This is “Rational Expressions and Equations”, chapter 7 from the book Beginning Algebra (v. 1.0). For details on it (including licensing), click here.

For more information on the source of this book, or why it is available for free, please see the project's home page. You can browse or download additional books there. You may also download a PDF copy of this book (81 MB) or just this chapter (8 MB), suitable for printing or most e-readers, or a .zip file containing this book's HTML files (for use in a web browser offline).

Has this book helped you? Consider passing it on:
Creative Commons supports free culture from music to education. Their licenses helped make this book available to you.
DonorsChoose.org helps people like you help teachers fund their classroom projects, from art supplies to books to calculators.

Chapter 7 Rational Expressions and Equations

 

7.1 Simplifying Rational Expressions

Learning Objectives

  1. Determine the restrictions to the domain of a rational expression.
  2. Simplify rational expressions.
  3. Simplify expressions with opposite binomial factors.
  4. Simplify and evaluate rational functions.

Rational Expressions, Evaluating, and Restrictions

A rational number, or fraction a b, is a real number defined as a quotient of two integers a and b, where b0. Similarly, we define a rational expressionThe quotient PQ of two polynomials P and Q, where Q ≠ 0., or algebraic fractionTerm used when referring to a rational expression. PQ, as the quotient of two polynomials P and Q, where Q0. Some examples of rational expressions follow:

The example x+3x5 consists of linear expressions in both the numerator and denominator. Because the denominator contains a variable, this expression is not defined for all values of x.

 

Example 1: Evaluate x+3x5 for the set of x-values {−3, 4, 5}.

Solution: Substitute the values in for x.

Answer: When x=3, the value of the rational expression is 0; when x=4, the value of the rational expression is −7; and when x=5, the value of the rational expression is undefined.

 

This example illustrates that variables are restricted to values that do not make the denominator equal to 0. The domain of a rational expressionThe set of real numbers for which the rational expression is defined. is the set of real numbers for which it is defined, and restrictionsThe set of real numbers for which a rational expression is not defined. are the real numbers for which the expression is not defined. We often express the domain of a rational expression in terms of its restrictions.

 

Example 2: Find the domain of the following: x+72x2+x6.

Solution: In this example, the numerator x+7 is a linear expression and the denominator 2x2+x6 is a quadratic expression. If we factor the denominator, then we will obtain an equivalent expression.

Because rational expressions are undefined when the denominator is 0, we wish to find the values for x that make it 0. To do this, apply the zero-product property. Set each factor in the denominator equal to 0 and solve.

We conclude that the original expression is defined for any real number except 3/2 and −2. These two values are the restrictions to the domain.

It is important to note that −7 is not a restriction to the domain because the expression is defined as 0 when the numerator is 0.

Answer: The domain consists of any real number x, where x32 and x2.

 

We can express the domain of the previous example using notation as follows:

The restrictions to the domain of a rational expression are determined by the denominator. Ignore the numerator when finding those restrictions.

 

Example 3: Determine the domain: x4+x32x2xx21.

Solution: To find the restrictions to the domain, set the denominator equal to 0 and solve:

These two values cause the denominator to be 0. Hence they are restricted from the domain.

Answer: The domain consists of any real number x, where x±1.

 

Example 4: Determine the domain: x2254.

Solution: There is no variable in the denominator and thus no restriction to the domain.

Answer: The domain consists of all real numbers, R.

Simplifying Rational Expressions

When simplifying fractions, look for common factors that cancel. For example,

We say that the fraction 12/60 is equivalent to 1/5. Fractions are in simplest form if the numerator and denominator share no common factor other than 1. Similarly, when working with rational expressions, look for factors to cancel. For example,

The resulting rational expression is equivalent if it shares the same domain. Therefore, we must make note of the restrictions and write

In words, x+4(x3)(x+4) is equivalent to 1x3, if x3 and x4. We can verify this by choosing a few values with which to evaluate both expressions to see if the results are the same. Here we choose x=7 and evaluate as follows:

It is important to state the restrictions before simplifying rational expressions because the simplified expression may be defined for restrictions of the original. In this case, the expressions are not equivalent. Here −4 is defined for the simplified equivalent but not for the original, as illustrated below:

 

Example 5: Simplify and state the restriction: 25x215x3.

Solution: In this example, the expression is undefined when x is 0.

Therefore, the domain consists of all real numbers x, where x0. With this understanding, we can cancel common factors.

Answer: 53x, where x0

 

Example 6: State the restrictions and simplify: 3x(x5)(2x+1)(x5).

Solution: To determine the restrictions, set the denominator equal to 0 and solve.

The domain consists of all real numbers except for −1/2 and 5. Next, we find an equivalent expression by canceling common factors.

Answer: 3x2x+1, where x12 and x5

 

Typically, rational expressions are not given in factored form. If this is the case, factor first and then cancel. The steps are outlined in the following example.

 

Example 7: State the restrictions and simplify: 3x+6x2+x2.

Solution:

Step 1: Completely factor the numerator and denominator.

Step 2: Determine the restrictions to the domain. To do this, set the denominator equal to 0 and solve.

The domain consists of all real numbers except −2 and 1.

Step 3: Cancel common factors, if any.

Answer: 3x1, where x1 and x2

 

Example 8: State the restrictions and simplify: x2+7x30x27x+12.

Solution: First, factor the numerator and denominator.

Any value of x that results in a value of 0 in the denominator is a restriction. By inspection, we determine that the domain consists of all real numbers except 4 and 3. Next, cancel common factors.

Answer: x+10x4, where x3 and x4

 

It is important to remember that we can only cancel factors of a product. A common mistake is to cancel terms. For example,

 

Try this! State the restrictions and simplify: x2165x220x.

Answer: x+45x, where x0 and x4

Video Solution

(click to see video)

In some examples, we will make a broad assumption that the denominator is nonzero. When we make that assumption, we do not need to determine the restrictions.

 

Example 9: Simplify: xy+y23x3yx2y2. (Assume all denominators are nonzero.)

Solution: Factor the numerator by grouping. Factor the denominator using the formula for a difference of squares.

Next, cancel common factors.

Answer: y3xy

Opposite Binomial Factors

Recall that the opposite of the real number a is −a. Similarly, we can define the opposite of a polynomial P to be −P. We first consider the opposite of the binomial ab:

This leads us to the opposite binomial propertyIf given a binomial ab, then the opposite is (ab)=ba.:

This is equivalent to factoring out a –1.

If ab, then we can divide both sides by (ab) and obtain the following:

 

Example 10: State the restrictions and simplify: 3xx3.

Solution: By inspection, we can see that the denominator is 0 if x=3. Therefore, 3 is the restriction to the domain. Apply the opposite binomial property to the numerator and then cancel.

Answer: 3xx3=1, where x3

 

Since addition is commutative, we have

or

Take care not to confuse this with the opposite binomial property. Also, it is important to recall that

In other words, show a negative fraction by placing the negative sign in the numerator, in front of the fraction bar, or in the denominator. Generally, negative denominators are avoided.

 

Example 11: Simplify and state the restrictions: 4x2x2+3x10.

Solution: Begin by factoring the numerator and denominator.

Answer: x+2x+5, where x2 and x5

 

Try this! Simplify and state the restrictions: 2x27x1525x2.

Answer: 2x+3x+5, where x±5

Video Solution

(click to see video)

Rational Functions

Rational functions have the form

where p(x) and q(x) are polynomials and q(x)0. The domain of a rational function consists of all real numbers x such that the denominator q(x)0.

 

Example 12:

a. Simplify: r(x)=2x2+5x36x2+18x.

b. State the domain.

c. Calculate r(2).

Solution:

a. To simplify the rational function, first factor and then cancel.

b. To determine the restrictions, set the denominator of the original function equal to 0 and solve.

The domain consists of all real numbers x, where x0 and x3.

c. Since −2 is not a restriction, substitute it for the variable x using the simplified form.

Answers:

a. r(x)=2x16x

b. The domain is all real numbers except 0 and −3.

c. r(2)=512

 

If a cost functionA function that represents the cost of producing a certain number of units. C(x) represents the cost of producing x units, then the average costThe total cost divided by the number of units produced, which can be represented by c(x)=C(x)x, where C(x) is a cost function. c(x) is the cost divided by the number of units produced.

 

Example 13: The cost in dollars of producing t-shirts with a company logo is given by C(x)=7x+200, where x represents the number of shirts produced. Determine the average cost of producing

a. 40 t-shirts

b. 250 t-shirts

c. 1,000 t-shirts

Solution: Set up a function representing the average cost.

Next, calculate c(40), c(250), and c(1000).

Answers:

a. If 40 t-shirts are produced, then the average cost per t-shirt is $12.00.

b. If 250 t-shirts are produced, then the average cost per t-shirt is $7.80.

c. If 1,000 t-shirts are produced, then the average cost per t-shirt is $7.20.

Key Takeaways

  • Rational expressions usually are not defined for all real numbers. The real numbers that give a value of 0 in the denominator are not part of the domain. These values are called restrictions.
  • Simplifying rational expressions is similar to simplifying fractions. First, factor the numerator and denominator and then cancel the common factors. Rational expressions are simplified if there are no common factors other than 1 in the numerator and the denominator.
  • Simplified rational expressions are equivalent for values in the domain of the original expression. Be sure to state the restrictions if the denominators are not assumed to be nonzero.
  • Use the opposite binomial property to cancel binomial factors that involve subtraction. Use (ab)=ba to replace factors that will then cancel. Do not confuse this with factors that involve addition, such as (a+b)=(b+a).

Topic Exercises

Part A: Rational Expressions

Evaluate for the given set of x-values.

1. 5x; {−1, 0, 1}

2. 4x3x2; {−1, 0, 1}

3. 1x+9; {−10, −9, 0}

4. x+6x5; {−6, 0, 5}

5. 3x(x2)2x1; {0, 1/2, 2}

6. 9x21x7; {0, 1/3, 7}

7. 5x29; {−3, 0, 3}

8. x225x23x10; {−5, −4, 5}

9. Fill in the following chart:

10. Fill in the following chart:

11. Fill in the following chart:

12. Fill in the following chart:

An object’s weight depends on its height above the surface of earth. If an object weighs 120 pounds on the surface of earth, then its weight in pounds, W, x miles above the surface is approximated by the formula W=12040002(4000+x)2

For each problem below, approximate the weight of a 120-pound object at the given height above the surface of earth. (1 mile = 5,280 feet)

13. 100 miles

14. 1,000 miles

15. 44,350 feet

16. 90,000 feet

The price to earnings ratio (P/E) is a metric used to compare the valuations of similar publicly traded companies. The P/E ratio is calculated using the stock price and the earnings per share (EPS) over the previous 12‑month period as follows: P/E=pricepershareearningspershare

If each share of a company stock is priced at $22.40, then calculate the P/E ratio given the following values for the earnings per share.

17. $1.40

18. $1.21

19. What happens to the P/E ratio when earnings decrease?

20. What happens to the P/E ratio when earnings increase?

State the restrictions to the domain.

21. 13x

22. 3x27x5

23. 3x(x+1)x+4

24. 2x2(x3)x1

25. 15x1

26. x23x2

27. x95x(x2)

28. 1(x3)(x+6)

29. x1x2

30. x29x236

31. 12x(x+3)(2x1)

32. x3(3x1)(2x+3)

33. 4x(2x+1)12x2+x1

34. x53x215x

Part B: Simplifying Rational Expressions

State the restrictions and then simplify.

35. 5x220x3

36. 12x660x

37. 3x2(x2)9x(x2)

38. 20(x3)(x5)6(x3)(x+1)

39. 6x2(x8)36x(x+9)(x8)

40. 16x21(4x+1)2

41. 9x26x+1(3x1)2

42. x7x249

43. x264x2+8x

44. x+10x2100

45. 2x312x25x230x

46. 30x5+60x42x38x

47. 2x12x2+x6

48. x2x63x28x3

49. 6x225x+253x2+16x35

50. 3x2+4x15x29

51. x210x+21x24x21

52. x31x21

53. x3+8x24

54. x416x24

Part C: Simplifying Rational Expressions with Opposite Binomial Factors

State the restrictions and then simplify.

