2. Chapter 6 Rational Expressions

3. § 6.1 Rational Functions and Multiplying and Dividing Rational Expressions

4. Martin-Gay, Intermediate Algebra, 5ed 33 Rational Expressions Rational expressions can be written in the form where P and Q are both polynomials and Q  0. Examples of Rational Expressions

5. Martin-Gay, Intermediate Algebra, 5ed 44 To evaluate a rational expression for a particular value(s), substitute the replacement value(s) into the rational expression and simplify the result. Evaluating Rational Expressions Example: Evaluate the following expression for y =  2.

6. Martin-Gay, Intermediate Algebra, 5ed 55 In the previous example, what would happen if we tried to evaluate the rational expression for y = 5? This expression is undefined! Evaluating Rational Expressions

7. Martin-Gay, Intermediate Algebra, 5ed 66 We have to be able to determine when a rational expression is undefined. A rational expression is undefined when the denominator is equal to zero. The numerator being equal to zero is okay (the rational expression simply equals zero). Undefined Rational Expressions

8. Martin-Gay, Intermediate Algebra, 5ed 77 Find any real numbers that make the following rational expression undefined. The expression is undefined when 15x + 45 = 0. So the expression is undefined when x =  3. Undefined Rational Expressions Example:

9. Martin-Gay, Intermediate Algebra, 5ed 88 Simplifying a rational expression means writing it in lowest terms or simplest form. Simplifying Rational Expressions Fundamental Principle of Rational Expressions For any rational expression and any polynomial R, where R ≠ 0, or, simply,

10. Martin-Gay, Intermediate Algebra, 5ed 99 Simplifying a Rational Expression 1) Completely factor the numerator and denominator of the rational expression. 2) Divide out factors common to the numerator and denominator. (This is the same thing as “removing the factor of 1.”) Warning! Only common FACTORS can be eliminated from the numerator and denominator. Make sure any expression you eliminate is a factor. Simplifying Rational Expressions

11. Martin-Gay, Intermediate Algebra, 5ed 10 Simplify the following expression. Simplifying Rational Expressions Example:

12. Martin-Gay, Intermediate Algebra, 5ed 11 Simplify the following expression. Simplifying Rational Expressions Example:

13. Martin-Gay, Intermediate Algebra, 5ed 12 Simplify the following expression. Simplifying Rational Expressions Example:

14. Martin-Gay, Intermediate Algebra, 5ed 13 Multiplying Rational Expressions The rule for multiplying rational expressions is Multiplying Rational Expressions as long as Q  0 and S  0. 1) Completely factor each numerator and denominator. 2) Use the rule above and multiply the numerators and denominators. 3) Simplify the product by dividing the numerator and denominator by their common factors.

15. Martin-Gay, Intermediate Algebra, 5ed 14 Multiply the following rational expressions. Multiplying Rational Expressions Example:

16. Martin-Gay, Intermediate Algebra, 5ed 15 Multiply the following rational expressions. Multiplying Rational Expressions Example:

17. Martin-Gay, Intermediate Algebra, 5ed 16 Dividing Rational Expressions The rule for dividing rational expressions is Dividing Rational Expressions as long as Q  0 and S  0 and R  0. To divide by a rational expression, use the rule above and multiply by its reciprocal. Then simplify if possible.

18. Martin-Gay, Intermediate Algebra, 5ed 17 Divide the following rational expression. Dividing Rational Expressions Example:

19. Martin-Gay, Intermediate Algebra, 5ed 18 § 6.2 Adding and Subtracting Rational Expressions

20. Martin-Gay, Intermediate Algebra, 5ed 19 Adding and Subtracting with Common Denominators The rule for multiplying rational expressions is Adding or Subtracting Rational Expressions with Common Denominators are rational expression, then

21. Martin-Gay, Intermediate Algebra, 5ed 20 Add the following rational expressions. Adding Rational Expressions Example:

22. Martin-Gay, Intermediate Algebra, 5ed 21 Subtract the following rational expressions. Subtracting Rational Expressions Example:

23. Martin-Gay, Intermediate Algebra, 5ed 22 Subtract the following rational expressions. Subtracting Rational Expressions Example:

24. Martin-Gay, Intermediate Algebra, 5ed 23 To add or subtract rational expressions with unlike denominators, you have to change them to equivalent forms that have the same denominator (a common denominator). This involves finding the least common denominator of the two original rational expressions. Least Common Denominators

25. Martin-Gay, Intermediate Algebra, 5ed 24 Finding the Least Common Denominator 1) Factor each denominator completely, 2) The LCD is the product of all unique factors each raised to a power equal to the greatest number of times that the factor appears in any factored denominator. Least Common Denominators

26. Martin-Gay, Intermediate Algebra, 5ed 25 Find the LCD of the following rational expressions. Least Common Denominators Example:

27. Martin-Gay, Intermediate Algebra, 5ed 26 Find the LCD of the following rational expressions. Least Common Denominators Example:

28. Martin-Gay, Intermediate Algebra, 5ed 27 Find the LCD of the following rational expressions. Least Common Denominators Example:

29. Martin-Gay, Intermediate Algebra, 5ed 28 Find the LCD of the following rational expressions. Both of the denominators are already factored. Since each is the opposite of the other, you can use either x – 3 or 3 – x as the LCD. Least Common Denominators Example:

30. Martin-Gay, Intermediate Algebra, 5ed 29 To change rational expressions into equivalent forms, we use the principal that multiplying by 1 (or any form of 1), will give you an equivalent expression. Multiplying by 1

31. Martin-Gay, Intermediate Algebra, 5ed 30 Rewrite the rational expression as an equivalent rational expression with the given denominator. Equivalent Expressions Example:

32. Martin-Gay, Intermediate Algebra, 5ed 31 As stated in the previous section, to add or subtract rational expressions with different denominators, we have to change them to equivalent forms first. Unlike Denominators

33. Martin-Gay, Intermediate Algebra, 5ed 32 Adding or Subtracting Rational Expressions with Unlike Denominators 1)Find the LCD of the rational expressions. 2)Write each rational expression as an equivalent rational expression whose denominator is the LCD found in Step 1. 3)Add or subtract numerators, and write the result over the denominator. 4)Simplify resulting rational expression, if possible. Unlike Denominators

34. Martin-Gay, Intermediate Algebra, 5ed 33 Add the following rational expressions. Adding with Unlike Denominators Example:

35. Martin-Gay, Intermediate Algebra, 5ed 34 Subtract the following rational expressions. Subtracting with Unlike Denominators Example:

36. Martin-Gay, Intermediate Algebra, 5ed 35 Subtract the following rational expressions. Subtracting with Unlike Denominators Example:

37. Martin-Gay, Intermediate Algebra, 5ed 36 Add the following rational expressions. Adding with Unlike Denominators Example:

38. Martin-Gay, Intermediate Algebra, 5ed 37 § 6.3 Simplifying Complex Fractions

39. Martin-Gay, Intermediate Algebra, 5ed 38 Complex Rational Fractions Complex rational expressions (complex fraction) are rational expressions whose numerator, denominator, or both contain one or more rational expressions. There are two methods that can be used when simplifying complex fractions.

40. Martin-Gay, Intermediate Algebra, 5ed 39 Simplifying a Complex Fraction (Method 1) 1)Simplify the numerator and the denominator of the complex fraction so that each is a single fraction. 2)Perform the indicated division by multiplying the numerator of the complex fraction by the reciprocal of the denominator of the complex fraction. 3)Simplify, if possible. Simplifying Complex Fractions

41. Martin-Gay, Intermediate Algebra, 5ed 40 Simplifying Complex Fractions Example:

42. Martin-Gay, Intermediate Algebra, 5ed 41 Simplifying a Complex Fraction (Method 2) 1)Multiply the numerator and the denominator of the complex fraction by the LCD of the fractions in both the numerator and denominator. 2)Simplify, if possible. Simplifying Complex Fractions

