminusthecost:00004x210x

200t

02t2

150t

03t2

50tthousanddollars,where0t12

61. Thedifferenceisgivenby12

05t2

3t54

075t385

1205t2

isapolynomialofdegree3,not2

66. False.Forexample, px3x1isapolynomialofdegree3andqx32isapolynomialofdegree3,but

pqx3x1

x32

x3isapolynomialofdegree1

1.3 FactoringPolynomials

ConceptQuestions page 18

1. Apolynomialiscompletelyfactoredoverthesetofintegersifitisexpressedasaproductofprimepolynomialswithintegralcoefficients.Anexampleis4x29y22x3y 2x3y.



32



2u5

9. 3ab2cd2a 2cd22cd[3ab2a 2cd]2cd3ab4ac2ad.

10. 4u22u6u22u

45 3ax  6ay  bx  2by  3a x  2y  b x  2y  x  2y 3a  b.

46 6ux  4uy  3x  2y  2u 3x  2y   3x  2y  3x  2y 2u  .

52 a2 b2 a  b  a  b a  b  a  b  a  b a  b  1.

53 au2 a  c u  c  au2 au  cu  c  au u  1  c u  1  u  1 au  c.

54 ax2 1  ab x y  by2 ax2 x y  abx y  by2 ax x  by  y x  by  x  by ax  y.

1.4 Rational Expressions

Concept Questions page 25

1 a Quotients of polynomials are rational expressions; 2x

2

 1

3x2

 3x  4.

b Any polynomial P can be written in the form P

1, but not all rational expressions can be written as a polynomial.

3. 4x  12 5x  15 

6x2 2x  1

8y24y3

 4y2 8y 

8y24y

 y  1 

2y  3 y  1

2y  1 y  1 

2y  3 2y  1.

3y  1 2y  3

2y  3 2y  3 

3y  1 2y  3.

3y25y3  54y2

6y  24 8y  12 

6r  12 4r  2 

 13y  6

6y2 13y  6 9y2

x  1 x



2x  2

x

x2 2x

y2



y  x y

Concept Questions page 30

1 If a is any real number and n is a natural number, then the expression a n is defined as the number

a n  a  a  a      a  

, where the number a is the base and the superscript n is the exponent, or power, to which the

base is raised For any real number a, a0 1 If n is a negative number and a  0, then a n 1

5

 35

4

16

81.

13 4

b  a ab

 b  a

b  a.

Concept Questions page 35

1 An equation is a statement that two mathematical expressions are equal A solution of an equation involving one

variable is a number that renders the equation a true statement when it is substituted for the variable The solutionset of an equation is the set of all solutions to the equation

One example: 2x  3 is an equation Its solution is x  32because 2

3 2

3 A linear equation in the variable x is an equation that can be written in the form ax  b  0, where a and b are

constants with a  0 Example: 3x  4  5 Solving for x, we have 3x  1, so x  13

3k  1



 12

1

4k  2



4k  12  3k  24 4k  12  12  3k  24  12

4k  3k  36 4k  3k  3k  36  3k

k  36.

9. 15p  3  13p  5

151

13. 35k  1  142k  4

12k  1  5 2k  4

12k  12  10k  20 2k  8

2[2x  3 x  4]



 6

2

5x2 16x  3  10x  5x2

16x  3  10x 6x  3  0 6x  3

52x  3  2 x  1

10x  15  2x  2 10x  2x  17 8x  17

q  1



 q  1 q  2

3

q  2



q  2 2  q  1 3 2q  4  3q  3

2 

k  3 k

3x  2 3x  1



2x  1 3x  2



 3x  2 3x  1



2x  1 3x  1

x x  2



 x x  2

2

1.7 Rational Exponents and Radicals

Concept Questions page 44

1 If n is a natural number and a and b are real numbers such that a n  b, then a is the nth root of b For example, 3 is

the 4th root of 81; that is481  3

2 The principal nth root of a positive real number b, when n is even, is the positive root of b If n is odd, it is the

unique nth root of b The principal 4th root of 16 is 2, and the principal (and only) 3rd root of 8 is 2.