55. x99x

56. 3x223x

57. x+66+x

58. 3x+11+3x

59. (2x5)(x7)(7x)(2x1)

60. (3x+2)(x+5)(x5)(2+3x)

61. x24(2x)2

62. 169x2(3x+4)2

63. 4x2(10x)3x3300x

64. 2x+14x349x

65. 2x27x414x2

66. 9x244x6x2

67. x25x14715x+2x2

68. 2x3+x22x11+x2x2

69. x3+2x3x262+x2

70. 27+x3x2+6x+9

71. 64x3x28x+16

72. x2+44x2

Simplify. (Assume all denominators are nonzero.)

73. 15x3y25xy2(x+y)

74. 14x7y2(x2y)47x8y(x2y)2

75. y+xx2y2

76. yxx2y2

77. x2y2(xy)2

78. a2ab6b2a26ab+9b2

79. 2a211a+1232+2a2

80. a2b3a23a23ab

81. xy2x+y3yxxy2

82. x3xy2x2y+y3x22xy+y2

83. x327x2+3x+9

84. x2x+1x3+1

Part D: Rational Functions

Calculate the following.

85. f(x)=5xx3; f(0), f(2), f(4)

86. f(x)=x+7x2+1; f(1), f(0), f(1)

87. g(x)=x3(x2)2; g(0), g(2), g(2)

88. g(x)=x299x2; g(2), g(0), g(2)

89. g(x)=x3x2+1; g(1), g(0), g(1)

90. g(x)=5x+1x225; g(1/5), g(1), g(5)

State the restrictions to the domain and then simplify.

91. f(x)=3x26xx2+4x+4

92. f(x)=x2+6x+92x2+5x3

93. g(x)=9xx281

94. g(x)=x3273x

95. g(x)=3x15102x

96. g(x)=255x4x20

97. The cost in dollars of producing coffee mugs with a company logo is given by C(x)=x+40, where x represents the number of mugs produced. Calculate the average cost of producing 100 mugs and the average cost of producing 500 mugs.

98. The cost in dollars of renting a moving truck for the day is given by C(x)=0.45x+90, where x represents the number of miles driven. Calculate the average cost per mile if the truck is driven 250 miles in one day.

99. The cost in dollars of producing sweat shirts with a custom design on the back is given by C(x)=1200+(120.05x)x, where x represents the number of sweat shirts produced. Calculate the average cost of producing 150 custom sweat shirts.

100. The cost in dollars of producing a custom injected molded part is given by C(x)=500+(30.001x)x, where x represents the number of parts produced. Calculate the average cost of producing 1,000 custom parts.

Part E: Discussion Board

101. Explain why baab=1 and illustrate this fact by substituting some numbers for the variables.

102. Explain why b+aa+b=1 and illustrate this fact by substituting some numbers for the variables.

103. Explain why we cannot cancel x in the expression xx+1.

Answers

1: −5, undefined, 5

3: −1, undefined, 1/9

5: 0, undefined, 0

7: Undefined, −5/9, undefined

9:

11:

13: 114 pounds

15: 119.5 pounds

17: 16

19: The P/E ratio increases.

21: x0

23: x4

25: x15

27: x0 and x2

29: x±1

31: x0, x3, and x12

33: x13 and x14

35: 14x; x0

37: x3; x0, 2

39: x6(x+9); x0,9, 8

41: 1 ; x13

43: x8x; x0,8

45: 2x5; x0, 6

47: 2x12x2+x6; x2,32

49: 2x5x+7; x7,53

51: x3x+3; x3, 7

53: x22x+4x2; x±2

55: −1; x9

57: 1; x6

59: 2x52x1; x12,7

61: x+2x2; x2

63: 4x3(x+10); x±10, 0

65: x412x; x±12

67: x+22x1; x12,7

69: x3; none

71: 16+4x+x2x4; x4

73: 3x2x+y

75: 1xy

77: x+yxy

79: 2a32(4+a)

81: x+yx

83: x3

85: f(0)=0, f(2)=10, f(4)=20

87: g(0)=0, g(2) undefined, g(2)=1/2

89: g(1)=1/2, g(0)=0, g(1)=1/2

91: f(x)=3xx+2; x2

93: g(x)=1x+9; x±9

95: g(x)=32; x5

97: The average cost of producing 100 mugs is $1.40 per mug. The average cost of producing 500 mugs is $1.08 per mug.

99: $12.50

7.2 Multiplying and Dividing Rational Expressions

Learning Objectives

  1. Multiply rational expressions.
  2. Divide rational expressions.
  3. Multiply and divide rational functions.

Multiplying Rational Expressions

When multiplying fractions, we can multiply the numerators and denominators together and then reduce, as illustrated:

Multiplying rational expressions is performed in a similar manner. For example,

In general, given polynomials P, Q, R, and S, where Q0 and S0, we have

In this section, assume that all variable expressions in the denominator are nonzero unless otherwise stated.

 

Example 1: Multiply: 12x25y320y46x3.

Solution: Multiply numerators and denominators and then cancel common factors.

Answer: 8yx

 

Example 2: Multiply: x3x+5x+5x+7.

Solution: Leave the product in factored form and cancel the common factors.

Answer: x3x+7

 

Example 3: Multiply: 15x2y3(2x1)x(2x1)3x2y(x+3).

Solution: Leave the polynomials in the numerator and denominator factored so that we can cancel the factors. In other words, do not apply the distributive property.

Answer: 5xy2x+3

 

Typically, rational expressions will not be given in factored form. In this case, first factor all numerators and denominators completely. Next, multiply and cancel any common factors, if there are any.

 

Example 4: Multiply: x+5x5x5x225.

Solution: Factor the denominator x225 as a difference of squares. Then multiply and cancel.

Keep in mind that 1 is always a factor; so when the entire numerator cancels out, make sure to write the factor 1.

Answer: 1x5

 

Example 5: Multiply: x2+3x+2x25x+6x27x+12x2+8x+7.

Solution:

It is a best practice to leave the final answer in factored form.

Answer: (x+2)(x4)(x2)(x+7)

 

Example 6: Multiply: 2x2+x+3x2+2x83x6x2+x.

Solution: The trinomial 2x2+x+3 in the numerator has a negative leading coefficient. Recall that it is a best practice to first factor out a −1 and then factor the resulting trinomial.

Answer: 3(2x3)x(x+4)

 

Example 7: Multiply: 7xx2+3xx2+10x+21x249.

Solution: We replace 7x with 1(x7) so that we can cancel this factor.

Answer: 1x

 

Try this! Multiply: x2648xx+x2x2+9x+8.

Answer: x

Video Solution

(click to see video)

Dividing Rational Expressions

To divide two fractions, we multiply by the reciprocal of the divisor, as illustrated:

Dividing rational expressions is performed in a similar manner. For example,

In general, given polynomials P, Q, R, and S, where Q0, R0, and S0, we have

 

Example 8: Divide: 8x5y25z6÷20xy415z3.

Solution: First, multiply by the reciprocal of the divisor and then cancel.

Answer: 6x425y3z3

 

Example 9: Divide: x+2x24÷x+3x2.

Solution: After multiplying by the reciprocal of the divisor, factor and cancel.

Answer: 1x+3

 

Example 10: Divide: x26x16x2+4x21÷x2+9x+14x28x+15.

Solution: Begin by multiplying by the reciprocal of the divisor. After doing so, factor and cancel.

Answer: (x8)(x5)(x+7)2

 

Example 11: Divide: 94x2x+2 ÷(2x3).

Solution: Just as we do with fractions, think of the divisor (2x3) as an algebraic fraction over 1.

Answer: 2x+3x+2

 

Try this! Divide: 4x2+7x225x2÷14x100x4.

Answer: 4x2(x+2)

Video Solution

(click to see video)

Multiplying and Dividing Rational Functions

The product and quotient of two rational functions can be simplified using the techniques described in this section. The restrictions to the domain of a product consist of the restrictions of each function.

 

Example 12: Calculate (fg)(x) and determine the restrictions to the domain.

Solution: In this case, the domain of f(x) consists of all real numbers except 0, and the domain of g(x) consists of all real numbers except 1/4. Therefore, the domain of the product consists of all real numbers except 0 and 1/4. Multiply the functions and then simplify the result.

Answer: (fg)(x)=4x+15x, where x0,14

 

The restrictions to the domain of a quotient will consist of the restrictions of each function as well as the restrictions on the reciprocal of the divisor.

 

Example 13: Calculate (f/g)(x) and determine the restrictions.

Solution:

In this case, the domain of f(x) consists of all real numbers except 3 and 8, and the domain of g(x) consists all real numbers except 3. In addition, the reciprocal of g(x) has a restriction of −8. Therefore, the domain of this quotient consists of all real numbers except 3, 8, and −8.

Answer: (f/g)(x)=1, where x3,8,8

Key Takeaways

  • After multiplying rational expressions, factor both the numerator and denominator and then cancel common factors. Make note of the restrictions to the domain. The values that give a value of 0 in the denominator are the restrictions.
  • To divide rational expressions, multiply by the reciprocal of the divisor.
  • The restrictions to the domain of a product consist of the restrictions to the domain of each factor.
  • The restrictions to the domain of a quotient consist of the restrictions to the domain of each rational expression as well as the restrictions on the reciprocal of the divisor.

Topic Exercises

Part A: Multiplying Rational Expressions

Multiply. (Assume all denominators are nonzero.)

1. 2x394x2

2. 5x3yy225x

3. 5x22y4y215x3

4. 16a47b249b32a3

5. x612x324x2x6

6. x+102x1x2x+10

7. (y1)2y+11y1

8. y29y+32y3y3

9. 2a5a52a+54a225

10. 2a29a+4a216(a2+4a)

11. 2x2+3x2(2x1)22xx+2

12. 9x2+19x+24x2x24x+49x28x1

13. x2+8x+1616x2x23x4x2+5x+4

14. x2x2x2+8x+7x2+2x15x25x+6

15. x+1x33xx+5

16. 2x1x1x+612x

17. 9+x3x+13x+9

18. 12+5x5x+25x

19. 100y2y1025y2y+10

20. 3y36y536y2255+6y

21. 3a2+14a5a2+13a+119a2

22. 4a216a4a1116a24a215a4

23. x+9x2+14x45(x281)

24. 12+5x(25x2+20x+4)

25. x2+x63x2+15x+182x28x24x+4

26. 5x24x15x26x+125x210x+1375x2

Part B: Dividing Rational Expressions

Divide. (Assume all denominators are nonzero.)

27. 5x8÷15x24

28. 38y÷152y2

29. 5x93y325x109y5

30. 12x4y221z56x3y27z3

31. (x4)230x4÷x415x

32. 5y410(3y5)2÷10y52(3y5)3

33. x295x÷(x3)

34. y2648y÷(8+y)

35. (a8)22a2+10a÷a8a

36. 24a2b3(a2b)÷12ab(a2b)5

37. x2+7x+10x2+4x+4÷1x24

38. 2x2x12x23x+1÷14x21

39. y+1y23y÷y21y26y+9

40. 9a2a28a+15÷2a210aa210a+25

41. a23a182a211a6÷a2+a62a2a1

42. y27y+10y2+5y142y29y5y2+14y+49

43. 6y2+y14y2+4y+13y2+2y12y27y4

44. x27x18x2+8x+12÷x281x2+12x+36

45. 4a2b2b+2a÷(b2a)2

46. x2y2y+x÷(yx)2

47. 5y2(y3)4x3÷25y(3y)2x2

48. 15x33(y+7)÷25x69(7+y)2

49. 3x+4x8÷7x8x

50. 3x22x+1÷23x3x

51. (7x1)24x+1÷28x211x+114x

52. 4x(x+2)2÷2xx24

53. a2b2a÷(ba)2

54. (a2b)22b÷(2b2+aba2)

55. x26x+9x2+7x+12÷9x2x2+8x+16

56. 2x29x525x2÷14x+4x22x29x+5

57. 3x216x+51004x29x26x+13x2+14x5

58. 10x225x15x26x+99x2x2+6x+9

Recall that multiplication and division are to be performed in the order they appear from left to right. Simplify the following.