43. Martin-Gay, Intermediate Algebra, 5ed 42 Simplifying Complex Fractions Example:

44. Martin-Gay, Intermediate Algebra, 5ed 43 When you have a rational expression where some of the variables have negative exponents, rewrite the expression using positive exponents. The resulting expression is a complex fraction, so use either method to simplify the expression. Simplifying with Negative Exponents

45. Martin-Gay, Intermediate Algebra, 5ed 44 Simplifying with Negative Exponents Example:

46. Martin-Gay, Intermediate Algebra, 5ed 45 § 6.4 Dividing Polynomials: Long Division and Synthetic Division

47. Martin-Gay, Intermediate Algebra, 5ed 46 Dividing a Polynomial by a Monomial Divide each term in the polynomial by the monomial. Example: Divide

48. Martin-Gay, Intermediate Algebra, 5ed 47 Dividing a Polynomial by a Monomial Example: Divide

49. Martin-Gay, Intermediate Algebra, 5ed 48 Dividing a polynomial by a polynomial other than a monomial uses a “long division” technique that is similar to the process known as long division in dividing two numbers, which is reviewed on the next slide. Dividing Polynomials

50. Martin-Gay, Intermediate Algebra, 5ed 49 Divide 43 into 72. Multiply 1 times 43. Subtract 43 from 72. Bring down 5. Divide 43 into 295. Multiply 6 times 43. Subtract 258 from 295. Bring down 6. Divide 43 into 376. Multiply 8 times 43. Subtract 344 from 376. Nothing to bring down. We then write our result as Dividing Polynomials

51. Martin-Gay, Intermediate Algebra, 5ed 50 As you can see from the previous example, there is a pattern in the long division technique. Divide Multiply Subtract Bring down Then repeat these steps until you can’t bring down or divide any longer. We will incorporate this same repeated technique with dividing polynomials. Dividing Polynomials

52. Martin-Gay, Intermediate Algebra, 5ed 51 35  x Divide 7x into 28x 2. Multiply 4x times 7x+3. Subtract 28x 2 + 12x from 28x 2 – 23x. Bring down – 15. Divide 7x into –35x. Multiply – 5 times 7x+3. Subtract –35x–15 from –35x–15. Nothing to bring down. 15  So our answer is 4x – 5. Dividing Polynomials

53. Martin-Gay, Intermediate Algebra, 5ed 52 86472 2  xxx x2 x x 14 4 2  20  x 10  7020  x 78 Divide 2x into 4x 2. Multiply 2x times 2x+7. Subtract 4x 2 + 14x from 4x 2 – 6x. Bring down 8. Divide 2x into –20x. Multiply -10 times 2x+7. Subtract –20x–70 from –20x+8. Nothing to bring down. 8   )72( 78  x x2 10  We write our final answer as Dividing Polynomials

54. Martin-Gay, Intermediate Algebra, 5ed 53 Dividing Polynomials Using Long Division Example: Divide: 1. Divide the leading term of the dividend, c 2, by the first term of the divisor, x. 3. Subtract c 2 + c from c 2 + 3c – 2. 2. Multiply c by c + 1. Continued.

55. Martin-Gay, Intermediate Algebra, 5ed 54 Dividing Polynomials Using Long Division Example continued: Bring down the next term to obtain a new polynomial. 4. Repeat the process until the degree of the remainder is less than the degree of the binomial divisor. Remainder 5. Check by verifying that (Quotient)(Divisor) + Remainder = Dividend. 2c

56. Martin-Gay, Intermediate Algebra, 5ed 55 Dividing Polynomials Using Long Division Example: Divide: (y 2 – 5y + 6) ÷ (y – 2) y y 2 – 2y – 3y + 6 – 3 – 3y + 6 0 No remainder (y 2 – 5y + 6) ÷ (y – 2) = y – 3 Check : (y – 2)(y – 3) = y 2 – 5y + 6

57. Martin-Gay, Intermediate Algebra, 5ed 56 Synthetic Division To find the quotient and remainder when a polynomial of degree 1 or higher is divided by x – c, a shortened version of long division called synthetic division may be used. Long DivisionSynthetic Division Subtraction signs are not necessary. The original terms in red are not needed because they are the same as the term directly above. Continued.