3 The process of eliminating a radical from the denominator of an algebraic expression is referred to as rationalizing

the denominator For example, 1



1 6



12

2

925

32



35

23



32

19 

278

13

 

827

2 3

3 

3

23



32 

3

18

3

81

4 

3

813



4 

3333



22 

3

23

2x2 

3

x2

2



2 

6



2y2 3



32 3

x12x  y135x  3y

6x x  y .

83.

1 2

x  1.

Check:31  1 2.?

Yes, x  1 is a solution.

86. 2x  3  3 2x  3  9 2x  12

87. k2

 4  4  k

k2 4  16  8k  k2

4  16  8k 8k  20

k 208 52

Check:



5 2

2

 4 4 ? 523

1

3 1 

1 3

?

 3

1 3

4

3

1 3

?

 2

1

3

1 33

1 3

?

 3

1

4

5 1 

4 5

5

4 5

?

 3

1

5 2

1 5

1 5

?



1

Concept Questions page 52

1 A quadratic equation in the variable x is any equation that can be written in the form ax2 bx  c  0 For

example, 4x2 3x  4  0 is a quadratic equation.

a x   c

a where the coefficient of x

2is 1 and the constant term is on the

right side of the equation For example, 3x2 2x  3  0 can be written as x223x  1.

232

Continuing our example, x2 23x  19  1  19, so 

x 132



10

9. 14x2 x  1  0 is equivalent to x2 4x  4  0, or x  22 0 So x  2 is a double root.

10. 12a2 a  12  0 is equivalent to a2 2a  24  0, or a  6 a  4  0, and so a  6 or a  4.

11 Rewrite the given equation in the form 2m2

 7m  6  0 Then 2m  3 m  2  0 and m  32 or m  2.

12 Rewrite the given equation in the form 6x2

 5x  6  0 Factoring, we have 3x  2 2x  3  0, and so x  23

16 Rewrite the given equation in the form 6m2 13m  5  0 Then 2m  1 3m  5  0, and so m  12or

2  12



6 Therefore, x  1 

 6

2 or x  1 

 6

2

 3 

1 2

2,

2

 4  2

3 4

2, 2

4 Therefore, x  34

 41

4 or x  34

 41

4 or x  54

 5

4

25 4x2 13, so x2134 and x  213

26 7 p2 20, so p2 207 and p  

20

7  2

5

29 Rewrite the given equation in the form m2

 4m  1  0 Then using the quadratic formula with a  1, b  4, and c  1, we obtain m   4 

31 Rewrite the given equation in the form 8x2  8x  3  0 Then using

the quadratic formula with a  8, b  8, and c  3, we obtain

33 Rewrite the given equation in the form 2x2 4x  3  0 Then using the quadratic formula with a  2, b  4, and

10

34 Rewrite the given equation in the form 2y2 7y  15  0 Then using the quadratic formula with a  2, b  7,

36 Using the quadratic formula with a  02, b  16, and c  12, we obtain

x  16 

162

37 x4 5x2 6  0 Let m  x2 Then the equation reads m2 5m  6  0 Now, factoring, we obtain

41 6x  22 7 x  2  3  0 Let y  x  2 Then we have 6y2 7y  3  0 Factoring, we obtain

2y  3 3y  1  0, and so y  32or 13 Therefore, x  2  32or 13, and so x  72or 53

43 6  13  6  0 Let x  Then 6x2

 13x  6  0, 2x  3 3x  2  0, and so x  32or x  23.Then the solutions are  x2

 94or 49.Check  4

9: 64 9



 13

4

9 6  249  13 23 6 0 Yes,? 49is a solution

Check  94: 6

9 4



 13

9

4 6  544  13 32 6 0 Yes,? 94is also a solution

3y  1 y  1  16  524 y  1

3y2 2y  1  16  524 y  1

3y2 2y  1  16  10y  10 3y2 8y  5  0

But because x  1 results in division by zero in the

original equation, we discard it Thus, the only solution