1. Section 7.1 Introduction to Rational Expressions Copyright © 2013, 2009, and 2005 Pearson Education, Inc.

2. Objectives Basic Concepts Simplifying Rational Expressions Applications

3. Basic Concepts Rational expressions can be written as quotients (fractions) of two polynomials. Examples include:

4. Example If possible, evaluate each expression for the given value of the variable. a.b.c. Solution a.b.c.

5. Example Find all values of the variable for which each expression is undefined. a.b.c. Solution a.b.c. Undefined when x 2 = 0 or when x = 0. Undefined when w – 4 = 0 or when w = 4. Undefined when w 2 – 4 = 0 or when w =  2.

6. Example Simplify each fraction by applying the basic principle of fractions. a. b. c. Solution a. The GCF of 9 and 15 is 3. b. The GCF of 20 and 28 is 4. c. The GCF of 45 and 135 is 45.

8. Example Simplify each expression. a.b.c. Solution a.b.c.

9. Example Simplify each expression. a.b. Solution a.b.

10. Example Suppose that n balls, numbered 1 to n, are placed in a container and two balls have the winning number. a. What is the probability of drawing the winning ball at random? b. Calculate this probability for n = 100, 1000 and 10,000. c. What happens to the probability of drawing the winning ball as the number of balls increases? Solution a. There are 2 chances of drawing the winning ball.

11. Example (cont) b. Calculate this probability for n = 100, 1000 and 10,000. c. What happens to the probability of drawing the winning ball as the number of balls increases? The probability decreases.

12. Section 7.2 Multiplication and Division of Rational Expressions Copyright © 2013, 2009, and 2005 Pearson Education, Inc.

13. Objectives Review of Multiplication and Division of Fractions Multiplication of Rational Expressions Division of Rational Expressions

14. Example Multiply and simplify your answers to lowest terms. a.b.c. Solution a. b. c.

15. Example Divide and simplify your answers to lowest terms. a.b.c. Solution a.b. c.

17. Example Multiply and simplify to lowest terms. Leave your answers in factored form. a.b. Solution a.b.

18. Example Multiply and simplify to lowest terms. Leave your answer in factored form. Solution

20. Example Divide and simplify to lowest terms. a.b. Solution a.b.

21. Section 7.3 Addition and Subtraction with Like Denominators Copyright © 2013, 2009, and 2005 Pearson Education, Inc.

22. Objectives Review of Addition and Subtraction of Fractions Rational Expressions Having Like Denominators

23. Example Simplify each expression to lowest terms. a.b. Solution a. b.

24. Example Simplify each expression to lowest terms. a.b. Solution a. b.

25. To add two rational expressions having like denominators, add their numerators. Keep the same denominator. C is nonzero SUMS OF RATIONAL EXPRESSIONS

26. Example Add and simplify to lowest terms. a.b. Solution a. b.

27. Example Add and simplify to lowest terms. a.b. Solution a. b.

28. To subtract two rational expressions having like denominators, subtract their numerators. Keep the same denominator. C is nonzero DIFFERENCES OF RATIONAL EXPRESSIONS

29. Example Subtract and simplify to lowest terms. a.b. Solution a. b.

30. Example Subtract and simplify to lowest terms. Solution

31. Section 7.4 Addition and Subtraction with Unlike Denominators Copyright © 2013, 2009, and 2005 Pearson Education, Inc.

32. Objectives Finding Least Common Multiples Review of Fractions Having Unlike Denominators Rational Expressions Having Unlike Denominators

33. The least common multiple (LCM) of two or more polynomials can be found as follows. Step 1: Factor each polynomial completely. Step 2: List each factor the greatest number of times that it occurs in either factorization. Step 3: Find the product of this list of factors. The result is the LCM. FINDING THE LEAST COMMON MULTIPLE

34. Example Find the least common multiple of each pair of expressions. a. 6x, 9x 4 b. x 2 + 7x + 12, x 2 + 8x + 16 Solution Step 1: Factor each polynomial completely. 6x = 3 ∙ 2 ∙ x9x 4 = 3 ∙ 3 ∙ x ∙ x ∙ x ∙ x Step 2: List each factor the greatest number of times. 3 ∙ 3 ∙ 2 ∙ x ∙ x ∙ x ∙ x Step 3: The LCM is 18x 4.