58. Martin-Gay, Intermediate Algebra, 5ed 57 Synthetic Division Continued. The variables are not necessary. The “boxed” numbers can be aligned horizontally. Continued.

59. Martin-Gay, Intermediate Algebra, 5ed 58 Synthetic Division The leading coefficient of the dividend can be brought down. The first two numbers in the last row are the coefficients of the quotient, the last number is the remainder. To simplify further, the top row can be removed and instead of subtracting, we can change the sign of each entry and add. The x + 1 is replaced with a – 1. Quotient Remainder Continued.

60. Martin-Gay, Intermediate Algebra, 5ed 59 § 6.5 Solving Equations Containing Rational Expressions

61. Martin-Gay, Intermediate Algebra, 5ed 60 Solving Equations First note that an equation contains an equal sign and an expression does not. To solve EQUATIONS containing rational expressions, clear the fractions by multiplying both sides of the equation by the LCD of all the fractions. Then solve as in previous sections. Note: this works for equations only, not simplifying expressions.

62. Martin-Gay, Intermediate Algebra, 5ed 61 true Solve the following rational equation. Check in the original equation. Solving Equations Example:

63. Martin-Gay, Intermediate Algebra, 5ed 62 Solve the following rational equation. Solving Equations Continued. Example:

64. Martin-Gay, Intermediate Algebra, 5ed 63 true Substitute the value for x into the original equation, to check the solution. So the solution is Solving Equations Example continued:

65. Martin-Gay, Intermediate Algebra, 5ed 64 Solve the following rational equation. Solving Equations Continued. Example:

66. Martin-Gay, Intermediate Algebra, 5ed 65 Substitute the value for x into the original equation, to check the solution. Solving Equations true So the solution is Example continued:

67. Martin-Gay, Intermediate Algebra, 5ed 66 Solve the following rational equation. Solving Equations Continued. Example:

68. Martin-Gay, Intermediate Algebra, 5ed 67 Substitute the value for x into the original equation, to check the solution. Solving Equations true So the solution is x = 3. Example continued:

69. Martin-Gay, Intermediate Algebra, 5ed 68 Solve the following rational equation. Solving Equations Continued. Example:

70. Martin-Gay, Intermediate Algebra, 5ed 69 Substitute the value for x into the original equation, to check the solution. Solving Equations Since substituting the suggested value of a into the equation produced undefined expressions, the solution is . Example continued:

71. Martin-Gay, Intermediate Algebra, 5ed 70 § 6.6 Rational Equations and Problem Solving

72. Martin-Gay, Intermediate Algebra, 5ed 71 Solving Equations for a Specified Variable 1)Clear the equation of fractions or rational expressions by multiplying each side of the equation by the LCD of the denominators in the equation. 2)Use the distributive property to remove grouping symbols such as parentheses. 3)Combine like terms on each side of the equation. 4)Use the addition property of equality to rewrite the equation as an equivalent equation with terms containing the specified variable on one side and all other terms on the other side. 5)Use the distributive property and the multiplication property of equality to get the specified variable alone. Solving Equations with Multiple Variables

73. Martin-Gay, Intermediate Algebra, 5ed 72 Solve the following equation for R 1 Solving Equations with Multiple Variables Example:

74. Martin-Gay, Intermediate Algebra, 5ed 73 Finding an Unknown Number Continued The quotient of a number and 9 times its reciprocal is 1. Find the number. Read and reread the problem. If we let n = the number, then = the reciprocal of the number 1.) Understand Example:

75. Martin-Gay, Intermediate Algebra, 5ed 74 Continued Finding an Unknown Number 2.) Translate The quotient of  a number n and 9 times its reciprocalis = 1 1 Example continued:

76. Martin-Gay, Intermediate Algebra, 5ed 75 3.) Solve Continued Finding an Unknown Number Example continued:

77. Martin-Gay, Intermediate Algebra, 5ed 76 4.) Interpret Finding an Unknown Number Check: We substitute the values we found from the equation back into the problem. Note that nothing in the problem indicates that we are restricted to positive values. State: The missing number is 3 or –3. true Example continued:

78. Martin-Gay, Intermediate Algebra, 5ed 77 Ratios and Rates Ratio is the quotient of two numbers or two quantities. The units associated with the ratio are important. The units should match. If the units do not match, it is called a rate, rather than a ratio. The ratio of the numbers a and b can also be written as a:b, or.

79. Martin-Gay, Intermediate Algebra, 5ed 78 Proportion is two ratios (or rates) that are equal to each other. We can rewrite the proportion by multiplying by the LCD, bd. This simplifies the proportion to ad = bc. This is commonly referred to as the cross product. Proportions

80. Martin-Gay, Intermediate Algebra, 5ed 79 Solve the proportion for x. Solving Proportions Continued. Example:

81. Martin-Gay, Intermediate Algebra, 5ed 80 true Substitute the value for x into the original equation, to check the solution. So the solution is Solving Proportions Example continued:

82. Martin-Gay, Intermediate Algebra, 5ed 81 If a 170-pound person weighs approximately 65 pounds on Mars, how much does a 9000-pound satellite weigh? Solving Proportions Example:

83. Martin-Gay, Intermediate Algebra, 5ed 82 Given the following prices charged for various sizes of picante sauce, find the best buy. 10 ounces for $0.99 16 ounces for $1.69 30 ounces for $3.29 Solving Proportions Continued. Example:

84. Martin-Gay, Intermediate Algebra, 5ed 83 Size Price Unit Price 10 ounces $0.99 $0.99/10 = $0.099 16 ounces $1.69 $1.69/16 = $0.105625 30 ounces $3.29 $3.29/30  $0.10967 The 10 ounce size has the lower unit price, so it is the best buy. Solving Proportions Example continued:

85. Martin-Gay, Intermediate Algebra, 5ed 84 Solving a Work Problem Continued An experienced roofer can roof a house in 26 hours. A beginner needs 39 hours to do the same job. How long will it take if the two roofers work together? Read and reread the problem. By using the times for each roofer to complete the job alone, we can figure out their corresponding work rates in portion of the job done per hour. 1.) Understand Experienced roofer 26 1/26 Beginner roofer 39 /39 Together t 1/t Time in hrsPortion job/hr Example:

86. Martin-Gay, Intermediate Algebra, 5ed 85 Continued Solving a Work Problem 2.) Translate Since the rate of the two roofers working together would be equal to the sum of the rates of the two roofers working independently, Example continued:

87. Martin-Gay, Intermediate Algebra, 5ed 86 3.) Solve Continued Solving a Work Problem Example continued:

88. Martin-Gay, Intermediate Algebra, 5ed 87 4.) Interpret Solving a Work Problem Check: We substitute the value we found from the proportion calculation back into the problem. State: The roofers would take 15.6 hours working together to finish the job. true Example continued:

89. Martin-Gay, Intermediate Algebra, 5ed 88 Solving a Rate Problem Continued The speed of Lazy River’s current is 5 mph. A boat travels 20 miles downstream in the same time as traveling 10 miles upstream. Find the speed of the boat in still water. Read and reread the problem. By using the formula d=rt, we can rewrite the formula to find that t = d/r. We note that the rate of the boat downstream would be the rate in still water + the water current and the rate of the boat upstream would be the rate in still water – the water current. 1.) Understand Down 20 r + 5 20/(r + 5) Up 10 r – 5 10/(r – 5) Distance rate time = d/r Example:

90. Martin-Gay, Intermediate Algebra, 5ed 89 Continued Solving a Rate Problem 2.) Translate Since the problem states that the time to travel downstairs was the same as the time to travel upstairs, we get the equation Example continued:

91. Martin-Gay, Intermediate Algebra, 5ed 90 3.) Solve Continued Solving a Rate Problem Example continued:

92. Martin-Gay, Intermediate Algebra, 5ed 91 4.) Interpret Solving a Rate Problem Check: We substitute the value we found from the proportion calculation back into the problem. true State: The speed of the boat in still water is 15 mph. Example continued:

93. Martin-Gay, Intermediate Algebra, 5ed 92 § 6.7 Variation and Problem Solving

94. Martin-Gay, Intermediate Algebra, 5ed 93 Direct Variation y varies directly as x, or y is directly proportional to x, if there is a nonzero constant k such that y = kx. The family of equations of the form y = kx are referred to as direct variation equations. The number k is called the constant of variation or the constant of proportionality.

95. Martin-Gay, Intermediate Algebra, 5ed 94 If y varies directly as x, find the constant of variation k and the direct variation equation, given that y = 5 when x = 30. y = kx 5 = k·30 k = 1/6 y = 1616 xSo the direct variation equation is Direct Variation

96. Martin-Gay, Intermediate Algebra, 5ed 95 If y varies directly as x, and y = 48 when x = 6, then find y when x = 15. y = kx 48 = k·6 8 = k So the equation is y = 8x. y = 8·15 y = 120 Direct Variation Example:

97. Martin-Gay, Intermediate Algebra, 5ed 96 At sea, the distance to the horizon is directly proportional to the square root of the elevation of the observer. If a person who is 36 feet above water can see 7.4 miles, find how far a person 64 feet above the water can see. Round your answer to two decimal places. Direct Variation Continued. Example:

98. Martin-Gay, Intermediate Algebra, 5ed 97 So our equation is We substitute our given value for the elevation into the equation. Direct Variation Example continued:

99. Martin-Gay, Intermediate Algebra, 5ed 98 y varies inversely as x, or y is inversely proportional to x, if there is a nonzero constant k such that y = k/x. The family of equations of the form y = k/x are referred to as inverse variation equations. The number k is still called the constant of variation or the constant of proportionality. Inverse Variation

100. Martin-Gay, Intermediate Algebra, 5ed 99 If y varies inversely as x, find the constant of variation k and the inverse variation equation, given that y = 63 when x = 3. y = k/x 63 = k/3 k = 63·3 k = 189 y = 189 x So the inverse variation equation is Inverse Variation Example:

101. Martin-Gay, Intermediate Algebra, 5ed 100 y can vary directly or inversely as powers of x, as well. y varies directly as a power of x if there is a nonzero constant k and a natural number n such that y = kx n y varies inversely as a power of x if there is a nonzero constant k and a natural number n such that Powers of x

102. Martin-Gay, Intermediate Algebra, 5ed 101 The maximum weight that a circular column can hold is inversely proportional to the square of its height. If an 8-foot column can hold 2 tons, find how much weight a 10-foot column can hold. Continued. Powers of x Example:

103. Martin-Gay, Intermediate Algebra, 5ed 102 So our equation is We substitute our given value for the height of the column into the equation. Powers of x Example continued:

104. Martin-Gay, Intermediate Algebra, 5ed 103 y varies jointly as, or is jointly proportional to two (or more) variables x and z, if there is a nonzero constant k such that y = kxz. k is STILL the constant of variation or the constant of proportionality. We can also use combinations of direct, inverse, and joint variation. These variations are referred to as combined variations. Combined Variation

105. Martin-Gay, Intermediate Algebra, 5ed 104 Write the following statements as equations. a varies jointly as b and c a = kbc P varies jointly as R and the square of S P = kRS 2 The weight (w) varies jointly with the width (d) and the square of the height (h) and inversely with the length (l) Combined Variation Example: