1. Warm Up Solve each proportion. 1. 2. 3. 4. 5. The value of y varies directly with x, and y = – 6 when x = 3. Find y when x = – 4. 6. The value of y varies directly with x, and y = 6 when x = 30. Find y when x = 45. 10 4.2 2.6252.5 8 9

2. Inverse Variation

3. A relationship that can be written in the form y =, where k is a nonzero constant and x ≠ 0, is an inverse variation. The constant k is the constant of variation. Inverse variation implies that one quantity will increase while the other quantity will decrease (the inverse, or opposite, of increase). Multiplying both sides of y = by x gives xy = k. So, for any inverse variation, the product of x and y is a nonzero constant.

4. A direct variation is an equation that can be written in the form y = kx, where k is a nonzero constant. Remember!

5. There are two methods to determine whether a relationship between data is an inverse variation. You can write a function rule in y = form, or you can check whether xy is a constant for each ordered pair.

6. Example 1A: Identifying an Inverse Variation Tell whether each relationship is an inverse variation. Explain. Method 1 Write a function rule. Can write in y = form. The relationship is an inverse variation. Method 2 Find xy for each ordered pair. 1(30) = 30, 2(15) = 30, 3(10) = 30 The product xy is constant, so the relationship is an inverse variation.

7. Example 1B: Identifying an Inverse Variation Tell whether each relationship is an inverse variation. Explain. Method 1 Write a function rule. Cannot write in y = form. The relationship is not an inverse variation. y = 5x Method 2 Find xy for each ordered pair. 1(5) = 5, 2(10) = 20, 4(20) = 80 The product xy is not constant, so the relationship is not an inverse variation.

8. Example 1C: Identifying an Inverse Variation Tell whether each relationship is an inverse variation. Explain. 2xy = 28 Find xy. Since xy is multiplied by 2, divide both sides by 2 to undo the multiplication. xy = 14 Simplify. xy equals the constant 14, so the relationship is an inverse variation.

9. Tell whether each relationship is an inverse variation. Explain. Method 1 Write a function rule. Cannot write in y = form. The relationship is not an inverse variation. y = – 2x Method 2 Find xy for each ordered pair. – 12 (24) = – 228, 1( – 2) = – 2, 8( – 16) = – 128 The product xy is not constant, so the relationship is not an inverse variation. Check It Out! Example 1a

10. Tell whether each relationship is an inverse variation. Explain. Check It Out! Example 1b Method 1 Write a function rule. Can write in y = form. The relationship is an inverse variation. Method 2 Find xy for each ordered pair. 3(3) = 9, 9(1) = 9, 18(0.5) = 9 The product xy is constant, so the relationship is an inverse variation.

11. 2x + y = 10 Tell whether each relationship is an inverse variation. Explain. Check It Out! Example 1c Cannot write in y = form. The relationship is not an inverse variation.

12. Since k is a nonzero constant, xy ≠ 0. Therefore, neither x nor y can equal 0, and no solution points will be on the x- or y-axes. Helpful Hint

13. An inverse variation can also be identified by its graph. Some inverse variation graphs are shown. Notice that each graph has two parts that are not connected. Also notice that none of the graphs contain (0, 0). This is because (0, 0) can never be a solution of an inverse variation equation.

14. Example 2: Graphing an Inverse Variation Write and graph the inverse variation in which y = 0.5 when x = – 12. Step 1 Find k. k = xy = – 12(0.5) Write the rule for constant of variation. Substitute –12 for x and 0.5 for y. = – 6 Step 2 Use the value of k to write an inverse variation equation. Write the rule for inverse variation. Substitute –6 for k.

15. Example 2 Continued Write and graph the inverse variation in which y = 0.5 when x = – 12. Step 3 Use the equation to make a table of values. y –2 –4 x –1–1 0 124 1.53 6 undef. –6–6 –3–3 – 1.5

16. Example 2 Continued Write and graph the inverse variation in which y = 0.5 when x = – 12. Step 4 Plot the points and connect them with smooth curves. ● ● ● ● ● ●

17. Check It Out! Example 2 Write and graph the inverse variation in which y = when x = 10. Step 1 Find k. k = xy Write the rule for constant of variation. = 5 Substitute 10 for x and for y. = 10 Step 2 Use the value of k to write an inverse variation equation. Write the rule for inverse variation. Substitute 5 for k.

18. Write and graph the inverse variation in which y = when x = 10. Step 3 Use the equation to make a table of values. Check It Out! Example 2 Continued x –4–2–10124 y –1.25–2.5–5undef.52.51.25

19. Check It Out! Example 2 Continued Step 4 Plot the points and connect them with smooth curves. Write and graph the inverse variation in which y = when x = 10. ● ● ● ● ● ●

20. Example 3: Transportation Application The inverse variation xy = 350 relates the constant speed x in mi/h to the time y in hours that it takes to travel 350 miles. Determine a reasonable domain and range and then graph this inverse variation. Use the graph to estimate how long it will take to travel 350 miles driving 55 mi/h. Step 1 Solve the function for y so you can graph it. xy = 350 Divide both sides by x.

21. Example 3 Continued Step 2 Decide on a reasonable domain and range. x > 0 y > 0 Length is never negative and x ≠ 0 Because x and xy are both positive, y is also positive. Step 3 Use values of the domain to generate reasonable ordered pairs. 4.385.838.7517.5y 80604020x

22. Example 3 Continued Step 4 Plot the points. Connect them with a smooth curve. ● ● ● ● Step 5 Find the y-value where x = 55. When the speed is 55 mi/h, the travel time is about 6 hours.

23. Recall that sometimes domain and range are restricted in real-world situations. Remember!

24. Check It Out! Example 3 The inverse variation xy = 100 represents the relationship between the pressure x in atmospheres (atm) and the volume y in mm ³ of a certain gas. Determine a reasonable domain and range and then graph this inverse variation. Use the graph to estimate the volume of the gas when the pressure is 40 atmospheric units. Step 1 Solve the function for y so you can graph it. xy = 100 Divide both sides by x.

25. Step 2 Decide on a reasonable domain and range. x > 0 y > 0 Pressure is never negative and x ≠ 0 Because x and xy are both positive, y is also positive. Step 3 Use values of the domain to generate reasonable pairs. 2.53.34510y 40302010x Check It Out! Example 3 Continued

26. Step 4 Plot the points. Connect them with a smooth curve. Check It Out! Example 3 Continued Step 5 Find the y-value where x = 40. When the pressure is 40 atm, the volume of gas is about 2.5 mm 3. ● ● ● ●

27. The fact that xy = k is the same for every ordered pair in any inverse variation can help you find missing values in the relationship.

29. Example 4: Using the Product Rule Let and Let y vary inversely as x. Find Write the Product Rule for Inverse Variation. Substitute 5 for 3 for and 10 for. Simplify. Solve for by dividing both sides by 5. Simplify.

30. Check It Out! Example 4 Write the Product Rule for Inverse Variation. Simplify. Substitute 2 for –4 for and –6 for Let and Let y vary inversely as x. Find Solve for by dividing both sides by –4.

31. Example 5: Physics Application Boyle ’ s law states that the pressure of a quantity of gas x varies inversely as the volume of the gas y. The volume of gas inside a container is 400 in 3 and the pressure is 25 psi. What is the pressure when the volume is compressed to 125 in 3 ? Use the Product Rule for Inverse Variation. Substitute 400 for 125 for and 25 for Simplify. Solve for by dividing both sides by 125. (400)(25) = (125)y 2

32. Example 5 Continued Boyle ’ s law states that the pressure of a quantity of gas x varies inversely as the volume of the gas y. The volume of gas inside a container is 400 in 3 and the pressure is 25 psi. What is the pressure when the volume is compressed to 125 in 3 ? When the gas is compressed to 125 in 3, the pressure increases to 80 psi.