35. Example (cont) b. x 2 + 7x + 12, x 2 + 8x + 16 Step 1: Factor each polynomial completely. x 2 + 7x + 12 = (x + 3)(x + 4) x 2 + 8x + 16 = (x+ 4)(x + 4) Step 2: List each factor the greatest number of times. (x + 3), (x + 4), and (x + 4) Step 3: The LCM is (x + 3)(x + 4) 2.

36. Example Simplify each expression. a.b. Solution a. The LCD is the LCM, 42. b. The LCD is 60.

37. Example Find each sum and leave your answer in factored form. a.b. Solution a. The LCD is x 2. b.

38. Example Simply the expression. Write your answer in lowest terms and leave it in factored form. Solution The LCD is x(x + 7).

39. Example Simplify the expression. Write your answer in lowest terms and leave it in factored form. Solution

40. Example Add and then find the reciprocal of the result. Solution The LCD is RS. The reciprocal is

41. Section 7.5 Complex Fractions Copyright © 2013, 2009, and 2005 Pearson Education, Inc.

42. Objectives Basic Concepts Simplifying Complex Fractions

43. Basic Concepts A complex fraction is a rational expression that contains fractions in its numerator, denominator, or both. Complex fractions involve division of fractions. When dividing two fractions, we can multiply the numerator by the reciprocal of the denominator.

44. For any real numbers a, b, c, and d, where b, c, and d are nonzero. SIMPLIFYING BASIC COMPLEX FRACTIONS

45. Simplifying Complex Fractions (Method 1) To simplify a complex fraction, perform the following steps. STEP 1: Write the numerator as a single fraction; write the denominator as a single fraction. STEP 2: Divide the denominator into the numerator by multiplying the numerator and the reciprocal of the denominator. STEP 3: Simplify the result to lowest terms.

46. Example Simplify each complex fraction. a.b. Solution a. b.

47. Example Simplify. Write your answer in lowest terms. Solution

48. Example Simplify. Write your answer in lowest terms. For the numerator, the LCD is (x + 2)(x – 2). For the denominator, the LCD is (x + 4)(x – 4).

49. Simplifying Complex Fractions (Method II) To simplify a complex fraction, perform the following steps. STEP 1: Find the LCD of all fractions within the complex fraction. STEP 2: Multiply the numerator and the denominator of the complex fraction by the LCD. STEP 3: Simplify the result to lowest terms.

50. Example Simplify. Solution

51. Example Simplify. Solution

52. Example Simplify. Solution

53. Section 7.6 Rational Equations and Formulas Copyright © 2013, 2009, and 2005 Pearson Education, Inc.

54. Objectives Solving Rational Equations Rational Expressions and Equations Graphical and Numerical Solutions Solving a Formula for a Variable Applications

55. Rational Equations If an equation contains one or more rational expressions, it is called a rational equation.

56. Example Solve each equation. a.b. Solution a.b. The solutions are The solution is

58. Example Determine whether you are given an expression or an equation. If it is an expression, simplify it and then evaluate it for x = 4. If it is an equation, solve it. a.b. Solution a. There is an equal sign, so it is an equation. The answer checks. The solution is −4.

59. Example (cont) b. There is no equals sign, so it is an expression. The common denominator is x – 2, so we can add the numerators. When x = 4, the expression evaluates 4 + 5 = 9.