59. 1x2x1x+3÷x1x3

60. x7x+91x3÷x7x

61. x+1x2÷xx5x2x+1

62. x+42x+5÷x32x+5x+4x3

63. 2x1x+1÷x4x2+1x42x1

64. 4x213x+2÷2x1x+53x+22x+1

Part C: Multiplying and Dividing Rational Functions

Calculate (fg)(x) and determine the restrictions to the domain.

65. f(x)=1x and g(x)=1x1

66. f(x)=x+1x1 and g(x)=x21

67. f(x)=3x+2x+2 and g(x)=x24(3x+2)2

68. f(x)=(13x)2x6 and g(x)=(x6)29x21

69. f(x)=25x21x2+6x+9 and g(x)=x295x+1

70. f(x)=x2492x2+13x7 and g(x)=4x24x+17x

Calculate (f/g)(x) and state the restrictions.

71. f(x)=1x and g(x)=x2x1

72. f(x)=(5x+3)2x2 and g(x)=5x+36x

73. f(x)=5x(x8)2 and g(x)=x225x8

74. f(x)=x22x15x23x10 and g(x)=2x25x3x27x+12

75. f(x)=3x2+11x49x26x+1 and g(x)=x22x+13x24x+1

76. f(x)=36x2x2+12x+36 and g(x)=x212x+36x2+4x12

Part D: Discussion Board Topics

77. In the history of fractions, who is credited for the first use of the fraction bar?

78. How did the ancient Egyptians use fractions?

79. Explain why x=7 is a restriction to 1x÷x7x2.

Answers

1: 32x

3: 2y3x

5: 2x

7: y1y+1

9: 1a5

11: 2x2x1

13: −1

15: x+1x+5

17: 33x+1

19: 25y2

21: a+5a2+1

23: (x+9)2x5

25: 2/3

27: 16x

29: 3y25x

31: x42x3

33: x+35x

35: a82(a+5)

37: (x+5)(x2)

39: y3y(y1)

41: a1a2

43: y4y+1

45: 12ab

47: y10x

49: 3x+47x

51: 7x14x+1

53: a+ba(ba)

55: (x3)(x+4)(x+3)2

57: −1/4

59: xx+3

61: x(x5)x2

63: x2+1x+1

65: (fg)(x)=1x(x1); x0,1

67: (fg)(x)=x23x+2; x2,23

69: (fg)(x)=(x3)(5x1)x+3; x3,15

71: (f/g)(x)=x1x(x2); x0,1,2

73: (f/g)(x)=1(x8)(x+5); x±5,8

75: (f/g)(x)=(x+4)(x1); x13,1

7.3 Adding and Subtracting Rational Expressions

Learning Objectives

  1. Add and subtract rational expressions with common denominators.
  2. Add and subtract rational expressions with unlike denominators.
  3. Add and subtract rational functions.

Adding and Subtracting with Common Denominators

Adding and subtracting rational expressions is similar to adding and subtracting fractions. Recall that if the denominators are the same, we can add or subtract the numerators and write the result over the common denominator.

When working with rational expressions, the common denominator will be a polynomial. In general, given polynomials P, Q, and R, where Q0, we have the following:

In this section, assume that all variable factors in the denominator are nonzero.

 

Example 1: Add: 3y+7y.

Solution: Add the numerators 3 and 7, and write the result over the common denominator, y.

Answer: 10y

 

Example 2: Subtract: x52x112x1.

Solution: Subtract the numerators x5 and 1, and write the result over the common denominator, 2x1.

Answer: x62x1

 

Example 3: Subtract: 2x+7(x+5)(x3)x+10(x+5)(x3).

Solution: We use parentheses to remind us to subtract the entire numerator of the second rational expression.

Answer: 1x+5

 

Example 4: Simplify: 2x2+10x+3x236x2+6x+5x236+x4x236.

Solution: Subtract and add the numerators. Make use of parentheses and write the result over the common denominator, x236.

Answer: x1x6

 

Try this! Subtract: x2+12x27x4x22x2x27x4.

Answer: 1x4

Video Solution

(click to see video)

Adding and Subtracting with Unlike Denominators

To add rational expressions with unlike denominators, first find equivalent expressions with common denominators. Do this just as you have with fractions. If the denominators of fractions are relatively prime, then the least common denominator (LCD) is their product. For example,

Multiply each fraction by the appropriate form of 1 to obtain equivalent fractions with a common denominator.

The process of adding and subtracting rational expressions is similar. In general, given polynomials P, Q, R, and S, where Q0 and S0, we have the following:

In this section, assume that all variable factors in the denominator are nonzero.

 

Example 5: Add: 1x+1y.

Solution: In this example, the LCD=xy. To obtain equivalent terms with this common denominator, multiply the first term by yy and the second term by xx.

Answer: y+xxy

 

Example 6: Subtract: 1y1y3.

Solution: Since the LCD=y(y3), multiply the first term by 1 in the form of (y3)(y3) and the second term by yy.

Answer: 3y(y3)

 

It is not always the case that the LCD is the product of the given denominators. Typically, the denominators are not relatively prime; thus determining the LCD requires some thought. Begin by factoring all denominators. The LCD is the product of all factors with the highest power. For example, given

there are three base factors in the denominator: x , (x+2), and (x3). The highest powers of these factors are x3, (x+2)2, and (x3)1. Therefore,

The general steps for adding or subtracting rational expressions are illustrated in the following example.

 

Example 7: Subtract: xx2+4x+33x24x5.

Solution:

Step 1: Factor all denominators to determine the LCD.

The LCDis(x+1)(x+3)(x5).

Step 2: Multiply by the appropriate factors to obtain equivalent terms with a common denominator. To do this, multiply the first term by (x5)(x5) and the second term by (x+3)(x+3).

Step 3: Add or subtract the numerators and place the result over the common denominator.

Step 4: Simplify the resulting algebraic fraction.

Answer: (x9)(x+3)(x5)

 

Example 8: Subtract: x29x+18x213x+36xx4.

Solution: It is best not to factor the numerator, x29x+18, because we will most likely need to simplify after we subtract.

Answer: 18(x4)(x9)

 

Example 9: Subtract: 1x2412x.

Solution: First, factor the denominators and determine the LCD. Notice how the opposite binomial property is applied to obtain a more workable denominator.

The LCD is (x+2)(x2). Multiply the second term by 1 in the form of (x+2)(x+2).

Now that we have equivalent terms with a common denominator, add the numerators and write the result over the common denominator.

Answer: x+3(x+2)(x2)

 

Example 10: Simplify: y1y+1y+1y1+y25y21.

Solution: Begin by factoring the denominator.

We can see that the LCD is (y+1)(y1). Find equivalent fractions with this denominator.

Next, subtract and add the numerators and place the result over the common denominator.

Finish by simplifying the resulting rational expression.

Answer: y5y1

 

Try this! Simplify: 2x21+x1+x51x.

Answer: x+3x1

Video Solution

(click to see video)

Rational expressions are sometimes expressed using negative exponents. In this case, apply the rules for negative exponents before simplifying the expression.

 

Example 11: Simplify: y2+(y1)1.

Solution: Recall that xn=1xn. We begin by rewriting the negative exponents as rational expressions.

Answer: y2+y1y2(y1)

Adding and Subtracting Rational Functions

We can simplify sums or differences of rational functions using the techniques learned in this section. The restrictions of the result consist of the restrictions to the domains of each function.

 

Example 12: Calculate (f+g)(x), given f(x)=1x+3 and g(x)=1x2, and state the restrictions.

Solution:

Here the domain of f consists of all real numbers except −3, and the domain of g consists of all real numbers except 2. Therefore, the domain of f + g consists of all real numbers except −3 and 2.

Answer: 2x+1(x+3)(x2), where x3,2

 

Example 13: Calculate (fg)(x), given f(x)=x(x1)x225 and g(x)=x3x5, and state the restrictions to the domain.

Solution:

The domain of f consists of all real numbers except 5 and −5, and the domain of g consists of all real numbers except 5. Therefore, the domain of fg consists of all real numbers except −5 and 5.

Answer: 3x+5, where x±5

Key Takeaways

  • When adding or subtracting rational expressions with a common denominator, add or subtract the expressions in the numerator and write the result over the common denominator.
  • To find equivalent rational expressions with a common denominator, first factor all denominators and determine the least common multiple. Then multiply numerator and denominator of each term by the appropriate factor to obtain a common denominator. Finally, add or subtract the expressions in the numerator and write the result over the common denominator.
  • The restrictions to the domain of a sum or difference of rational functions consist of the restrictions to the domains of each function.

Topic Exercises

Part A: Adding and Subtracting with Common Denominators

Simplify. (Assume all denominators are nonzero.)

1. 3x+7x

2. 9x10x

3. xy3y

4. 4x3+6x3

5. 72x1x2x1

6. 83x83x3x8

7. 2x9+x11x9

8. y+22y+3y+32y+3

9. 2x34x1x44x1

10. 2xx13x+4x1+x2x1

11. 13y2y93y135y3y

12. 3y+25y10+y+75y103y+45y10

13. x(x+1)(x3)3(x+1)(x3)

14. 3x+5(2x1)(x6)x+6(2x1)(x6)

15. xx236+6x236

16. xx2819x281

17. x2+2x2+3x28+x22x2+3x28

18. x2x2x33x2x2x3

Part B: Adding and Subtracting with Unlike Denominators

Simplify. (Assume all denominators are nonzero.)

19. 12+13x

20. 15x21x

21. 112y2+310y3

22. 1x12y

23. 1y2

24. 3y+24

25. 2x+4+2

26. 2y1y2

27. 3x+1+1x

28. 1x12x

29. 1x3+1x+5

30. 1x+21x3

31. xx+12x2

32. 2x3x+5xx3

33. y+1y1+y1y+1

34. 3y13yy+4y2

35. 2x52x+52x+52x5

36. 22x12x+112x

37. 3x+4x828x

38. 1y1+11y

39. 2x2x29+x+159x2

40. xx+3+1x315x(x+3)(x3)

41. 2x3x113x+1+2(x1)(3x1)(3x+1)

42. 4x2x+1xx5+16x3(2x+1)(x5)

43. x3x+2x2+43x(x2)

44. 2xx+63x6x18(x2)(x+6)(x6)

45. xx+51x7257x(x+5)(x7)

46. xx22x3+2x3

47. 1x+5x2x225

48. 5x2x242x2

49. 1x+16x3x27x8

50. 3x9x21613x+4

51. 2xx21+1x2+x

52. x(4x1)2x2+7x4x4+x

53. 3x23x2+5x22x3x1

54. 2xx411x+4x22x8

55. x2x+1+6x242x27x4

56. 1x2x6+1x23x10

57. xx2+4x+33x24x5

58. y+12y2+5y3y4y21

59. y1y2252y210y+25

60. 3x2+24x22x812x4

61. 4x2+28x26x728x7

62. a4a+a29a+18a213a+36

63. 3a12a28a+16a+24a

64. a2142a27a451+2a

65. 1x+3xx26x+9+3x29

66. 3xx+72xx2+23x10x2+5x14

67. x+3x1+x1x+2x(x+11)x2+x2

68. 2x3x+14x2+4(x+5)3x25x2

69. x14x1x+32x+33(x+5)8x2+10x3

70. 3x2x322x+36x25x94x29

71. 1y+1+1y+2y21

72. 1y1y+1+1y1

73. 52+21

74. 61+42

75. x1+y1

76. x2y1

77. (2x1)1x2

78. (x4)1(x+1)1

79. 3x2(x1)12x

80. 2(y1)2(y1)1

Part C: Adding and Subtracting Rational Functions

Calculate (f+g)(x) and (fg)(x) and state the restrictions to the domain.

81. f(x)=13x and g(x)=1x2

82. f(x)=1x1 and g(x)=1x+5

83. f(x)=xx4 and g(x)=14x

84. f(x)=xx5 and g(x)=12x3

85. f(x)=x1x24 and g(x)=4x26x16

86. f(x)=5x+2 and g(x)=3x+4

Calculate (f+f)(x) and state the restrictions to the domain.

87. f(x)=1x

88. f(x)=12x

89. f(x)=x2x1

90. f(x)=1x+2

Part D: Discussion Board

91. Explain to a classmate why this is incorrect: 1x2+2x2=32x2.

92. Explain to a classmate how to find the common denominator when adding algebraic expressions. Give an example.

Answers

1: 10x

3: x3y

5: 7x2x1

7: 1

9: x+14x1

11: y1y

13: 1x+1

15: 1x6

17: x+5x+7

19: 3x+26x

21: 5y+1860y3

23: 12yy

25: 2(x+5)x+4

27: 4x+1x(x+1)

29: 2(x+1)(x3)(x+5)

31: x24x2(x2)(x+1)

33: 2(y2+1)(y+1)(y1)

35: 40x(2x+5)(2x5)

37: 3(x+2)x8

39: 2x+5x+3

41: 2x+13x+1

43: x2+4x+43x(x2)

45: x6x7

47: x2+x5(x+5)(x5)

49: 5x8

51: 2x1x(x1)

53: x(x4)(x+2)(3x1)

55: x+62x+1

57: x9(x5)(x+3)

59: y28y5(y+5)(y5)2

61: 4xx+1

63: a+5a4

65: 6x(x+3)(x3)2

67: x7x+2

69: x54x1

71: 2y1y(y1)

73: 2750

75: x+yxy

77: (x1)2x2(2x1)

79: x(x+2)x1

81: (f+g)(x)=2(2x1)3x(x2); (fg)(x)=2(x+1)3x(x2); x0,2

83: (f+g)(x)=x1x4; (fg)(x)=x+1x4; x4

85: (f+g)(x)=x(x5)(x+2)(x2)(x8); (fg)(x)=x213x+16(x+2)(x2)(x8); x2,2,8

87: (f+f)(x)=2x; x0

89: (f+f)(x)=2x2x1; x12

7.4 Complex Rational Expressions

Learning Objectives

  1. Simplify complex rational expressions by multiplying the numerator by the reciprocal of the divisor.
  2. Simplify complex rational expressions by multiplying numerator and denominator by the least common denominator (LCD).

Definitions

A complex fractionA fraction where the numerator or denominator consists of one or more fractions. is a fraction where the numerator or denominator consists of one or more fractions. For example,

Simplifying such a fraction requires us to find an equivalent fraction with integer numerator and denominator. One way to do this is to divide. Recall that dividing fractions involves multiplying by the reciprocal of the divisor.

An alternative method for simplifying this complex fraction involves multiplying both the numerator and denominator by the LCD of all the given fractions. In this case, the LCD = 4.

A complex rational expressionA rational expression where the numerator or denominator consists of one or more rational expressions. is defined as a rational expression that contains one or more rational expressions in the numerator or denominator or both. For example,

We simplify a complex rational expression by finding an equivalent fraction where the numerator and denominator are polynomials. As illustrated above, there are two methods for simplifying complex rational expressions, and we will outline the steps for both methods. For the sake of clarity, assume that variable expressions used as denominators are nonzero.

Method 1: Simplify Using Division

We begin our discussion on simplifying complex rational expressions using division. Before we can multiply by the reciprocal of the divisor, we must simplify the numerator and denominator separately. The goal is to first obtain single algebraic fractions in the numerator and the denominator. The steps for simplifying a complex algebraic fraction are illustrated in the following example.

 

Example 1: Simplify: 12+1x141x2.

Solution:

Step 1: Simplify the numerator and denominator. The goal is to obtain a single algebraic fraction divided by another single algebraic fraction. In this example, find equivalent terms with a common denominator in both the numerator and denominator before adding and subtracting.

At this point we have a single algebraic fraction divided by a single algebraic fraction.

Step 2: Multiply the numerator by the reciprocal of the divisor.

Step 3: Factor all numerators and denominators completely.

Step 4: Cancel all common factors.

Answer: 2xx2

 

Example 2: Simplify: 1x1x24x22x.

Solution:

Answer: 12

 

Example 3: Simplify 14x21x212x15x2.

Solution: The LCD of the rational expressions in both the numerator and denominator is x2. Multiply by the appropriate factors to obtain equivalent terms with this as the denominator and then subtract.

We now have a single rational expression divided by another single rational expression. Next, multiply the numerator by the reciprocal of the divisor and then factor and cancel.

Answer: x7x5

 

Example 4: Simplify: 11x21x1.

Solution:

Answer: x+1x

 

Try this! Simplify: 1811x219+1x.

Answer: x99x

Video Solution

(click to see video)

Method 2: Simplify Using the LCD

An alternative method for simplifying complex rational expressions involves clearing the fractions by multiplying the expression by a special form of 1. In this method, multiply the numerator and denominator by the least common denominator (LCD) of all given fractions.

 

Example 5: Simplify: 12+1x141x2.

Solution:

Step 1: Determine the LCD of all the fractions in the numerator and denominator. In this case, the denominators of the given fractions are 2, x , 4, and x2. Therefore, the LCD is 4x2.

Step 2: Multiply the numerator and denominator by the LCD. This step should clear the fractions in both the numerator and denominator.

This leaves us with a single algebraic fraction with a polynomial in the numerator and in the denominator.

Step 3: Factor the numerator and denominator completely.

Step 4: Cancel all common factors.

Answer: 2xx2

Note

This was the same problem that we began this section with, and the results here are the same. It is worth taking the time to compare the steps involved using both methods on the same problem.

 

Example 6: Simplify: 12x15x2314x5x2.

Solution: Considering all of the denominators, we find that the LCD is x2. Therefore, multiply the numerator and denominator by x2:

At this point, we have a rational expression that can be simplified by factoring and then canceling the common factors.

Answer: x+33x+1

 

It is important to point out that multiplying the numerator and denominator by the same nonzero factor is equivalent to multiplying by 1 and does not change the problem. Because x2x2=1, we can multiply the numerator and denominator by x2 in the previous example and obtain an equivalent expression.

 

Example 7: Simplify: 1x+1+3x32x31x+1.

Solution: The LCM of all the denominators is (x+1)(x3). Begin by multiplying the numerator and denominator by these factors.

Answer: 4xx+5

 

Try this! Simplify: 1y141161y2.

Answer: 4yy+4

Video Solution

(click to see video)

Key Takeaways

  • Complex rational expressions can be simplified into equivalent expressions with a polynomial numerator and polynomial denominator.
  • One method of simplifying a complex rational expression requires us to first write the numerator and denominator as a single algebraic fraction. Then multiply the numerator by the reciprocal of the divisor and simplify the result.
  • Another method for simplifying a complex rational expression requires that we multiply it by a special form of 1. Multiply the numerator and denominator by the LCM of all the denominators as a means to clear the fractions. After doing this, simplify the remaining rational expression.
  • An algebraic fraction is reduced to lowest terms if the numerator and denominator are polynomials that share no common factors other than 1.

Topic Exercises

Part A: Complex Rational Expressions

Simplify. (Assume all denominators are nonzero.)

1. 1254

2. 7854

3. 103209

4. 42187

5. 2356

6. 74143

7. 1325413

8. 12512+13

9. 1+32114

10. 2121+34

11. 5x2x+125xx+1

12. 7+x7xx+714x2

13. 3yxy2x1

14. 5a2b115a3( b1)2

15. 1+1x21x

16. 2x+131x

17. 23y461y

18. 5y1210yy2

19. 151x1251x2

20. 1x+151251x2

21. 1x13191x2

22. 14+1x1x2116

23. 161x21x4

24. 21y114y2

25. 1x+1y1y21x2

26. 12x4314x2169

27. 22512x21512x

28. 42514x215+14x

29. 1y1x42xy

30. 1ab+21a+1b

31. 1y+1xxy

32. 3x131x

33. 14x21x212x15x2

34. 13x4x2116x2

35. 312x12x222x+12x2

36. 125x+12x2126x+18x2

37. 1x43x238x+163x2

38. 1+310x110x235110x15x2

39. x11+4x5x2

40. 252x3x24x+3

41. 1x3+2x1x3x3

42. 14x5+1x21x2+13x10

43. 1x+5+4x2 2x21x+5 

44. 3x12x+3 2x+3+1x3 

45. xx+12x+3 x3x+4+1x+1 

46. xx9+2x+1x7x91x+1

47. x3x+21x+2 xx+22x+2 

48. xx4+1x+2 x3x+4+1x+2 

49. a38b327a2b

50. 27a3+b3ab3a+b

51. 1b3+1a31b+1a

52. 1b31a31a1b

53. x2+y2xy+2x2y22xy

54. xy+4+4yxxy+3+2yx

55. 1+11+12

56. 211+13

57. 11+11+x

58. x+1x11x+1

59. 11xx1x

60. 1xxx1x2

Part B: Discussion Board Topics

61. Choose a problem from this exercise set and clearly work it out on paper, explaining each step in words. Scan your page and post it on the discussion board.

62. Explain why we need to simplify the numerator and denominator to a single algebraic fraction before multiplying by the reciprocal of the divisor.

63. Two methods for simplifying complex rational expressions have been presented in this section. Which of the two methods do you feel is more efficient, and why?

Answers

1: 25

3: 32

5: 4/5

7: −6/11

9: 103

11: x5

13: 3(x1)xy

15: x+12x1

17: 23

19: 5xx+5

21: 3xx+3

23: 4x+1x

25: xyxy

27: 2x+55x

29: xy4xy2

31: x+yx2y2

33: x7x5

35: 3x+12x1

37: 13x4

39: x2x+5

41: 3(x2)2x+3

43: 5x+18x+12

45: (x1)(3x+4)(x+2)(x+3)

47: x+13x+2

49: a2+2ab+4b227

51: a2ab+b2a2b2

53: 2(x+y)xy

55: 53

57: x+1x+2

59: 1x+1

7.5 Solving Rational Equations

Learning Objectives

  1. Solve rational equations.
  2. Solve literal equations, or formulas, involving rational expressions.

Solving Rational Equations

A rational equationAn equation containing at least one rational expression. is an equation containing at least one rational expression. Rational expressions typically contain a variable in the denominator. For this reason, we will take care to ensure that the denominator is not 0 by making note of restrictions and checking our solutions.

Solve rational equations by clearing the fractions by multiplying both sides of the equation by the least common denominator (LCD).

 

Example 1: Solve: 5x13=1x.

Solution: We first make a note that x0 and then multiply both sides by the LCD, 3x:

Check your answer by substituting 12 for x to see if you obtain a true statement.

Answer: The solution is 12.

 

After multiplying both sides of the previous example by the LCD, we were left with a linear equation to solve. This is not always the case; sometimes we will be left with a quadratic equation.

 

Example 2: Solve: 21x(x+1)=3x+1.

Solution: In this example, there are two restrictions, x0 and x1. Begin by multiplying both sides by the LCD, x(x+1).

After distributing and dividing out the common factors, a quadratic equation remains. To solve it, rewrite it in standard form, factor, and then set each factor equal to 0.

Check to see if these values solve the original equation.

Answer: The solutions are −1/2 and 1.

 

Up to this point, all of the possible solutions have solved the original equation. However, this may not always be the case. Multiplying both sides of an equation by variable factors may lead to extraneous solutionsA solution that does not solve the original equation., which are solutions that do not solve the original equation. A complete list of steps for solving a rational equation is outlined in the following example.

 

Example 3: Solve: xx+2+2x2+5x+6=5x+3.

Solution:

Step 1: Factor all denominators and determine the LCD.

The LCD is (x+2)(x+3).

Step 2: Identify the restrictions. In this case, they are x2 and x3.

Step 3: Multiply both sides of the equation by the LCD. Distribute carefully and then simplify.

Step 4: Solve the resulting equation. Here the result is a quadratic equation. Rewrite it in standard form, factor, and then set each factor equal to 0.

Step 5: Check for extraneous solutions. Always substitute into the original equation, or the factored equivalent. In this case, choose the factored equivalent to check:

Here −2 is an extraneous solution and is not included in the solution set. It is important to note that −2 is a restriction.

Answer: The solution is 4.

 

If this process produces a solution that happens to be a restriction, then disregard it as an extraneous solution.

 

Try this! Solve: xx5+3x+2=7xx23x10.

Answer: −3

Video Solution

(click to see video)

Sometimes all potential solutions are extraneous, in which case we say that there is no solution to the original equation. In the next two examples, we demonstrate two ways in which a rational equation can have no solutions.

 

Example 4: Solve: 3xx242x+2=1x+2.

Solution: To identify the LCD, first factor the denominators.

Multiply both sides by the least common denonominator (LCD), (x+2)(x2), distributing carefully.

The equation is a contradiction and thus has no solution.

Answer: No solution,

 

Example 5: Solve: xx44x+5=36x2+x20.

Solution: First, factor the denominators.

Take note that the restrictions are x4 and x5. To clear the fractions, multiply by the LCD, (x4)(x+5).

Both of these values are restrictions of the original equation; hence both are extraneous.

Answer: No solution,

 

Try this! Solve: 1x+1+xx3=4xx22x3.

Answer:

Video Solution

(click to see video)

It is important to point out that this technique for clearing algebraic fractions only works for equations. Do not try to clear algebraic fractions when simplifying expressions. As a reminder, we have

Expressions are to be simplified and equations are to be solved. If we multiply the expression by the LCD, x(2x+1), we obtain another expression that is not equivalent.

Literal Equations

Literal equations, or formulas, are often rational equations. Hence the techniques described in this section can be used to solve for particular variables. Assume that all variable expressions in the denominator are nonzero.

 

Example 6: Solve for x:   z=x5y.

Solution: The goal is to isolate x. Assuming that y is nonzero, multiply both sides by y and then add 5 to both sides.

Answer: x=yz+5

 

Example 7: Solve for c:   1c=1a+1b.

Solution: In this example, the goal is to isolate c. We begin by multiplying both sides by the LCD, abc, distributing carefully.

On the right side of the equation, factor out c.

Next, divide both sides of the equation by the quantity (b+a).

Answer: c=abb+a

 

Try this! Solve for y:   x=y+1y1.

Answer: y=x+1x1

Video Solution

(click to see video)

Key Takeaways

  • Begin solving rational equations by multiplying both sides by the LCD. The resulting equivalent equation can be solved using the techniques learned up to this point.
  • Multiplying both sides of a rational equation by a variable expression introduces the possibility of extraneous solutions. Therefore, we must check the solutions against the set of restrictions. If a solution is a restriction, then it is not part of the domain and is extraneous.
  • When multiplying both sides of an equation by an expression, distribute carefully and multiply each term by that expression.
  • If all of the resulting solutions are extraneous, then the original equation has no solutions.

Topic Exercises

Part A: Rational Equations

Solve.

1. 12+1x=18

2. 131x=29

3. 13x23=1x

4. 25x1x=310

5. 12x+1=5

6. 33x1+4=5

7. 2x3x+5=2x+5

8. 5x2x1=x12x1

9. 5x7=6x9

10. 5x+5=3x+1

11. x66x=0

12. 5x+x5=2

13. xx+12=2x

14. 2xx+5=16x

15. 1x+x2x+1=0

16. 9x3x14x=0

17. 12x=48x2

18. 29x=5x2

19. 1+12x=12x2

20. 13x5x(3x4)=1x

21. x2=14x+3

22. 3x2=x+13x

23. 6=3x+3x1

24. 12x2=2+6(4x)x2

25. 2+2xx3=3(x1)x3

26. xx1+16x1=x(x1)(6x1)

27. 12x281=1x+92x9

28. 14x249=2x73x+7

29. 6xx+3+4x3=3xx29

30. 3xx+217x2=48x24

31. x1+3=0

32. 4y1=0

33. y24=0

34. 9x21=0

35. 3(x1)1+5=0

36. 52(3x+1)1=0

37. 3+2x3=2x3

38. 1x=1x+1

39. xx+1=x+1x

40. 3x13x=xx+3

41. 4x7x5=3x2x5

42. xx29=1x3

43. 3x+4x828x=1

44. 1x=6x(x+3)

45. 3x=1x+1+13x(x+1)

46. xx134x1=9x(4x1)(x1)

47. 1x4+xx2=2x26x+8

48. xx5+x1x211x+30=5x6

49. xx+165x2+4x1=55x1

50. 8x24x12+2(x+2)x2+4x60=1x+2

51. xx+220x2x6=4x3

52. x+7x1+x1x+1=4x21

53. x1x3+x3x1=x+5x3

54. x2x5x5x2=8xx5

55. x+7x281x2+5x14=9x+7

56. xx6+1=5x+3036x2

57. 2xx+144x3=74x2+x3

58. x5x10+5x5=5xx215x+50

59. 5x2+5x+4+x+1x2+3x4=5x21

60. 1x22x63+x9x2+10x+21=1x26x27

61. 4x24+2(x2)x24x12=x+2x28x+12

62. x+2x25x+4+x+2x2+x2=x1x22x8

63. 6xx111x+12x2x1=6x2x+1

64. 8x2x3+4x2x27x+6=1x2

Part B: Literal Equations

Solve for the indicated variable.

65. Solve for r:   t=Dr.

66. Solve for b:   h=2Ab.

67. Solve for P:   t=IPr.

68. Solve for π :   r=C2π.

69. Solve for c:   1a=1b+1c.

70. Solve for y:   m=yy1xx1.

71. Solve for w:   P=2(l+w).

72. Solve for t:   A=P(1+rt).

73. Solve for m:   s=1n+m.

74. Solve for S:   h=S2πrr.

75. Solve for x:   y=xx+2.

76. Solve for x:   y=2x+15x.

77. Solve for R:   1R=1R1+1R2.

78. Solve for S1:   1f=1S1+1S2.

Part C: Discussion Board

79. Explain why multiplying both sides of an equation by the LCD sometimes produces extraneous solutions.

80. Explain the connection between the technique of cross multiplication and multiplying both sides of a rational equation by the LCD.

81. Explain how we can tell the difference between a rational expression and a rational equation. How do we treat them differently?

Answers

1: −8/3

3: −1

5: −2/5

7: 5/2

9: −3

11: −6, 6

13: −4, 6

15: −1

17: −6, 8

19: −4, 6

21: −7, 4

23:

25:

27: −39

29: 4/3, 3/2

31: −1/3

33: −1/2, 1/2

35: 2/5

37:

39: −1/2

41:

43: −7

45: 5

47: −1

49:

51: −4

53: 5/3

55:

57: 1/2

59: −6, 4

61: 10

63: 1/3

65: r=Dt

67: P=Itr

69: c=abba

71: w=P2l2

73: m=1sns

75: x=2y1y

77: R=R1R2R1+R2

7.6 Applications of Rational Equations

Learning Objectives

  1. Solve applications involving relationships between real numbers.
  2. Solve applications involving uniform motion (distance problems).
  3. Solve work-rate applications.

Number Problems

Recall that the reciprocalThe reciprocal of a nonzero number n is 1/n. of a nonzero number n is 1/n. For example, the reciprocal of 5 is 1/5 and 5 ⋅ 1/5 = 1. In this section, the applications will often involve the key word “reciprocal.” When this is the case, we will see that the algebraic setup results in a rational equation.

 

Example 1: A positive integer is 4 less than another. The sum of the reciprocals of the two positive integers is 10/21. Find the two integers.

Solution: Begin by assigning variables to the unknowns.

Next, use the reciprocals 1n and 1n4 to translate the sentences into an algebraic equation.

We can solve this rational expression by multiplying both sides of the equation by the least common denominator (LCD). In this case, the LCD is 21n(n4).

Solve the resulting quadratic equation.

The question calls for integers and the only integer solution is n=7. Hence disregard 6/5. Use the expression n4 to find the smaller integer.

Answer: The two positive integers are 3 and 7. The check is left to the reader.

 

Example 2: A positive integer is 4 less than another. If the reciprocal of the smaller integer is subtracted from twice the reciprocal of the larger, then the result is 1/30. Find the two integers.

Solution:

Set up an algebraic equation.

Solve this rational expression by multiplying both sides by the LCD. The LCD is 30n(n4).

Here we have two viable possibilities for the larger integer. For this reason, we will we have two solutions to this problem.

As a check, perform the operations indicated in the problem.

Answer: Two sets of positive integers solve this problem: {6, 10} and {20, 24}.

 

Try this! The difference between the reciprocals of two consecutive positive odd integers is 2/15. Find the integers.

Answer: The integers are 3 and 5.

Video Solution

(click to see video)

Uniform Motion Problems

Uniform motionDescribed by the formula D=rt, where the distance, D, is given as the product of the average rate, r, and the time, t, traveled at that rate. problems, also referred to as distance problems, involve the formula

where the distance, D, is given as the product of the average rate, r, and the time, t, traveled at that rate. If we divide both sides by the average rate, r, then we obtain the formula

For this reason, when the unknown quantity is time, the algebraic setup for distance problems often results in a rational equation. Similarly, when the unknown quantity is the rate, the setup also may result in a rational equation.

We begin any uniform motion problem by first organizing our data with a chart. Use this information to set up an algebraic equation that models the application.

 

Example 5: Mary spent the first 120 miles of her road trip in traffic. When the traffic cleared, she was able to drive twice as fast for the remaining 300 miles. If the total trip took 9 hours, then how fast was she moving in traffic?

Solution: First, identify the unknown quantity and organize the data.

To avoid introducing two more variables for the time column, use the formula t=Dr. Here the time for each leg of the trip is calculated as follows:

Use these expressions to complete the chart.

The algebraic setup is defined by the time column. Add the times for each leg of the trip to obtain a total of 9 hours:

We begin solving this equation by first multiplying both sides by the LCD, 2x.

Answer: Mary averaged 30 miles per hour in traffic.

 

Example 6: A passenger train can travel, on average, 20 miles per hour faster than a freight train. If the passenger train covers 390 miles in the same time it takes the freight train to cover 270 miles, then how fast is each train?

Solution: First, identify the unknown quantities and organize the data.

Next, organize the given data in a chart.

Use the formula t=Dr to fill in the time column for each train.

Because the trains travel the same amount of time, finish the algebraic setup by equating the expressions that represent the times:

Solve this equation by first multiplying both sides by the LCD, x(x+20).

Use x + 20 to find the speed of the passenger train.

Answer: The speed of the passenger train is 65 miles per hour and the speed of the freight train is 45 miles per hour.

 

Example 7: Brett lives on the river 8 miles upstream from town. When the current is 2 miles per hour, he can row his boat downstream to town for supplies and back in 3 hours. What is his average rowing speed in still water?

Solution:

Rowing downstream, the current increases his speed, and his rate is x + 2 miles per hour. Rowing upstream, the current decreases his speed, and his rate is x − 2 miles per hour. Begin by organizing the data in the following chart:

Use the formula t=Dr to fill in the time column for each leg of the trip.

The algebraic setup is defined by the time column. Add the times for each leg of the trip to obtain a total of 3 hours:

Solve this equation by first multiplying both sides by the LCD, (x+2)(x2).

Next, solve the resulting quadratic equation.

Use only the positive solution, x=6 miles per hour.

Answer: His rowing speed is 6 miles per hour.

 

Try this! Dwayne drove 18 miles to the airport to pick up his father and then returned home. On the return trip he was able to drive an average of 15 miles per hour faster than he did on the trip there. If the total driving time was 1 hour, then what was his average speed driving to the airport?

Answer: His average speed driving to the airport was 30 miles per hour.

Video Solution

(click to see video)

Work-Rate Problems

The rate at which a task can be performed is called a work rateThe rate at which a task can be performed.. For example, if a painter can paint a room in 8 hours, then the task is to paint the room, and we can write

In other words, the painter can complete 18 of the task per hour. If he works for less than 8 hours, then he will perform a fraction of the task. For example,

Obtain the amount of the task completed by multiplying the work rate by the amount of time the painter works. Typically, work-rate problems involve people working together to complete tasks. When this is the case, we can organize the data in a chart, just as we have done with distance problems.

Suppose an apprentice painter can paint the same room by himself in 10 hours. Then we say that he can complete 110 of the task per hour. Let t represent the time it takes both of the painters, working together, to paint the room.

To complete the chart, multiply the work rate by the time for each person. The portion of the room each can paint adds to a total of 1 task completed. This is represented by the equation obtained from the first column of the chart:

This setup results in a rational equation that can be solved for t by multiplying both sides by the LCD, 40.

Therefore, the two painters, working together, complete the task in 449 hours.

In general, we have the following work-rate formula1t1t+1t2t=1, where 1t1 and 1t2 are the individual work rates and t is the time it takes to complete the task working together.:

Here 1t1 and 1t2 are the individual work rates and t is the time it takes to complete one task working together. If we factor out the time, t, and then divide both sides by t, we obtain an equivalent work-rate formula:

In summary, we have the following equivalent work-rate formulas:

 

Example 3: Working alone, Billy’s dad can complete the yard work in 3 hours. If Billy helps his dad, then the yard work takes 2 hours. How long would it take Billy working alone to complete the yard work?

Solution: The given information tells us that Billy’s dad has an individual work rate of 13 task per hour. If we let x represent the time it takes Billy working alone to complete the yard work, then Billy’s individual work rate is 1x, and we can write

Working together, they can complete the task in 2 hours. Multiply the individual work rates by 2 hours to fill in the chart.

The amount of the task each completes will total 1 completed task. To solve for x, we first multiply both sides by the LCD, 3x.

Answer: It takes Billy 6 hours to complete the yard work alone.

 

Of course, the unit of time for the work rate need not always be in hours.

 

Example 4: Working together, two construction crews can build a shed in 5 days. Working separately, the less experienced crew takes twice as long to build a shed than the more experienced crew. Working separately, how long does it take each crew to build a shed?

Solution:

Working together, the job is completed in 5 days. This gives the following setup:

The first column in the chart gives us an algebraic equation that models the problem:

Solve the equation by multiplying both sides by 2x.

To determine the time it takes the less experienced crew, we use 2x:

Answer: Working separately, the experienced crew takes 7½ days to build a shed, and the less experienced crew takes 15 days to build a shed.

 

Try this! Joe’s garden hose fills the pool in 12 hours. His neighbor has a thinner hose that fills the pool in 15 hours. How long will it take to fill the pool using both hoses?

Answer: It will take both hoses 623 hours to fill the pool.

Video Solution

(click to see video)

Key Takeaways

  • In this section, all of the steps outlined for solving general word problems apply. Look for the new key word “reciprocal,” which indicates that you should write the quantity in the denominator of a fraction with numerator 1.
  • When solving distance problems where the time element is unknown, use the equivalent form of the uniform motion formula, t=Dr, to avoid introducing more variables.
  • When solving work-rate problems, multiply the individual work rate by the time to obtain the portion of the task completed. The sum of the portions of the task results in the total amount of work completed.

Topic Exercises

Part A: Number Problems

Use algebra to solve the following applications.

1. A positive integer is twice another. The sum of the reciprocals of the two positive integers is 3/10. Find the two integers.

2. A positive integer is twice another. The sum of the reciprocals of the two positive integers is 3/12. Find the two integers.

3. A positive integer is twice another. The difference of the reciprocals of the two positive integers is 1/8. Find the two integers.

4. A positive integer is twice another. The difference of the reciprocals of the two positive integers is 1/18. Find the two integers.

5. A positive integer is 2 less than another. If the sum of the reciprocal of the smaller and twice the reciprocal of the larger is 5/12, then find the two integers.

6. A positive integer is 2 more than another. If the sum of the reciprocal of the smaller and twice the reciprocal of the larger is 17/35, then find the two integers.

7. The sum of the reciprocals of two consecutive positive even integers is 11/60. Find the two even integers.

8. The sum of the reciprocals of two consecutive positive odd integers is 16/63. Find the integers.

9. The difference of the reciprocals of two consecutive positive even integers is 1/24. Find the two even integers.

10. The difference of the reciprocals of two consecutive positive odd integers is 2/99. Find the integers.

11. If 3 times the reciprocal of the larger of two consecutive integers is subtracted from 2 times the reciprocal of the smaller, then the result is 1/2. Find the two integers.

12. If 3 times the reciprocal of the smaller of two consecutive integers is subtracted from 7 times the reciprocal of the larger, then the result is 1/2. Find the two integers.

13. A positive integer is 5 less than another. If the reciprocal of the smaller integer is subtracted from 3 times the reciprocal of the larger, then the result is 1/12. Find the two integers.

14. A positive integer is 6 less than another. If the reciprocal of the smaller integer is subtracted from 10 times the reciprocal of the larger, then the result is 3/7. Find the two integers.

Part B: Uniform Motion Problems

Use algebra to solve the following applications.

15. James can jog twice as fast as he can walk. He was able to jog the first 9 miles to his grandmother’s house, but then he tired and walked the remaining 1.5 miles. If the total trip took 2 hours, then what was his average jogging speed?

16. On a business trip, an executive traveled 720 miles by jet aircraft and then another 80 miles by helicopter. If the jet averaged 3 times the speed of the helicopter and the total trip took 4 hours, then what was the average speed of the jet?

17. Sally was able to drive an average of 20 miles per hour faster in her car after the traffic cleared. She drove 23 miles in traffic before it cleared and then drove another 99 miles. If the total trip took 2 hours, then what was her average speed in traffic?

18. Harry traveled 15 miles on the bus and then another 72 miles on a train. If the train was 18 miles per hour faster than the bus and the total trip took 2 hours, then what was the average speed of the train?

19. A bus averages 6 miles per hour faster than a trolley. If the bus travels 90 miles in the same time it takes the trolley to travel 75 miles, then what is the speed of each?

20. A passenger car averages 16 miles per hour faster than the bus. If the bus travels 56 miles in the same time it takes the passenger car to travel 84 miles, then what is the speed of each?

21. A light aircraft travels 2 miles per hour less than twice as fast as a passenger car. If the passenger car can travel 231 miles in the same time it takes the aircraft to travel 455 miles, then what is the average speed of each?

22. Mary can run 1 mile per hour more than twice as fast as Bill can walk. If Bill can walk 3 miles in the same time it takes Mary to run 7.2 miles, then what is Bill’s average walking speed?

23. An airplane traveling with a 20-mile-per-hour tailwind covers 270 miles. On the return trip against the wind, it covers 190 miles in the same amount of time. What is the speed of the airplane in still air?

24. A jet airliner traveling with a 30-mile-per-hour tailwind covers 525 miles in the same amount of time it is able to travel 495 miles after the tailwind eases to 10 miles per hour. What is the speed of the airliner in still air?

25. A boat averages 16 miles per hour in still water. With the current, the boat can travel 95 miles in the same time it travels 65 miles against it. What is the speed of the current?

26. A river tour boat averages 7 miles per hour in still water. If the total 24-mile tour downriver and 24 miles back takes 7 hours, then how fast is the river current?

27. If the river current flows at an average 3 miles per hour, then a tour boat makes the 9-mile tour downstream with the current and back the 9 miles against the current in 4 hours. What is the average speed of the boat in still water?

28. Jane rowed her canoe against a 1-mile-per-hour current upstream 12 miles and then returned the 12 miles back downstream. If the total trip took 5 hours, then at what speed can Jane row in still water?

29. Jose drove 15 miles to pick up his sister and then returned home. On the return trip, he was able to average 15 miles per hour faster than he did on the trip to pick her up. If the total trip took 1 hour, then what was Jose’s average speed on the return trip?

30. Barry drove the 24 miles to town and then back in 1 hour. On the return trip, he was able to average 14 miles per hour faster than he averaged on the trip to town. What was his average speed on the trip to town?

31. Jerry paddled his kayak upstream against a 1-mile-per-hour current for 12 miles. The return trip downstream with the 1-mile-per-hour current took 1 hour less time. How fast can Jerry paddle the kayak in still water?

32. It takes a light aircraft 1 hour more time to fly 360 miles against a 30-mile-per-hour headwind than it does to fly the same distance with it. What is the speed of the aircraft in calm air?

Part C: Work-Rate Problems

Use algebra to solve the following applications.

33. James can paint the office by himself in 7 hours. Manny paints the office in 10 hours. How long will it take them to paint the office working together?

34. Barry can lay a brick driveway by himself in 12 hours. Robert does the same job in 10 hours. How long will it take them to lay the brick driveway working together?

35. Jerry can detail a car by himself in 50 minutes. Sally does the same job in 1 hour. How long will it take them to detail a car working together?

36. Jose can build a small shed by himself in 26 hours. Alex builds the same small shed in 2 days. How long would it take them to build the shed working together?

37. Allison can complete a sales route by herself in 6 hours. Working with an associate, she completes the route in 4 hours. How long would it take her associate to complete the route by herself?

38. James can prepare and paint a house by himself in 5 days. Working with his brother, Bryan, they can do it in 3 days. How long would it take Bryan to prepare and paint the house by himself?

39. Joe can assemble a computer by himself in 1 hour. Working with an assistant, he can assemble a computer in 40 minutes. How long would it take his assistant to assemble a computer working alone?

40. The teacher’s assistant can grade class homework assignments by herself in 1 hour. If the teacher helps, then the grading can be completed in 20 minutes. How long would it take the teacher to grade the papers working alone?

41. A larger pipe fills a water tank twice as fast as a smaller pipe. When both pipes are used, they fill the tank in 5 hours. If the larger pipe is left off, then how long would it take the smaller pipe to fill the tank?

42. A newer printer can print twice as fast as an older printer. If both printers working together can print a batch of flyers in 45 minutes, then how long would it take the newer printer to print the batch working alone?

43. Working alone, Henry takes 9 hours longer than Mary to clean the carpets in the entire office. Working together, they clean the carpets in 6 hours. How long would it take Mary to clean the office carpets if Henry were not there to help?

44. Working alone, Monique takes 4 hours longer than Audrey to record the inventory of the entire shop. Working together, they take inventory in 1.5 hours. How long would it take Audrey to record the inventory working alone?

45. Jerry can lay a tile floor in 3 hours less time than Jake. If they work together, the floor takes 2 hours. How long would it take Jerry to lay the floor by himself?

46. Jeremy can build a model airplane in 5 hours less time than his brother. Working together, they need 6 hours to build the plane. How long would it take Jeremy to build the model airplane working alone?

47. Harry can paint a shed by himself in 6 hours. Jeremy can paint the same shed by himself in 8 hours. How long will it take them to paint two sheds working together?

48. Joe assembles a computer by himself in 1 hour. Working with an assistant, he can assemble 10 computers in 6 hours. How long would it take his assistant to assemble 1 computer working alone?

49. Jerry can lay a tile floor in 3 hours, and his assistant can do the same job in 4 hours. If Jerry starts the job and his assistant joins him 1 hour later, then how long will it take to lay the floor?

50. Working alone, Monique takes 6 hours to record the inventory of the entire shop, while it takes Audrey only 4 hours to do the same job. How long will it take them working together if Monique leaves 2 hours early?

Answers

1: {5, 10}

3: {4, 8}

5: {6, 8}

7: {10, 12}

9: {6, 8}

11: {1, 2} or {−4, −3}

13: {4, 9} or {15, 20}

15: 6 miles per hour

17: 46 miles per hour

19: Trolley: 30 miles per hour; bus: 36 miles per hour

21: Passenger car: 66 miles per hour; aircraft: 130 miles per hour

23: 115 miles per hour

25: 3 miles per hour

27: 6 miles per hour

29: 40 miles per hour

31: 5 miles per hour

33: 4217 hours

35: 27311 minutes

37: 12 hours

39: 2 hours

41: 15 hours

43: 9 hours

45: 3 hours

47: 667 hours

49: 217 hours

7.7 Variation

Learning Objectives

  1. Solve applications involving direct variation.
  2. Solve applications involving inverse variation.
  3. Solve applications involving joint variation.

Direct Variation

Consider a freight train moving at a constant speed of 30 miles per hour. The equation that expresses the distance traveled at that speed in terms of time is given by

After 1 hour the train has traveled 30 miles, after 2 hours the train has traveled 60 miles, and so on. We can construct a chart and graph this relation.

In this example, we can see that the distance varies over time as the product of the constant rate, 30 miles per hour, and the variable, t. This relationship is described as direct variationDescribes two quantities x and y that are constant multiples of each other: y=kx. and 30 is called the variation constant. In addition, if we divide both sides of D=30t by t we have

In this form, it is reasonable to say that D is proportional to t, where 30 is the constant of proportionality. In general, we have

Key words Translation

y varies directly as x

y=kx

y is directly proportionalUsed when referring to direct variation. to x

y is proportional to x

Here k is nonzero and is called the constant of variationThe nonzero multiple k, when quantities vary directly or inversely. or the constant of proportionalityUsed when referring to the constant of variation..