33. Check It Out! Example 5 On a balanced lever, weight varies inversely as the distance from the fulcrum to the weight. The diagram shows a balanced lever. How much does the child weigh?

34. Check It Out! Example 5 Continued Use the Product Rule for Inverse Variation. Substitute 3.2 for, 60 for and 4.3 for Simplify. Solve for by dividing both sides by 3.2. Simplify. The child weighs 80.625 lb.

35. Lesson Quiz: Part I 1. Write and graph the inverse variation in which y = 0.25 when x = 12.

36. Lesson Quiz: Part II 2. The inverse variation xy = 210 relates the length y in cm to the width x in cm of a rectangle with an area of 210 cm 2. Determine a reasonable domain and range and then graph this inverse variation. Use the graph to estimate the length when the width is 14 cm.

37. Lesson Quiz: Part III 3. Let x 1 = 12, y 1 = – 4, and y 2 = 6, and let y vary inversely as x. Find x 2. –8–8

38. Identify excluded values of rational functions. Graph rational functions. New Objectives

39. Rational function Excluded value Discontinuous function Asymptote Vocabulary

40. A rational function is a function whose rule is a quotient of polynomials in which the denominator has a degree of at least 1. In other words, there must be a variable in the denominator. The inverse variations you studied in the previous lesson are a special type of rational function. Rational functions: Not rational functions:

41. For any function involving x and y, an excluded value is any x-value that makes the function value y undefined. For a rational function, an excluded value is any value that makes the denominator equal to 0.

42. Example 1: Identifying Excluded Values Identify the excluded value for each rational function. A. x = 0 Set the denominator equal to 0. The excluded value is 0. B. x – 2 = 0 x = 2 Set the denominator equal to 0. Solve for x. The excluded value is 2.

43. Check It Out! Example 1 Identify the excluded value for each rational function. a. x = 0 Set the denominator equal to 0. The excluded value is 0. b. x – 1 = 0 x = 1 Set the denominator equal to 0. Solve for x. The excluded value is 1.

44. Check It Out! Example 1 Identify the excluded value for each rational function. x + 4 = 0 x = – 4 Set the denominator equal to 0. Solve for x. The excluded value is – 4. c.

45. Most rational functions are discontinuous functions, meaning their graphs contain one or more jumps, breaks, or holes. This occurs at an excluded value. One place that a graph of a rational function is discontinuous is at an asymptote. An asymptote is a line that a graph gets closer to as the absolute value of a variable increases. In the graph shown, both the x- and y-axes are asymptotes. The graphs of rational functions will get closer and closer to but never touch the asymptotes.

46. For rational functions, vertical asymptotes will occur at excluded values. Look at the graph of y = The denominator is 0 when x = 0 so 0 is an excluded value. This means there is a vertical asymptote at x = 0. Notice the horizontal asymptote at y = 0.

47. Vertical lines are written in the form x = b, and horizontal lines are written in the form y = c. Writing Math

48. Look at the graph of y = Notice that the graph of the parent function y= has been translated 3 units right and there is a vertical asymptote at x = 3. The graph has also been translated 2 units up and there is a horizontal asymptote at y = 2.

49. These translations lead to the following formulas for identifying asymptotes in rational functions.

50. Example 2A: Identifying Asymptotes Identify the asymptotes. Step 2 Identify the asymptotes. vertical: x = – 7 horizontal: y = 0 Step 1 Write in y = form.

51. Example 2B: Identifying Asymptotes Identify the asymptotes. Step 1 Identify the vertical asymptote. 2x – 3 = 0 +3 2x = 3 Find the excluded value. Set the denominator equal to 0. Add 3 to both sides. Solve for x. Is an excluded value.

52. Example 2B Continued Identify the asymptotes. Step 2 Identify the horizontal asymptote. c = 8 y = 8 Vertical asymptote: x = ; horizontal asymptote: y = 8 y = c

53. Check It Out! Example 2a Identify the asymptotes. Step 1 Identify the vertical asymptote. x – 5 = 0 +5 x = 5 Find the excluded value. Set the denominator equal to 0. Add 5 to both sides. Solve for x. 5 is an excluded value. x = 5

54. Identify the asymptotes. Step 2 Identify the horizontal asymptote. c = 0 y = 0 y = c Check It Out! Example 2a Continued Vertical asymptote: x = 5; horizontal asymptote: y = 0

55. Check It Out! Example 2b Identify the asymptotes. Step 1 Identify the vertical asymptote. 4x + 16 = 0 – 16 4x = – 16 Find the excluded value. Set the denominator equal to 0. Subtract 16 from both sides. Solve for x. –4 is an excluded value. x = – 4

56. Check It Out! Example 2b Continued Identify the asymptotes. Step 2 Identify the horizontal asymptote. c = 5 y = 5 Vertical asymptote: x = – 4; horizontal asymptote: y = 5 y = c

57. Check It Out! Example 2c Identify the asymptotes. Step 1 Identify the vertical asymptote. x + 77 = 0 – 77 x = – 77 Find the excluded value. Set the denominator equal to 0. Subtract 77 from both sides. Solve for x. –77 is an excluded value. x = – 77

58. Check It Out! Example 2c Continued Identify the asymptotes. Step 2 Identify the horizontal asymptote. c = – 15 y = – 15 Vertical asymptote: x = – 77; horizontal asymptote: y = – 15 y = c

59. To graph a rational function in the form y = when a = 1, you can graph the asymptotes and then translate the parent function y =. However, if a ≠ 1, the graph is not a translation of the parent function. In this case, you can use the asymptotes and a table of values.

60. Example 3A: Graphing Rational Functions Using Asymptotes Graph the function. Since the numerator is not 1, use the asymptotes and a table of values. Step 1 Identify the vertical and horizontal asymptotes. vertical: x = 3 horizontal: y = 0 Use x = b. x – 3 = 0, so b = 3. Use y = c. c = 0

61. Example 3A Continued Step 2 Graph the asymptotes using dashed lines. Step 3 Make a table of values. Choose x-values on both sides of the vertical asymptote. y 654210x x = 3 y = 0 Step 4 Plot the points and connect them with smooth curves. The curves will get very close to the asymptotes, but will not touch them. ● ● ● ● ● ●

62. Example 3B: Graphing Rational Functions Using Asymptotes Graph the function. Step 1 Since the numerator is 1, use the asymptotes and translate y =. vertical: x = – 4 horizontal: y = – 2 Use x = b. b = –4 Use y = c. c = –2

63. Step 2 Graph the asymptotes using dashed lines. Example 3B Continued Step 3 Draw smooth curves to show the translation.

64. Check It Out! Example 3a Graph each function. Step 1 Since the numerator is 1, use the asymptotes and translate y =. vertical: x = – 7 horizontal: y = 3 Use x = b. b = –7 Use y = c. c = 3

65. Check It Out! Example 3a Continued Graph each function. Step 2 Graph the asymptotes using dashed lines. Step 3 Draw smooth curves to show the translation.

66. Check It Out! Example 3b Graph each function. Since the numerator is not 1, use the asymptotes and a table of values. Step 1 Identify the vertical and horizontal asymptotes. vertical: x = 3 horizontal: y = 2 Use x = b. x – 3 = 0, so b = 3. Use y = c. c = 2

67. Check It Out! Example 3b Step 2 Graph the asymptotes using dashed lines. Step 3 Make a table of values. Choose x-values on both sides of the vertical asymptote. y 654210x 4 Step 4 Plot the points and connect them with smooth curves. The curves will get very close to the asymptotes, but will not touch them.

68. Example 4: Application Your club has $75 with which to purchase snacks to sell at an afterschool game. The number of snacks y that you can buy, if the average price of the snacks is x-dollars, is given by y = a. Describe the reasonable domain and range values. Both the number of snacks purchased and their cost will be positive values so nonnegative values are reasonable for both domain and range.

69. Example 4 Continued b. Graph the function. Step 1 Identify the vertical and horizontal asymptotes. vertical: x = 0 horizontal: y = 0 Use x = b. b = 0 Use y = c. c = 0 Step 2 Graph the asymptotes using dashed lines. The asymptotes will be the x- and y- axes.

70. Example 4 Continued Step 3 Since the domain is restricted to nonnegative values, only choose x-values on the right side of the vertical asymptote. Number of snacks 2468 Cost of snacks($) 37.518.7512.59.38

71. Example 4 Continued Step 4 Plot the points and connect them with a smooth curve.

72. Check It Out! Example 4 a. Describe the reasonable domain and range values. The domain would be all values greater than 0 up to $500 dollars and the range would be all natural numbers greater than 10. A librarian has a budget of $500 to buy copies of a software program. She will receive 10 free copies when she sets up an account with the supplier. The number of copies y of the program that she can buy is given by y = + 10, where x is the price per copy.

73. Check It Out! Example 4 Continued b. Graph the function. Step 1 Identify the vertical and horizontal asymptotes. vertical: x = 0 horizontal: y = 10 Use x = b. b = 0 Use y = c. c = 10 Step 2 Graph the asymptotes using dashed lines. The asymptotes will be the x- and y- axes.

74. Step 3 Since the domain is restricted to nonnegative values, only choose x-values on the right side of the vertical asymptote. Check It Out! Example 4 Continued Number of copies 20406080 Price ($) 35231816

75. Step 4 Plot the points and connect them with a smooth curve. Check It Out! Example 4 Continued

76. The table shows some of the properties of the four families of functions you have studied and their graphs.

79. Lesson Quiz: Part I Identify the exceeded value for each rational function. 1. 2. 3. Identify the asymptotes of and then graph the function. x = – 4; y = 0 0 5

80. Lesson Quiz: Part II You have $ 100 to spend on CDs. A CD club advertises 6 free CDs for anyone who becomes a member. The number of CDs y that you can receive is given by y =, where x is the average price per CD. a. Describe the reasonable domain and range values. D: x > 0 R: natural numbers > 6

81. Lesson Quiz: Part III b. Graph the function.

82. Simplify rational expressions. Identify excluded values of rational expressions. New Objectives

83. A rational expression is an algebraic expression whose numerator and denominator are polynomials. The value of the polynomial expression in the denominator cannot be zero since division by zero is undefined. This means that rational expressions may have excluded values.

84. Example 1A: Identifying Excluded Values Find any excluded values of each rational expression. g + 4 = 0 g = – 4 The excluded value is – 4. Set the denominator equal to 0. Solve for g by subtracting 4 from each side.

85. Example 1B: Identifying Excluded Values Find any excluded values of each rational expression. x 2 – 15x = 0 x(x – 15) = 0 x = 0 or x – 15 = 0 x = 15 Set the denominator equal to 0. Factor. Use the Zero Product Property. Solve for x. The excluded values are 0 and 15.

86. Example 1C: Identifying Excluded Values Find any excluded values of each rational expression. y 2 + 5y + 4 = 0 (y + 4)(y + 1) = 0 y + 4 = 0 or y + 1 = 0 y = – 4 or y = – 1 The excluded values are – 4 and – 1. Set the denominator equal to 0. Factor Use the Zero Product Property. Solve each equation for y.

87. Check It Out! Example 1a Find any excluded values of each rational expression. t + 5 = 0 t = – 5 The excluded value is – 5. Set the denominator equal to 0. Solve for t by subtracting 5 from each side.

88. Check It Out! Example 1b Find any excluded values of each rational expression. b 2 + 5b = 0 b(b + 5) = 0 b = 0 or b + 5 = 0 b = – 5 Set the denominator equal to 0. Factor. Use the Zero Product Property. Solve for b. The excluded values are 0 and – 5.

89. Check It Out! Example 1c Find any excluded values of each rational expression. k 2 + 7k + 12 = 0 (k + 4)(k + 3) = 0 k + 4 = 0 or k + 3 = 0 k = – 4 or k = – 3 The excluded values are – 4 and – 3. Set the denominator equal to 0. Factor Use the Zero Product Property. Solve each equation for k.

90. A rational expression is in its simplest form when the numerator and denominator have no common factors except 1. Remember that to simplify fractions you can divide out common factors that appear in both the numerator and the denominator. You can do the same to simplify rational expressions.

91. Example 2A: Simplifying Rational Expressions Simplify each rational expression, if possible. Identify any excluded values. Factor 14. Divide out common factors. Note that if r = 0, the expression is undefined. Simplify. The excluded value is 0. 4

92. Example 2B: Simplifying Rational Expressions Simplify each rational expression, if possible. Identify any excluded values. 3n; n ≠ Factor 6n² + 3n. Divide out common factors. Note that if n =, the expression is undefined. Simplify. The excluded value is.

93. Example 2C: Simplifying Rational Expressions Simplify each rational expression, if possible. Identify any excluded values. 3p – 2 = 0 3p = 2 There are no common factors. Add 2 to both sides. Divide both sides by 3. The excluded value is

94. Be sure to use the original denominator when finding excluded values. The excluded values may not be “seen” in the simplified denominator. Caution

95. Check It Out! Example 2a Simplify each rational expression, if possible. Identify any excluded values. Factor 15. Divide out common factors. Note that if m = 0, the expression is undefined. Simplify. The excluded value is 0.

96. Check It Out! Example 2b Simplify each rational expression, if possible. Identify any excluded values. Factor the numerator. Divide out common factors. Note that the expression is not undefined. Simplify. There is no excluded value.

97. Check It Out! Example 2c Simplify each rational expression, if possible. Identify any excluded values. The numerator and denominator have no common factors. The excluded value is 2.

98. From now on in this chapter, you may assume that the values of the variables that make the denominator equal to 0 are excluded values. You do not need to include excluded values in your answers unless they are asked for.

99. Example 3: Simplifying Rational Expressions with Trinomials Simplify each rational expression, if possible. Factor the numerator and the denominator when possible. Divide out common factors. Simplify. A.B.

100. Check It Out! Example 3 Simplify each rational expression, if possible. Factor the numerator and the denominator when possible. Divide out common factors. Simplify. a. b.

101. Opposite binomials can help you factor polynomials. Recognizing opposite binomials can also help you simplify rational expressions. Consider The numerator and denominator are opposite binomials. Therefore,

102. Example 4: Simplifying Rational Expressions Using Opposite Binomials Simplify each rational expression, if possible. Factor. Rewrite one opposite binomial. Identify opposite binomials. A.B.

103. Example 4 Continued Simplify each rational expression, if possible. Divide out common factors. Simplify.

104. Check It Out! Example 4 Simplify each rational expression, if possible. Factor. Rewrite one opposite binomial. Identify opposite binomials. a.b.

105. Check It Out! Example 4 Continued Simplify each rational expression, if possible. Divide out common factors. Simplify.

106. Check It Out! Example 4 Continued Simplify each rational expression, if possible. Factor. Divide out common factors. c.

107. Example 5: Application A theater at an amusement park is shaped like a sphere. The sphere is held up with support rods. a. What is the ratio of the theater ’ s volume to its surface area? (Hint: For a sphere, V = and S = 4r 2.) Write the ratio of volume to surface area. Divide out common factors.

108. Example 5 Continued Multiply by the reciprocal of 4. Divide out common factors. Simplify. Use the Property of Exponents.

109. Example 5 Continued b. Use this ratio to find the ratio of the theater ’ s volume to its surface area when the radius is 45 feet. Write the ratio of volume to surface area. Substitute 45 for r.

110. Check It Out! Example 5 Which barrel cactus has less of a chance to survive in the desert, one with a radius of 6 inches or one with a radius of 3 inches? Explain. The barrel cactus with a radius of 3 inches has less of a chance to survive. Its surface- area-to-volume ratio is greater than for a cactus with a radius of 6 inches.

111. For two fractions with the same numerator, the value of the fraction with a greater denominator is less than the value of the other fraction. 9 > 3 Remember!

112. Lesson Quiz: Part I Find any excluded values of each rational expression. 1. 2. 0 0, 2 Simplify each rational expression, if possible. 3.4. 5.

113. 6. Calvino is building a rectangular tree house. The length is 10 feet longer than the width. His friend Fabio is also building a tree house, but his is square. The sides of Fabio ’ s tree house are equal to the width of Calvino ’ s tree house. a. What is the ratio of the area of Calvino ’ s tree house to the area of Fabio ’ s tree house? b. Use this ratio to find the ratio of the areas if the width of Calvino ’ s tree house is 14 feet.

114. Warm Up Multiply. 1. 2x 2 (x + 3) 2. (x – 5)(3x + 7) 3. 3x(x 2 +2x + 2) 4. Simplify. 2x 3 + 6x 2 3x 2 – 8x – 35 3x 3 + 6x 2 + 6x

115. Warm Up Divide. Simplify your answer. 5. 6. 7. 8.

116. Multiply and divide rational expressions. New Objective

117. The rules for multiplying rational expressions are the same as the rules for multiplying fractions. You multiply the numerators, and you multiply the denominators.

119. Example 1A: Multiplying Rational Expressions Multiply. Simplify your answer. Multiply the numerators and denominators. Factor. Divide out the common factors. Simplify.

120. Example 1B: Multiplying Rational Expressions Multiply. Simplify your answer. Multiply the numerators and the denominators. Arrange the expression so like variables are together. Simplify. Divide out common factors. Use properties of exponents. Simplify. Remember that z 0 = 1.

121. Example 1C: Multiplying Rational Expressions Multiply. Simplify your answer. Multiply. There are no common factors, so the product cannot be simplified.

122. The Quotient of Powers Property Remember!

123. Check It Out! Example 1a Multiply. Simplify your answer. Multiply the numerators and the denominators. Arrange the expression so like variables are together. Simplify. Divide out common factors. Use properties of exponents.

124. Check It Out! Example 1b Multiply. Simplify your answer. Multiply the numerators and the denominators. Arrange the expression so like variables are together. Simplify. Divide out common factors. Use properties of exponents.

125. Example 2: Multiplying a Rational Expression by a Polynomial. Multiply. Simplify your answer. Write the polynomial over 1. Factor the numerator and denominator. Divide out common factors. Multiply remaining factors.

126. Check It Out! Example 2 Multiply Simplify your answer. Write the polynomial over 1. Factor the numerator and denominator. Divide out common factors. Multiply remaining factors.

127. Just as you can write an integer as a fraction, you can write any expression as a rational expression by writing it with a denominator of 1. Remember!

128. There are two methods for simplifying rational expressions. You can simplify first by dividing out and then multiply the remaining factors. You can also multiply first and then simplify. Using either method will result in the same answer.

129. Example 3: Multiplying a Rational Expression Containing Polynomial Multiply. Simplify your answer. Method 1 Simplify first. Then multiply. Factor. Divide out common factors. Simplify.

130. Method 2 Multiply first. Example 3 Continued Multiply. Distribute.

131. Then simplify. Factor. Divide out common factors. Simplify. Example 3 Continued

132. Multiply. Simplify your answer. Simplify first. Check It Out! Example 3a Factor. Divide out common factors. Simplify. Then multiply.

133. Multiply. Simplify your answer. Simplify first. Check It Out! Example 3b Factor. Divide out common factors. Simplify. Then multiply. p

134. The rules for dividing rational expressions are the same as the rules for dividing fractions. To divide by a rational expression, multiply by its reciprocal.

135. Example 4A: Dividing by Rational Expressions and Polynomials Divide. Simplify your answer. Write as multiplication by the reciprocal. Multiply the numerators and the denominators. Divide out common factors. Simplify.

136. Example 4B: Dividing by Rational Expressions and Polynomials Divide. Simplify your answer. Write as multiplication by the reciprocal. Factor. Rewrite one opposite binomial.

137. Example 4B Continued Divide. Simplify your answer. Divide out common factors. Simplify.

138. Example 4C: Dividing by Rational Expressions and Polynomials Divide. Simplify your answer. Write the binomial over 1. Write as multiplication by the reciprocal. Multiply the numerators and the denominators.

139. Example 4C Continued Divide. Simplify your answer. Divide out common factors. Simplify.

140. Divide. Simplify your answer. Check It Out! Example 4a Write as multiplication by the reciprocal. Multiply the numerators and the denominators. Simplify. There are no common factors.

141. Divide. Simplify your answer. Check It Out! Example 4b Write as multiplication by the reciprocal. Multiply the numerators and the denominators and cancel common factors. Simplify.

142. Divide. Simplify your answer. Check It Out! Example 4c Write the trinomial over 1. Write as multiplication by the reciprocal. Multiply.

143. Divide. Simplify your answer. Check It Out! Example 4c Continued Factor. Divide out common factors. Simplify.

144. Tanya is playing a carnival game. She needs to pick 2 cards out of a deck without looking. The deck has cards with numbers and cards with letters. There are 6 more letter cards than number cards. a. Write and simplify an expression that represents the probability that Tanya will pick 2 number cards. Let x = the number cards. number + letter = total x +x + 6 = 2x + 6 Write expressions for the number of each kind of card and for the total number of items.

145. Example 5 Continued The probability of picking a letter card and then another letter card is the product of the probabilities of the individual events. 2nd pick: total items 1st pick letter2nd pick letter 1st pick: total items

146. b. What is the probability that Tanya picks 2 number cards if there are 25 number cards in the deck before her first pick? Round your answer to the nearest hundredth. Use the probability of picking two number cards. Since x represents the number of number cards, substitute 25 for x. Substitute. Use the order of operations to simplify. P(number, letter)

147. Example 5 Continued The probability of picking 2 number cards if there are 25 number cards in the deck of 51 before the first pick is approximately 0.19.

148. Check It Out! Example 5 What if … ? There are 50 blue items in the bag before Marty ’ s first pick. What is the probability that Marty picks two blue items? Use the probability of picking two blue items. Since x represents the number of blue items, substitute 50 for x. Substitute. The probability of picking two blue items is about 0.23. P(blue, blue) Use the order of operations to simplify.

149. Lesson Quiz: Part I Multiply. Simplify your answer. 1. 2. 2 Divide. Simplify your answer. 4.5. 3.

150. Lesson Quiz: Part II 6. A bag contains purple and green toy cars. There are 9 more purple cars than green cars. a. Write and simplify an expression to represent the probability that someone will pick a purple car and a green car. b. What is the probability of someone picking a purple car and a green car if there are 12 green cars before the first pick? Round to the nearest hundredth. 0.24

151. Warm Up Add. Simplify your answer. 1. 2. 3.4. Subtract. Simplify your answer. 5. 7. 6. 8.

152. Add and subtract rational expressions with like denominators. Add and subtract rational expressions with unlike denominators. Objectives

153. The rules for adding rational expressions are the same as the rules for adding fractions. If the denominators are the same, you add the numerators and keep the common denominator.

155. Example 1A: Adding Rational Expressions with Like Denominators Add. Simplify your answer. Combine like terms in the numerator. Divide out common factors. Simplify.

156. Example 1B: Adding Rational Expressions with Like Denominators Add. Simplify your answer. Combine like terms in the numerator. Factor. Divide out common factors. Simplify.

157. Example 1C: Adding Rational Expressions with Like Denominators Add. Simplify your answer. Combine like terms in the numerator. Factor. Divide out common factors. Simplify.

158. Check It Out! Example 1a Add. Simplify your answer. = 2 Combine like terms in the numerator. Divide out common factors. Simplify.

159. Check It Out! Example 1b Add. Simplify your answer. Combine like terms in the numerator. Factor. Divide out common factors. Simplify.

160. To subtract rational expressions with like denominators, remember to add the opposite of each term in the second numerator.

161. Example 2: Subtracting Rational Expressions with Like Denominators Subtract. Simplify your answer. Subtract numerators. Combine like terms. Factor. Divide out common factors. Simplify.

162. Make sure you add the opposite of all the terms in the numerator of the second expression when subtracting rational expressions. Caution

163. Check It Out! Example 2a Subtract. Simplify your answer. Subtract numerators. Combine like terms. Factor. Divide out common factors. Simplify.

164. Check It Out! Example 2b Subtract. Simplify your answer. Subtract numerators. Combine like terms. Factor. There are no common factors.

165. As with fractions, rational expressions must have a common denominator before they can be added or subtracted. If they do not have a common denominator, you can use the least common multiple, or LCM, of the denominators to find one. To find the LCM, write the prime factorization of both expressions. Use each factor the greatest number of times it appears in either expression.

166. Example 3A: Identifying the Least Common Multiple Find the LCM of the given expressions. 12x 2 y, 9xy 3 12x 2 y = 2  2  3  x  x  y  9xy 3 = 3  3  x  y  y  y LCM = 2  2  3  3  x  x  y  y  y Write the prime factorization of each expression. Use every factor of both expressions the greatest number of times it appears in either expression. = 36x 2 y 3

167. Example 3B: Identifying the Least Common Multiple Find the LCM of the given expressions. c 2 + 8c + 15, 3c 2 + 18c + 27 c 2 + 8c + 15 = (c + 3)(c + 5) 3c 2 + 18c + 27 = 3(c 2 +6c +9) = 3(c + 3)(c + 3) LCM = 3(c + 3) 2 (c + 5) Factor each expression. Use every factor of both expressions the greatest number of times it appears in either expression.

168. Check It Out! Example 3a Find the LCM of the given expressions. 5f 2 h, 15fh 2 5f 2 h = 5  f  f  h 15fh 2 = 3  5  f  h  h LCM = 3  5  f  f  h  h = 15f 2 h 2 Write the prime factorization of each expression. Use every factor of both expressions the greatest number of times it appears in either expression.

169. Check It Out! Example 3b Find the LCM of the given expressions. x 2 – 4x – 12, (x – 6)(x +5) x 2 – 4x – 12 = (x – 6)(x + 2) (x – 6)(x +5) = (x – 6)(x + 5) LCM = (x – 6)(x + 5)(x + 2) Factor each expression. Use every factor of both expressions the greatest number of times it appears in either expression.

170. The LCM of the denominators of fractions or rational expressions is also called the least common denominator, or LCD. You use the same method to add or subtract rational expressions.

171. Example 4A: Adding and Subtracting with Unlike Denominators Add or subtract. Simplify your answer. Step 1 5n 3 = 5  n  n  n 2n 2 = 2  n  n LCD = 2  5  n  n  n = 10n 3 Identify the LCD. Step 2 Multiply each expression by an appropriate form of 1. Write each expression using the LCD. Step 3

172. Example 4A Continued Add or subtract. Simplify your answer. Add the numerators. Factor and divide out common factors. Step 6 Simplify. Step 4 Step 5

173. Example 4B: Adding and Subtracting with Unlike Denominators. Add or subtract. Simplify your answer. Step 1 The denominators are opposite binomials. The LCD can be either w – 5 or 5 – w. Identify the LCD. Step 2 Step 3 Multiply the first expression by to get an LCD of w – 5. Write each expression using the LCD.

174. Example 4B Continued Add or Subtract. Simplify your answer. Step 4 Step 5, 6 Subtract the numerators. No factoring needed, so just simplify.

175. Add or subtract. Simplify your answer. Identify the LCD. 3d 3  d 2d 3 = 2  d  d  d LCD = 2  3 d  d  d = 6d 3 Step 1 Multiply each expression by an appropriate form of 1. Write each expression using the LCD. Check It Out! Example 4a Step 2 Step 3

176. Add or subtract. Simplify your answer. Check It Out! Example 4a Continued Subtract the numerators. Factor and divide out common factors. Step 4 Simplify. Step 5 Step 6

177. Add or subtract. Simplify your answer. Check It Out! Example 4b Factor the first term. The denominator of second term is a factor of the first. Add the two fractions. Divide out common factors. Step 1 Step 2 Step 3 Step 4 Simplify.

178. Example 5: Recreation Application Roland needs to take supplies by canoe to some friends camping 2 miles upriver and then return to his own campsite. Roland ’ s average paddling rate is about twice the speed of the river ’ s current. a. Write and simplify an expression for how long it will take Roland to canoe round trip. Step 1 Write expressions for the distances and rates in the problem. The distance in both directions is 2 miles.

179. Example 5 Continued Roland ’ s rate against the current is 2x – x, or x. Roland ’ s rate with the current is 2x + x, or 3x. Step 2 Use a table to write expressions for time. Downstream (with current) Upstream (against current) Rate (mi/h) Distance (mi) Direction Time (h) = Distance rate 2 2 x 3x3x Let x represent the rate of the current, and let 2x represent Roland ’ s paddling rate.

180. Example 5 Continued Step 3 Write and simplify an expression for the total time. total time = time upstream + time downstream total time = Substitute known values. Multiply the first fraction by a appropriate form of 1. Write each expression using the LCD 3x. Add the numerators. Step 4 Step 5 Step 6

181. Example 5 Continued b. If the speed of the river ’ s current is 2.5 miles per hour, about how long will it take Roland to make the round trip? Substitute 2.5 for x. Simplify. It will take Roland of an hour or 64 minutes to make the round trip.

182. Check It Out! Example 5 What if?...Katy ’ s average paddling rate increases to 5 times the speed of the current. Now how long will it take Katy to kayak the round trip? Step 1 Let x represent the rate of the current, and let 5x represent Katy ’ s paddling rate. Katy ’ s rate against the current is 5x – x, or 4x. Katy ’ s rate with the current is 5x + x, or 6x.

183. Step 2 Use a table to write expressions for time. Check It Out! Example 5 Continued Downstream (with current) Upstream (against current) Rate (mi/h) Distance (mi) Direction Time (h) = distance rate 1 1 4x4x 6x6x

184. Check It Out! Example 5 Continued Step 3 Write and simplify an expression for the total time. total time = time upstream + time downstream Substitute known values. Multiply each fraction by a appropriate form of 1. Write each expression using the LCD 24x. Add the numerators. Step 4 Step 5Step 6 total time =

185. b. If the speed of the river ’ s current is 2 miles per hour, about how long will it take Katy to make the round trip? Substitute 2 for x. Simplify. Check It Out! Example 5 Continued It will take Katy of an hour or 12.5 minutes to make the round trip.

186. Lesson Quiz: Part I Add or subtract. Simplify your answer. 1. 2. 5. 3. 4.

187. Lesson Quiz: Part II 6. Vong drove 98 miles on interstate highways and 80 miles on state roads. He drove 25% faster on the interstate highways than on the state roads. Let r represent his rate on the state roads in miles per hour. a. Write and simplify an expression that represents the number of hours Vong drove in terms of r. b. Find Vong ’ s driving time if he averaged 55 miles per hour on the state roads. about 2 h 53 min

188. Warm Up Divide. 1. m 2 n ÷ mn 4 2. 2x 3 y 2 ÷ 6xy 3. (3a + 6a 2 ) ÷ 3a 2 b Factor each expression. 4. 5x 2 + 16x + 12 5. 16p 2 – 72p + 81

189. Divide a polynomial by a monomial or binomial. New Objective

190. To divide a polynomial by a monomial, you can first write the division as a rational expression. Then divide each term in the polynomial by the monomial.

191. Example 1: Dividing a Polynomial by a Monomial Divide (5x 3 – 20x 2 + 30x) ÷ 5x x 2 – 4x + 6 Write as a rational expression. Divide each term in the polynomial by the monomial 5x. Divide out common factors. Simplify.

192. Check It Out! Example 1a Divide. (8p 3 – 4p 2 + 12p) ÷ ( – 4p 2 ) Write as a rational expression. Divide each term in the polynomial by the monomial –4p 2. Divide out common factors. Simplify.

193. Check It Out! Example 1b Divide. (6x 3 + 2x – 15) ÷ 6x Write as a rational expression. Divide each term in the polynomial by the monomial 6x. Divide out common factors. Simplify.

194. Division of a polynomial by a binomial is similar to division of whole numbers.

195. Example 2A: Dividing a Polynomial by a Binomial Divide. x + 5 Factor the numerator. Divide out common factors. Simplify.

196. Example 2B: Dividing a Polynomial by a Binomial Divide. Factor both the numerator and denominator. Divide out common factors. Simplify.

197. Put each term of the numerator over the denominator only when the denominator is a monomial. If the denominator is a polynomial, try to factor first. Helpful Hint

198. Check It Out! Example 2a Divide. k + 5 Factor the numerator. Divide out common factors. Simplify.

199. Check It Out! Example 2b Divide. b – 7 Factor the numerator. Divide out common factors. Simplify.

200. Check It Out! Example 2c Divide. s + 6 Factor the numerator. Divide out common factors. Simplify.

201. Recall how you used long division to divide whole numbers as shown at right. You can also use long division to divide polynomials. An example is shown below. ) x 2 + 3x + 2 x + 1 x 2 + 2x x + 2 0 (x 2 + 3x + 2) ÷ (x + 2) Divisor Quotient Dividend

203. Example 3A: Polynomial Long Division Divide using long division. (x 2 +10x + 21) ÷ (x + 3) x 2 + 10x + 21 ) Step 1x + 3 Write in long division form with expressions in standard form. Divide the first term of the dividend by the first term of the divisor to get the first term of the quotient. x 2 + 10x + 21 ) Step 2x + 3 x

204. Example 3A Continued Divide using long division. (x 2 +10x + 21) ÷ (x + 3) Multiply the first term of the quotient by the binomial divisor. Place the product under the dividend, aligning like terms. x 2 + 10x + 21 ) Step 3x + 3 x x 2 + 3x x 2 + 10x + 21 ) Step 4x + 3 – (x 2 + 3x) x 0 + 7x Subtract the product from the dividend.

205. Example 3A Continued Divide using long division. x 2 + 10x + 21 ) Step 5x + 3 – (x 2 + 3x) x + 21 Bring down the next term in the dividend. Repeat Steps 2-5 as necessary. x 2 + 10x + 21 ) Step 6x + 3 – (x 2 + 3x) x + 7 7x + 21 – (7x + 21) 0 The remainder is 0. 7x7x

206. Example 3A Continued Check: Multiply the answer and the divisor. (x + 3)(x + 7) x 2 + 3x + 7x + 21 x 2 + 10x + 21

207. Example 3B: Polynomial Long Division Divide using long division. x 2 – 2x – 8 ) x – 4 Write in long division form. – (x 2 – 4x) 2x2x x 2 – 2x – 8 ) x – 4 – (2x – 8) 0 x 2 ÷ x = x Multiply x  (x – 4 ). Subtract. Bring down the 8. 2x ÷ x =2. Multiply 2(x – 4). Subtract. The remainder is 0. x + 2 – 8

208. Example 3B Continued Check: Multiply the answer and the divisor. (x + 2)(x – 4) x 2 + 2x – 4x – 8 x 2 – 2x + 8

209. Check It Out! Example 3a Divide using long division. (2y 2 – 5y – 3) ÷ (y – 3) 2y 2 – 5y – 3 ) Step 1 y – 3 Write in long division form with expressions in standard form. Divide the first term of the dividend by the first term of the divisor to get the first term of the quotient. 2y 2 – 5y – 3 ) Step 2 y – 3 2y2y

210. Check It Out! Example 3a Continued Divide using long division. (2y 2 – 5y – 3) ÷ (y – 3) Multiply the first term of the quotient by the binomial divisor. Place the product under the dividend, aligning like terms. Subtract the product from the dividend. 2y 2 – 5y – 3 ) Step 3 y – 3 2y2y 2y 2 – 6y – (2y 2 – 6y) 0 + y 2y 2 – 5y – 3 ) Step 4 y – 3 2y2y

211. Check It Out! Example 3a Continued Divide using long division. ) Step 5 y – 3 2y2y – 3– 3 Bring down the next term in the dividend. Repeat Steps 2–5 as necessary. 2y2y 2y 2 – 5y – 3 ) Step 6 y – 3 – (2y 2 – 6y) y – 3 – (y – 3) 0 The remainder is 0. 2y 2 – 5y – 3 – (2y 2 – 6y) y + 1

212. Check: Multiply the answer and the divisor. (y – 3)(2y + 1) 2y 2 + y – 6y – 3 2y 2 – 5y – 3 Check It Out! Example 3a Continued

213. Check It Out! Example 3b Divide using long division. (a 2 – 8a + 12) ÷ (a – 6) a 2 – 8a + 12 ) a – 6 Write in long division form. – (a 2 – 6a) –2a–2a a a 2 – 8a + 12 ) a – 6 – ( – 2a + 12) 0 a 2 ÷ a = a Multiply a  (a – 6 ). Subtract. Bring down the 12. –2a ÷ a = –2. Multiply –2(a – 6). Subtract. The remainder is 0. – 2 + 12

214. Check It Out! Example 3b Continued Check: Multiply the answer and the divisor. (a – 6)(a – 2) a 2 – 2a – 6a + 12 a 2 – 8a + 12

215. Sometimes the divisor is not a factor of the dividend, so the remainder is not 0. Then the remainder can be written as a rational expression.

216. Example 4: Long Division with a Remainder Divide (3x 2 + 19x + 26) ÷ (x + 5) 3x 2 + 19x + 26 ) x + 5 Write in long division form. 3x 2 + 19x + 26 ) x + 5 3x3x – (3x 2 + 15x) 4x4x 3x 2 ÷ x = 3x. Multiply 3x(x + 5). Subtract. Bring down the 26. 4x ÷ x = 4. Multiply 4(x + 5). Subtract. – (4x + 20) 6 The remainder is 6. Write the remainder as a rational expression using the divisor as the denominator. + 4 + 26

217. Example 4 Continued Divide (3x 2 + 19x + 26) ÷ (x + 5) Write the quotient with the remainder.

218. Check It Out! Example 4a Divide. 3m 2 + 4m – 2 ) m + 3 Write in long division form. 3m 2 + 4m – 2 ) m + 3 3m3m – (3m 2 + 9m) 3m 2 ÷ m = 3m. Multiply 3m(m + 3). Subtract. Bring down the –2. –5m ÷ m = –5. Multiply –5(m + 3). Subtract. –5m–5m The remainder is 13. 13 – ( – 5m – 15) – 5 – 2

219. Check It Out! Example 4a Continued Divide. Write the remainder as a rational expression using the divisor as the denominator.

220. Check It Out! Example 4b Divide. y 2 + 3y + 2 ) y – 3 Write in long division form. – (y 2 – 3y) y 2 ÷ y = y. Multiply y(y – 3). Subtract. Bring down the 2. 6y ÷ y = 6. y y 2 + 3y + 2 ) y – 3 Multiply 6(y – 3). Subtract. The remainder is 20. 20 6y6y – (6y – 18) Write the quotient with the remainder. + 6 +2+2 y + 6 +

221. Sometimes you need to write a placeholder for a term using a zero coefficient. This is best seen if you write the polynomials in standard form.

222. Example 5: Dividing Polynomials That Have a Zero Coefficient Divide (x 3 – 7 – 4x) ÷ (x – 3). (x 3 – 4x – 7) ÷ (x – 3) Write in standard format. x 3 + 0x 2 – 4x – 7 ) x – 3 Write in long division form. Use 0x 2 as a placeholder for the x 2 term.

223. Example 5: Dividing Polynomials That Have a Zero Coefficient Divide (x 3 – 7 – 4x) ÷ (x – 3). x 3 + 0x 2 – 4x – 7 ) x – 3 x 3 ÷ x = x 2 Multiply x 2 (x – 3). Subtract. (x 3 – 4x – 7) ÷ (x – 3) Write the polynomials in standard form. Write in long division form. Use 0x 2 as a placeholder for the x 2 term. x2x2 x 3 + 0x 2 – 4x – 7 ) x – 3 –(x3 – 3x2)–(x3 – 3x2) 3x23x2 – 4x Bring down –4x.

224. Example 5 Continued x 3 + 0x 2 – 4x – 7 ) x – 3 3x 3 ÷ x = 3x Multiply x 2 (x – 3). Subtract. x2x2 –(x3 – 3x2)–(x3 – 3x2) 3x23x2 – 4xBring down – 4x. – (3x 2 – 9x) 5x5x – (5x – 15) 8 Bring down – 7. Multiply 3x(x – 3). Subtract. The remainder is 8. + 3x – 7 Multiply 5(x – 3). Subtract. + 5 (x 3 – 4x – 7) ÷ (x – 3) =

225. Recall from Chapter 7 that a polynomial in one variable is written in standard form when the degrees of the terms go from greatest to least. Remember!

226. Divide (1 – 4x 2 + x 3 ) ÷ (x – 2). Check It Out! Example 5a (x 3 – 4x 2 + 1) ÷ (x – 2) x 3 – 4x 2 + 0x + 1x – 2 ) Write in standard format. Write in long division form. Use 0x as a placeholder for the x term. x 3 – 4x 2 + 0x + 1x – 2 ) x2x2 x 3 ÷ x = x 2 – ( – 2x 2 + 4x) – 4x – ( – 4x + 8) –7–7 Bring down 0x. – 2x 2 ÷ x = –2x. Multiply –2x(x – 2). Subtract. Bring down 1. Multiply –4(x – 2). Subtract. – (x 3 – 2x 2 ) – 2x 2 Multiply x 2 (x – 2). Subtract. – 2x + 0x + 1 – 4

227. Divide (1 – 4x 2 + x 3 ) ÷ (x – 2). Check It Out! Example 5a Continued (1 – 4x 2 + x 3 ) ÷ (x – 2) =

228. Divide (4p – 1 + 2p 3 ) ÷ (p + 1). Check It Out! Example 5b (2p 3 + 4p – 1) ÷ (p + 1) 2p 3 + 0p 2 + 4p – 1 p + 1 ) Write in standard format. Write in long division form. Use 0p 2 as a placeholder for the p 2 term. 2p 3 – 0p 2 + 4p – 1 p + 1 ) 2p22p2 p 3 ÷ p = p 2 – ( – 2p 2 – 2p) 6p – (6p + 6) –7–7 Bring down 4p. – 2p 2 ÷ p = –2p. Multiply –2p(p + 1). Subtract. Bring down –1. Multiply 6(p + 1). Subtract. – (2p 3 + 2p 2 ) – 2p 2 Multiply 2p 2 (p + 1). Subtract. – 2p + 4p –1 –1 + 6

229. Check It Out! Example 5b Continued (2p 3 + 4p – 1) ÷ (p + 1) =

230. Lesson Quiz: Part I Add or Subtract. Simplify your answer. 1. 3. 2. (12x 2 – 4x 2 + 20x) ÷ 4x) 3x 2 – x + 5 x – 2 4. x + 3 2x + 3

231. Lesson Quiz: Part II Divide using long division. 5. 6. (8x 2 + 2x 3 + 7)  (x + 3) (x 2 + 4x + 7)  (x + 1)

232. Warm Up 1. Find the LCM of x, 2x 2, and 6. 2. Find the LCM of p 2 – 4p and p 2 – 16. Multiply. Simplify your answer. 3. 4. 5.

233. Solve rational equations. Identify extraneous solutions. New Objectives

234. A rational equation is an equation that contains one or more rational expressions. If a rational equation is a proportion, it can be solved using the Cross Product Property.

235. Example 1: Solving Rational Equations by Using Cross Products Use cross products. 5x = (x – 2)(3) 5x = 3x – 6 2x = – 6 Solve. Check your answer. x = – 3 Check Distribute 3 on the right side. Subtract 3x from both sides. – 1 – 1

236. Check It Out! Example 1a Solve. Check your answer. Use cross products. Distribute 1 on the right side. Subtract n from both sides. 3n = (n + 4)(1) 3n = n + 4 2n = 4 n = 2 Check Divide both sides by 2.

237. Check It Out! Example 1b Solve. Check your answer. 4h = (h + 1)(2) 4h = 2h + 2 2h = 2 h = 1 Use cross products. Distribute 2 on the right side. Subtract 2h from both sides. Check Divide both sides by 2.

238. Check It Out! Example 1c Solve. Check your answer. 21x = (x – 7)(3) 21x = 3x – 21 18x = – 21 x = Use cross products. Distribute 3 on the right side. Subtract 3x from both sides. Check Divide both sides by 18.

239. Some rational equations contain sums or differences of rational expressions. To solve these, you must find the LCD of all the rational expressions in the equation.

240. Example 2A: Solving Rational Equations by Using the LCD Solve each equation. Check your answer. Step 1 Find the LCD 2x(x + 1) Include every factor of the denominator. Step 2 Multiply both sides by the LCD Distribute on the left side.

241. Example 2A Continued Step 3 Simplify and solve. (2x)(2) +6(x +1) = 5(x +1) 4x + 6x + 6 = 5x + 5 10x + 6 = 5x + 5 5x = – 1 Divide out common factors. Simplify. Distribute and multiply. Combine like terms. Subtract 5x and 6 from both sides. Divide both sides by 5.

242. Example 2A Continued Check Verify that your solution is not extraneous.

243. Example 2B: Solving Rational Equations by Using the LCD Solve each equation. Check your answer. Step 1 Find the LCD (x 2 ) Include every factor of the denominator. Step 2 Multiply both sides by the LCD Distribute on the left side.

244. Example 2B Continued Step 3 Simplify and solve. Divide out common factors. 4x – 3 = x 2 – x 2 + 4x – 3 = 0 x 2 – 4x + 3 = 0 (x – 3)(x – 1) = 0 x = 3, 1 Simplify. Subtract x 2 from both sides. Factor. Solve. Multiply by – 1.

245. Example 2B Continued Check Verify that your solution is not extraneous.

246. Check It Out! Example 2a Solve each equation. Check your answer. Step 1 Find the LCD a(a +1) Include every factor of the denominator. Step 2 Multiply both sides by the LCD Distribute on the left side.

247. Check It Out! Example 2a Continued Step 3 Simplify and solve. Divide out common factors. 3a = 4(a + 1) 3a = 4a + 4 – 4 = a Simplify. Distribute the 4. Subtract the 4 and 3a from both sides.

248. Check It Out! Example 2a Continued Check Verify that your solution is not extraneous.

249. Check It Out! Example 2b Solve each equation. Check your answer. Step 1 Find the LCD 2j(j +2) Include every factor of the denominator. Step 2 Multiply both sides by the LCD Distribute on the left side.

250. Check It Out! Example 2b Continued Solve each equation. Check your answer. 12j – 10(2j + 4) = 4j + 8 12j – 20j – 40 = 4j + 8 – 12j = 48 j = – 4 Simplify. Distribute 10. Combine like terms. Divide out common terms.

251. Check It Out! Example 2b Continued Check Verify that your solution is not extraneous.

252. Check It Out! Example 2c Solve each equation. Check your answer. Step 1 Find the LCD t(t +3) Include every factor of the denominator. Step 2 Multiply both sides by the LCD Distribute on the right side.

253. Check It Out! Example 2c Continued Solve each equation. Check your answer. Divide out common terms. 8t = (t + 3) + t(t + 3) 8t = t + 3 + t 2 + 3t 0 = t 2 – 4t + 3 0 = (t – 3)(t – 1) t = 3, 1 Simplify. Distribute t. Combine like terms. Factor.

254. Check It Out! Example 2c Continued Check Verify that your solution is not extraneous.

255. Example 3: Problem-Solving Application Copy machine A can make 200 copies in 60 minutes. Copy machine B can make 200 copies in 10 minutes. How long will it take both machines working together to make 200 copies?

256. List the important information: Machine A can print the copies in 60 minutes, which is of the job in 1 minute. Machine B can print the copies in 10 minutes, which is of the job in 1 minute. 1 Understand the Problem The answer will be the number of minutes m machine A and machine B need to print the copies.

257. The part of the copies that machine A can print plus the part that machine B can print equals the complete job. Machine A ’ s rate times the number of minutes plus machine B ’ s rate times the number of minutes will give the complete time to print the copies. 2 Make a Plan (machine A ’ s rate) m (machine B ’ s rate) m +complet e job = mm + = 1

258. Solve 3 Multiply both sides by the LCD, 60. 1m + 6m = 60 Distribute 60 on the left side. 7m = 60 Combine like terms. Divide both sides by 7. Machine A and Machine B working together can print the copies in a little more than 8.5 minutes.

259. Machine A prints of the copies per minute and machine B prints of the copies per minute. So in minutes, machine A prints of the copies and machine B prints of the copies. Together, they print

260. Check It Out! Example 3 Cindy mows a lawn in 50 minutes. It takes Sara 40 minutes to mow the same lawn. How long will it take them to mow the lawn if they work together? 1 Understand the Problem The answer will be the number of minutes m Sara and Cindy need to mow the lawn.

261. List the important information: Cindy can mow the lawn in 50 minutes, which is of the job in 1 minute. Sara can mow the lawn in 40 minutes, which is of the job in 1 minute. 1 Understand the Problem The answer will be the number of minutes m Sara and Cindy need to mow the lawn.

262. The part of the lawn Cindy can mow plus the part of the lawn Sara can mow equals the complete job. Cindy ’ s rate times the number of minutes plus Sara ’ s rate times the number of minutes will give the complete time to mow the lawn. 2 Make a Plan (Cindy ’ s rate) m (Sara ’ s rate) m + lawn mowed = mm+= 1

263. Solve 3 Multiply both sides by the LCD, 200. Distribute 200 on the left side. Combine like terms. Divide both sides by 9. Cindy and Sara working together can mow the lawn in a little more than 22.2 minutes. 5m + 4m = 200 9m = 200

264. Look Back 4 Cindy mows of the lawn in 1 minute and Sara mows of the lawn in 1 minute. So, in minutes Cindy mows of the lawn and Sara mows of the lawn. Together they mow lawn.

265. When you multiply each side of an equation by the LCD, you may get an extraneous solution. Recall from Chapter 11 that an extraneous solution is a solution to a resulting equation that is not a solution to the original equation.

266. Extraneous solutions may be introduced by squaring both sides of an equation or by multiplying both sides of an equation by a variable expression. Helpful Hint

267. Example 4: Extraneous Solutions Solve. Identify any extraneous solutions. Step 1 Solve. 2(x 2 – 1) = (x + 1)(x – 6) 2x 2 – 2 = x 2 – 5x – 6 x 2 + 5x + 4 = 0 (x + 1)(x + 4) = 0 x = – 1 or x = – 4 Use cross products. Distribute 2 on the left side. Multiply the right side. Subtract x 2 from both sides. Add 5x and 6 to both sides. Factor the quadratic expression. Use the Zero Product Property. Solve.

268. Example 4 Continued Solve. Identify any extraneous solutions. Step 2 Find extraneous solutions.  Because and are undefined –1 is not a solution. The only solution is – 4, so – 1 is an extraneous solution.

269. Check It Out! Example 4a Solve. Identify any extraneous solutions. Step 1 Solve. (x – 2)(x – 7) = 3(x – 7) Use cross products. Distribute 3 on the right side. Multiply the left side. 2x 2 – 9x + 14 = 3x – 21 X 2 – 12x + 35 = 0 Subtract 3x from both sides. Add 21 to both sides. (x – 7)(x – 5) = 0 x = 7 or x = 5 Factor the quadratic expression. Use the Zero Product Property. Solve.

270. Check It Out! Example 4a Continued Step 2 Find extraneous solutions. The only solution is 5, so 7 is an extraneous solution. Because and are undefined 7 is not a solution. 

271. Check It Out! Example 4b Solve. Identify any extraneous solutions. Step 1 Solve. (x + 1)(x – 3) = 4(x – 2) Use cross products. Distribute 4 on the right side. Multiply the left side. x 2 – 2x – 3 = 4x – 8 X 2 – 6x + 5 = 0 Subtract 4x from both sides. Add 8 to both sides. (x – 1)(x – 5) = 0 x = 1 or x = 5 Factor the quadratic expression. Use the Zero Product Property. Solve.

272. Check It Out! Example 4b Continued Step 2 Find extraneous solutions. The solutions are 1 and 5, there are no extraneous solutions. 1 and 5 are solutions.

273. Check It Out! Example 4c Solve. Identify any extraneous solutions. Step 1 Solve. 6(x 2 + 2x) = 9(x 2 ) Use cross products. Distribute 6 on the left side. Multiply the right side. 3x 2 – 12x = 0 3x(x – 4) = 0 3x = 0, or x – 4 = 0 x = 0 or x = 4 Factor the quadratic expression. Use the Zero Product Property. Solve. Subtract 9x 2 from both sides. Multiply through with – 1.

274. Check It Out! Example 4c Continued Step 2 Find extraneous solutions. The only solution is 4, so 0 is an extraneous solutions. Because and are undefined 0 is not a solution. 

275. Lesson Quiz: Part I Solve each equation. Check your answer. 3. 2. 1.24 – 4, 3 4. Pipe A can fill a tank with water in 4 hours. Pipe B can fill the same tank in 5 hours. How long will it take both pipes working together to fill the tank?

276. Lesson Quiz: Part II 5. Solve. Identify any extraneous solutions. – 5; 3 is extraneous.