60. Example Solve graphically and numerically. Solution Graph and The solutions are −2 and 1. x−3−3−2−2−1−10123 −1−2−−21 (−2, −2) (1, 1)

61. Example (cont) Solve graphically and numerically. Solution Numerical Solution The solutions are −2 and 1. x−3−3−2−2−1−10123 −1−2−−21 −3−3−2−2−1−10123

62. Example Solve the equation for the specified variable. Solution

63. Example Solve the equation for the specified variable. Solution

64. Example Solve the equation for the specified variable. Solution

65. Example A pump can fill a swimming pool ¾ full in 6 hours, another can fill the pool ¾ full in 9 hours. How long would it take the pumps to fill the pool ¾ full, working together? Solution The two pumps can fill the pool ¾ full in hours.

66. Section 7.7 Proportions and Variation Copyright © 2013, 2009, and 2005 Pearson Education, Inc.

67. Objectives Proportions Direct Variation Inverse Variation Analyzing Data Joint Variation

68. Proportions A proportion is a statement (equation) that two ratios (fractions) are equal. The following property is a convenient way to solve proportions: is equivalent to provided b ≠ 0 and d ≠ 0.

69. Example On an elliptical machine, Francis can burn 370 calories in 25 minutes. If he increases his work time to 30 minutes, how many calories will he burn? Solution Let x be the equivalent amount of calories. Thus, in 30 minutes, Francis will burn 444 calories.

70. Example A 6-foot tall person casts a shadow that is 8-foot long. If a nearby tree casts a 32-foot long shadow, estimate the height of the tree. Solution The triangles are similar because the measures of its corresponding angles are equal. Therefore corresponding sides are proportional. 6 fth 8 ft 32 ft The tree is 24 feet tall.

73. Example Let y be directly proportional to x, or vary directly with x. Suppose y = 9 when x = 6. Find y when x = 13. Solution Step 1 The general equation is y = kx. Step 2 Substitute 9 for y and 6 for x in y = kx. Solve for k. Step 3 Replace k with 9/6 in the equation y = 9x/6. Step 4 To find y, let x = 13.

74. Example The table lists the amount of pay for various hours worked. a. Find the constant of proportionality. b. Predict the pay for 19 hours of work. HoursPay 6$138 11$253 15$345 23$529 31$713

75. Example (cont) The slope of the line equals the proportionality, k. If we use the first and last data points (6, 138) and (31, 713), the slope is The amount of pay per hour is $23. The graph of the line y = 23x, models the given graph. To find the pay for 19 hours, substitute 19 for x. y = 23x, y = 23(19) y = 437 19 hours of work would pay $437.00

77. Example Let y be inversely proportional to x, or vary inversely with x. Suppose y = 6 when x = 4. Find y when x = 8. Solution Step 1 The general equation is y = k/x. Step 2 Substitute 6 for y and 4 for x in Solve for k. Step 3 Replace k with 24 in the equation y = k/x. Step 4 To find y, let x = 8.

78. Example Determine whether the data in each table represent direct variation, inverse variation, or neither. For direct and inverse variation, find the equation. a. b. c. x37912 y 283248 x5101215 y12654 x8111421 y486684126 Neither the product xy nor the ratio y/x are constant in the data in the table. Therefore there is neither direct variation nor indirect variation in this table. As x increases, y decreases. Because xy = 60 for each data point, the equation y = 60/x models the data. This represents an inverse variation. The equation y = 6x models the data. The data represents direct variation.

79. Let x, y, and z denote three quantities. Then z varies jointly with x and y if there is a nonzero number k such that JOINT VARIATION

80. Example The strength S of a rectangular beam varies jointly as its width w and the square of its thickness t. If a beam 5 inches wide and 2 inches thick supports 280 pounds, how much can a similar beam 4 inches wide and 3 inches thick support? Solution The strength of the beam is modeled by S = kwt 2.

81. Example (cont) Thus S = 14wt 2 models the strength of this type of beam. When w = 4 and t = 3, the beam can support S = 14 ∙ 4 ∙ 3 2 = 504 pounds