 

Example 1: The circumference of a circle is directly proportional to its diameter, and the constant of proportionality is π . If the circumference is measured to be 20 inches, then what is the radius of the circle?

Solution:

Use the fact that “the circumference is directly proportional to the diameter” to write an equation that relates the two variables.

We are given that “the constant of proportionality is π ,” or k=π. Therefore, we write

Now use this formula to find d when the circumference is 20 inches.

The radius of the circle, r, is one-half of its diameter.

Answer: The radius is 10π inches, or approximately 3.18 inches.

 

Typically, we will not be given the constant of variation. Instead, we will be given information from which it can be determined.

 

Example 2: An object’s weight on earth varies directly to its weight on the moon. If a man weighs 180 pounds on earth, then he will weigh 30 pounds on the moon. Set up an algebraic equation that expresses the weight on earth in terms of the weight on the moon and use it to determine the weight of a woman on the moon if she weighs 120 pounds on earth.

Solution:

We are given that the “weight on earth varies directly to the weight on the moon.”

To find the constant of variation k, use the given information. A 180-pound man on earth weighs 30 pounds on the moon, or y=180 when x=30.

Solve for k.

Next, set up a formula that models the given information.

This implies that a person’s weight on earth is 6 times her weight on the moon. To answer the question, use the woman’s weight on earth, y=120pounds, and solve for x.

Answer: The woman weighs 20 pounds on the moon.

Inverse Variation

Next, consider the relationship between time and rate,

If we wish to travel a fixed distance, then we can determine the average speed required to travel that distance in a given amount of time. For example, if we wish to drive 240 miles in 4 hours, we can determine the required average speed as follows:

The average speed required to drive 240 miles in 4 hours is 60 miles per hour. If we wish to drive the 240 miles in 5 hours, then determine the required speed using a similar equation:

In this case, we would only have to average 48 miles per hour. We can make a chart and view this relationship on a graph.

This is an example of an inverse relationship. We say that r is inversely proportional to the time t, where 240 is the constant of proportionality. In general, we have

Key words Translation

y varies inverselyDescribes two quantities x and y, where one variable is directly proportional to the reciprocal of the other: y=kx. as x

y=kx

y is inversely proportionalUsed when referring to inverse variation. to x

Again, k is nonzero and is called the constant of variation or the constant of proportionality.

 

Example 3: If y varies inversely as x and y=5 when x=2, then find the constant of proportionality and an equation that relates the two variables.

Solution: If we let k represent the constant of proportionality, then the statement “y varies inversely as x” can be written as follows:

Use the given information, y=5 when x=2, to find k.

Solve for k.

Therefore, the formula that models the problem is

Answer: The constant of proportionality is 10, and the equation is y=10x.

 

Example 4: The weight of an object varies inversely as the square of its distance from the center of earth. If an object weighs 100 pounds on the surface of earth (approximately 4,000 miles from the center), then how much will it weigh at 1,000 miles above earth’s surface?

Solution:

Since “w varies inversely as the square of d,” we can write

Use the given information to find k. An object weighs 100 pounds on the surface of earth, approximately 4,000 miles from the center. In other words, w = 100 when d = 4,000:

Solve for k.

Therefore, we can model the problem with the following formula:

To use the formula to find the weight, we need the distance from the center of earth. Since the object is 1,000 miles above the surface, find the distance from the center of earth by adding 4,000 miles:

To answer the question, use the formula with d = 5,000.

Answer: The object will weigh 64 pounds at a distance 1,000 miles above the surface of earth.

Joint Variation

Lastly, we define relationships between multiple variables. In general, we have

Vocabulary Translation

y varies jointlyDescribes a quantity y that varies directly as the product of two other quantities x and z: y=kxz. as x and z

y=kxz

y is jointly proportionalUsed when referring to joint variation. to x and z

Here k is nonzero and is called the constant of variation or the constant of proportionality.

 

Example 5: The area of an ellipse varies jointly as a, half of the ellipse’s major axis, and b, half of the ellipse’s minor axis. If the area of an ellipse is 300πcm2, where a=10cm and b=30cm, then what is the constant of proportionality? Give a formula for the area of an ellipse.

Solution: If we let A represent the area of an ellipse, then we can use the statement “area varies jointly as a and b” to write

To find the constant of variation, k, use the fact that the area is 300π when a=10 and b=30.

Therefore, the formula for the area of an ellipse is

Answer: The constant of proportionality is π , and the formula for the area is A=abπ.

 

Try this! Given that y varies directly as the square of x and inversely to z, where y = 2 when x = 3 and z = 27, find y when x = 2 and z = 16.

Answer: 3/2

Video Solution

(click to see video)

Key Takeaway

  • The setup of variation problems usually requires multiple steps. First, identify the key words to set up an equation and then use the given information to find the constant of variation k. After determining the constant of variation, write a formula that models the problem. Once a formula is found, use it to answer the question.

Topic Exercises

Part A: Variation Problems

Translate the following sentences into a mathematical formula.

1. The distance, D, an automobile can travel is directly proportional to the time, t, that it travels at a constant speed.

2. The extension of a hanging spring, d, is directly proportional to the weight, w, attached to it.

3. An automobile’s breaking distance, d, is directly proportional to the square of the automobile’s speed, v.

4. The volume, V, of a sphere varies directly as the cube of its radius, r.

5. The volume, V, of a given mass of gas is inversely proportional to the pressure, p, exerted on it.

6. The intensity, I, of light from a light source is inversely proportional to the square of the distance, d, from the source.

7. Every particle of matter in the universe attracts every other particle with a force, F, that is directly proportional to the product of the masses, m1 and m2, of the particles and inversely proportional to the square of the distance, d, between them.

8. Simple interest, I, is jointly proportional to the annual interest rate, r, and the time, t, in years a fixed amount of money is invested.

9. The period, T, of a pendulum is directly proportional to the square root of its length, L.

10. The time, t, it takes an object to fall is directly proportional to the square root of the distance, d, it falls.

Construct a mathematical model given the following.

11. y varies directly as x, and y = 30 when x = 6.

12. y varies directly as x, and y = 52 when x = 4.

13. y is directly proportional to x, and y = 12 when x = 3.

14. y is directly proportional to x, and y = 120 when x = 20.

15. y varies directly as x, and y = 14 when x = 10.

16. y varies directly as x, and y = 2 when x = 8.

17. y varies inversely as x, and y = 5 when x = 7.

18. y varies inversely as x, and y = 12 when x = 2.

19. y is inversely proportional to x, and y = 3 when x = 9.

20. y is inversely proportional to x, and y = 21 when x = 3.

21. y varies inversely as x, and y = 2 when x = 1/8.

22. y varies inversely as x, and y = 3/2 when x = 1/9.

23. y varies jointly as x and z, where y = 8 when x = 4 and z = 1/2.

24. y varies jointly as x and z, where y = 24 when x = 1/3 and z = 9.

25. y is jointly proportional to x and z, where y = 2 when x = 1 and z = 3.

26. y is jointly proportional to x and z, where y = 15 when x = 3 and z = 7.

27. y varies jointly as x and z, where y = 2/3 when x = 1/2 and z = 12.

28. y varies jointly as x and z, where y = 5 when x = 3/2 and z = 2/9.

29. y varies directly as the square of x, where y = 45 when x = 3.

30. y varies directly as the square of x, where y = 3 when x = 1/2.

31. y is inversely proportional to the square of x, where y = 27 when x = 1/3.

32. y is inversely proportional to the square of x, where y = 9 when x = 2/3.

33. y varies jointly as x and the square of z, where y = 54 when x = 2 and z = 3.

34. y varies jointly as x and the square of z, where y = 6 when x = 1/4 and z = 2/3.

35. y varies jointly as x and z and inversely as the square of w, where y = 30 when x = 8, z = 3, and w = 2.

36. y varies jointly as x and z and inversely as the square of w, where y = 5 when x = 1, z = 3, and w = 1/2.

37. y varies directly as the square root of x and inversely as z, where y = 12 when x = 9 and z = 5.

38. y varies directly as the square root of x and inversely as the square of z, where y = 15 when x = 25 and z = 2.

39. y varies directly as the square of x and inversely as z and the square of w, where y = 14 when x = 4, w = 2, and z = 2.

40. y varies directly as the square root of x and inversely as z and the square of w, where y = 27 when x = 9, w = 1/2, and z = 4.

Part B: Variation Problems

Applications involving variation.

41. Revenue in dollars is directly proportional to the number of branded sweat shirts sold. If the revenue earned from selling 25 sweat shirts is $318.75, then determine the revenue if 30 sweat shirts are sold.

42. The sales tax on the purchase of a new car varies directly as the price of the car. If an $18,000 new car is purchased, then the sales tax is $1,350. How much sales tax is charged if the new car is priced at $22,000?

43. The price of a share of common stock in a company is directly proportional to the earnings per share (EPS) of the previous 12 months. If the price of a share of common stock in a company is $22.55 and the EPS is published to be $1.10, then determine the value of the stock if the EPS increases by $0.20.

44. The distance traveled on a road trip varies directly with the time spent on the road. If a 126-mile trip can be made in 3 hours, then what distance can be traveled in 4 hours?

45. The circumference of a circle is directly proportional to its radius. If the circumference of a circle with radius 7 centimeters is measured as 14π centimeters, then find the constant of proportionality.

46. The area of circle varies directly as the square of its radius. If the area of a circle with radius 7 centimeters is determined to be 49π square centimeters, then find the constant of proportionality.

47. The surface area of a sphere varies directly as the square of its radius. When the radius of a sphere measures 2 meters, the surface area measures 16π square meters. Find the surface area of a sphere with radius 3 meters.

48. The volume of a sphere varies directly as the cube of its radius. When the radius of a sphere measures 3 meters, the volume is 36π cubic meters. Find the volume of a sphere with radius 1 meter.

49. With a fixed height, the volume of a cone is directly proportional to the square of the radius at the base. When the radius at the base measures 10 centimeters, the volume is 200 cubic centimeters. Determine the volume of the cone if the radius of the base is halved.

50. The distance, d, an object in free fall drops varies directly with the square of the time, t, that it has been falling. If an object in free fall drops 36 feet in 1.5 seconds, then how far will it have fallen in 3 seconds?

Hooke’s law suggests that the extension of a hanging spring is directly proportional to the weight attached to it. The constant of variation is called the spring constant.

Figure 7.1 Robert Hooke (1635–1703)

51. If a hanging spring is stretched 5 inches when a 20‑pound weight is attached to it, then determine its spring constant.

52. If a hanging spring is stretched 3 centimeters when a 2-kilogram weight is attached to it, then determine the spring constant.

53. If a hanging spring is stretched 3 inches when a 2‑pound weight is attached, then how far will it stretch with a 5-pound weight attached?

54. If a hanging spring is stretched 6 centimeters when a 4-kilogram weight is attached to it, then how far will it stretch with a 2-kilogram weight attached?

The breaking distance of an automobile is directly proportional to the square of its speed.

55. If it takes 36 feet to stop a particular automobile moving at a speed of 30 miles per hour, then how much breaking distance is required if the speed is 35 miles per hour?

56. After an accident, it was determined that it took a driver 80 feet to stop his car. In an experiment under similar conditions, it takes 45 feet to stop the car moving at a speed of 30 miles per hour. Estimate how fast the driver was moving before the accident.

Boyle’s law states that if the temperature remains constant, the volume, V, of a given mass of gas is inversely proportional to the pressure, p, exerted on it.

Figure 7.2 Robert Boyle (1627–1691)

57. A balloon is filled to a volume of 216 cubic inches on a diving boat under 1 atmosphere of pressure. If the balloon is taken underwater approximately 33 feet, where the pressure measures 2 atmospheres, then what is the volume of the balloon?

58. If a balloon is filled to 216 cubic inches under a pressure of 3 atmospheres at a depth of 66 feet, then what would the volume be at the surface, where the pressure is 1 atmosphere?

59. To balance a seesaw, the distance from the fulcrum that a person must sit is inversely proportional to his weight. If a 72-pound boy is sitting 3 feet from the fulcrum, then how far from the fulcrum must a 54-pound boy sit to balance the seesaw?

60. The current, I, in an electrical conductor is inversely proportional to its resistance, R. If the current is 1/4 ampere when the resistance is 100 ohms, then what is the current when the resistance is 150 ohms?

61. The number of men, represented by y, needed to lay a cobblestone driveway is directly proportional to the area, A, of the driveway and inversely proportional to the amount of time, t, allowed to complete the job. Typically, 3 men can lay 1,200 square feet of cobblestone in 4 hours. How many men will be required to lay 2,400 square feet of cobblestone given 6 hours?

62. The volume of a right circular cylinder varies jointly as the square of its radius and its height. A right circular cylinder with a 3-centimeter radius and a height of 4 centimeters has a volume of 36π cubic centimeters. Find a formula for the volume of a right circular cylinder in terms of its radius and height.

63. The period, T, of a pendulum is directly proportional to the square root of its length, L. If the length of a pendulum is 1 meter, then the period is approximately 2 seconds. Approximate the period of a pendulum that is 0.5 meter in length.

64. The time, t, it takes an object to fall is directly proportional to the square root of the distance, d, it falls. An object dropped from 4 feet will take 1/2 second to hit the ground. How long will it take an object dropped from 16 feet to hit the ground?

Newton’s universal law of gravitation states that every particle of matter in the universe attracts every other particle with a force, F, that is directly proportional to the product of the masses, m1 and m2, of the particles and inversely proportional to the square of the distance, d, between them. The constant of proportionality is called the gravitational constant.

Figure 7.3 Sir Isaac Newton (1643–1724)

65. If two objects with masses 50 kilograms and 100 kilograms are 1/2 meter apart, then they produce approximately 1.34×106 newtons (N) of force. Calculate the gravitational constant.

66. Use the gravitational constant from the previous exercise to write a formula that approximates the force, F, in newtons between two masses m1 and m2, expressed in kilograms, given the distance d between them in meters.

67. Calculate the force in newtons between earth and the moon, given that the mass of the moon is approximately 7.3×1022 kilograms, the mass of earth is approximately 6.0×1024 kilograms, and the distance between them is on average 1.5×1011 meters.

68. Calculate the force in newtons between earth and the sun, given that the mass of the sun is approximately 2.0×1030 kilograms, the mass of earth is approximately 6.0×1024 kilograms, and the distance between them is on average 3.85×108 meters.

69. If y varies directly as the square of x, then how does y change if x is doubled?

70. If y varies inversely as square of t, then how does y change if t is doubled?

71. If y varies directly as the square of x and inversely as the square of t, then how does y change if both x and t are doubled?

Answers

1: D=kt

3: d=kv2

5: V=kp

7: F=km1m2d2

9: T=kL

11: y=5x

13: y=4x

15: y=75x

17: y=35x

19: y=27x

21: y=14x

23: y=4xz

25: y=23xz

27: y=19xz

29: y=5x2

31: y=3x2

33: y=3xz2

35: y=5xzw2

37: y=20xz

39: y=7x2w2z

41: $382.50

43: $26.65

45: 2π

47: 36π square meters

49: 50 cubic centimeters

51: 1/4

53: 7.5 inches

55: 49 feet

57: 108 cubic inches

59: 4 feet

61: 4 men

63: 1.4 seconds

65: 6.7×1011Nm2/kg2

67: 1.98×1020N

69: y changes by a factor of 4

71: y remains unchanged

7.8 Review Exercises and Sample Exam

Review Exercises

Simplifying Rational Expressions

Evaluate for the given set of x-values.

1. 252x2; {−5, 0, 5}

2. x42x1; {1/2, 2, 4}

3. 1x2+9; {−3, 0, 3}

4. x+3x29; {−3, 0, 3}

State the restrictions to the domain.

5. 5x

6. 1x(3x+1)

7. x+2x225

8. x1(x1)(2x3)

State the restrictions and simplify.

9. x8x264

10. 3x2+9x2x318x

11. x25x24x23x40

12. 2x2+9x54x21

13. x214412x

14. 8x210x394x2

15. Given f(x)=x3x2+9, find f(3), f(0), and f(3).

16. Simplify g(x)=x22x242x29x18 and state the restrictions.

Multiplying and Dividing Rational Expressions

Multiply. (Assume all denominators are nonzero.)

17. 3x5x3x39x2

18. 12y2y3(2y1)(2y1)3y

19. 3x2x2x24x+45x3

20. x28x+159x512x2x3

21. x236x2x302x2+10xx2+5x6

22. 9x2+11x+2481x29x2(x+1)2

Divide. (Assume all denominators are nonzero.)

23. 9x2255x3÷3x+515x4

24. 4x24x21÷2x2x1

25. 3x213x10x2x20÷9x2+12x+4x2+8x+16

26. 2x2+xyy2x2+2xy+y2÷4x2y23x2+2xyy2

27. 2x26x208x2+17x+2÷(8x239x5)

28. 12x227x415x4+10x3÷(3x2+x2)

29. 25y215y4(y2)15y1÷10y2(y2)2

30. 10x4136x2÷5x26x27x+1x12x

31. Given f(x)=16x29x+5 and g(x)=x2+3x104x2+5x6, calculate (fg)(x) and state the restrictions.

32. Given f(x)=x+75x1 and g(x)=x24925x25x, calculate (f/g)(x) and state the restrictions.

Adding and Subtracting Rational Expressions

Simplify. (Assume all denominators are nonzero.)

33. 5xy3y

34. xx2x63x2x6

35. 2x2x+1+1x5

36. 3x7+12xx2

37. 7x4x29x+22x2

38. 5x5+209x2x215x+25

39. xx52x35(x3)x28x+15

40. 3x2x1x4x+4+12(2x)2x2+7x4

41. 1x2+8x91x2+11x+18

42. 4x2+13x+36+3x2+6x27

43. y+1y+212y+2yy24

44. 1y11y2y21

45. Given f(x)=x+12x5 and g(x)=xx+1, calculate (f+g)(x) and state the restrictions.

46. Given f(x)=x+13x and g(x)=2x8, calculate (fg)(x) and state the restrictions.

Complex Fractions

Simplify.

47. 42x2x13x

48. 1313y1515y

49. 16+1x1361x2

50. 11001x21101x

51. xx+32x+1xx+4+1x+3

52. 3x1x55x+22x

53. 112x+35x2125x2

54. 215x+25x22x5

Solving Rational Equations

Solve.

55. 6x6=22x1

56. xx6=x+2x2

57. 13x29=1x

58. 2x5+35=1x5

59. xx5+4x+5=10x225

60. 2x122x+3=23x22x2+3x

61. x+12(x2)+x6x=1

62. 5x+2x+1xx+4=4

63. xx+5+1x4=4x7x2+x20

64. 23x1+x2x+1=2(34x)6x2+x1

65. xx1+1x+1=2xx21

66. 2xx+512x3=47x2x2+7x15

67. Solve for a:   1a=1b+1c.

68. Solve for y:   x=2y13y.

Applications of Rational Equations

Use algebra to solve the following applications.

69. A positive integer is twice another. The sum of the reciprocals of the two positive integers is 1/4. Find the two integers.

70. If the reciprocal of the smaller of two consecutive integers is subtracted from three times the reciprocal of the larger, the result is 3/10. Find the integers.

71. Mary can jog, on average, 2 miles per hour faster than her husband, James. James can jog 6.6 miles in the same amount of time it takes Mary to jog 9 miles. How fast, on average, can Mary jog?

72. Billy traveled 140 miles to visit his grandmother on the bus and then drove the 140 miles back in a rental car. The bus averages 14 miles per hour slower than the car. If the total time spent traveling was 4.5 hours, then what was the average speed of the bus?

73. Jerry takes twice as long as Manny to assemble a skateboard. If they work together, they can assemble a skateboard in 6 minutes. How long would it take Manny to assemble the skateboard without Jerry’s help?

74. Working alone, Joe completes the yard work in 30 minutes. It takes Mike 45 minutes to complete work on the same yard. How long would it take them working together?

Variation

Construct a mathematical model given the following.

75. y varies directly with x, and y = 12 when x = 4.

76. y varies inversely as x, and y = 2 when x = 5.

77. y is jointly proportional to x and z, where y = 36 when x = 3 and z = 4.

78. y is directly proportional to the square of x and inversely proportional to z, where y = 20 when x = 2 and z = 5.

79. The distance an object in free fall drops varies directly with the square of the time that it has been falling. It is observed that an object falls 16 feet in 1 second. Find an equation that models the distance an object will fall and use it to determine how far it will fall in 2 seconds.

80. The weight of an object varies inversely as the square of its distance from the center of earth. If an object weighs 180 pounds on the surface of earth (approximately 4,000 miles from the center), then how much will it weigh at 2,000 miles above earth’s surface?

Sample Exam

Simplify and state the restrictions.

1. 15x3(3x1)23x(3x1)

2. x2144x2+12x

3. x2+x122x2+7x4

4. 9x2(x3)2

Simplify. (Assume all variables in the denominator are positive.)

5. 5xx225x525x2

6. x2+x6x24x+43x25x2x29

7. x24x1212x2÷x66x

8. 2x27x46x224x÷2x2+7x+310x2+30x

9. 1x5+1x+5

10. xx+182x12xx2x2

11. 1y+1x1y21x2

12. 16x+9x225x3x2

13. Given f(x)=x281(4x3)2 and g(x)=4x3x9, calculate (fg)(x) and state the restrictions.

14. Given f(x)=xx5 and g(x)=13x5, calculate (fg)(x) and state the restrictions.

Solve.

15. 13+1x=2

16. 1x5=32x3

17. 19x+20x2=0

18. x+2x2+1x+2=4(x+1)x24

19. xx21x3=3x10x25x+6

20. 5x+4x4x=9x4x216

21. Solve for r:   P=1201+3r.

Set up an algebraic equation and then solve.

22. An integer is three times another. The sum of the reciprocals of the two integers is 1/3. Find the two integers.

23. Working alone, Joe can paint the room in 6 hours. If Manny helps, then together they can paint the room in 2 hours. How long would it take Manny to paint the room by himself?

24. A river tour boat averages 6 miles per hour in still water. With the current, the boat can travel 17 miles in the same time it can travel 7 miles against the current. What is the speed of the current?

25. The breaking distance of an automobile is directly proportional to the square of its speed. Under optimal conditions, a certain automobile moving at 35 miles per hour can break to a stop in 25 feet. Find an equation that models the breaking distance under optimal conditions and use it to determine the breaking distance if the automobile is moving 28 miles per hour.

Review Exercises Answers

1: 1/2, undefined, 1/2

3: 1/18, 1/9, 1/18

5: x0

7: x±5

9: 1x+8; x±8

11: x+3x+5; x5,8

13: (x+12); x12

15: f(3)=13, f(0)=13, f(3)=0

17: x33

19: 3(x2)5x

21: 2xx1

23: 3x(3x5)

25: x+43x+2

27: 2(8x+1)2

29: (5y+1)(y2)50y6

31: (fg)(x)=(4x+3)(x2)x+2; x5,2,34

33: 5x3y

35: 2x28x+1(2x+1)(x5)

37: 14x1

39: x5x3

41: 3(x1)(x+2)(x+9)

43: yy2

45: (f+g)(x)=3x23x+1(2x5)(x+1); x1,52

47: 6

49: 6xx6

51: (x3)(x+4)(x+1)(x+2)

53: x7x+5

55: −3/5

57: −3

59: −10, 1

61: 3, 8

63: 3

65: Ø

67: a=bcb+c

69: 6, 12

71: 7.5 miles per hour

73: 9 minutes

75: y=3x

77: y=3xz

79: d=16t2; 64 feet

Sample Exam Answers

1: 5x2(3x1); x0,13

3: x32x1; x4,12

5: 15x(x+5)

7: x+22x

9: 2x(x5)(x+5)

11: xyxy

13: (fg)(x)=x+94x3; x34,9

15: 3/5

17: 4, 5

19: 4

21: r=40P13

23: 3 hours

25: y=149x2; 16 feet