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Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Rational Expressions and Equations CHAPTER 7.1Simplifying Rational Expressions 7.2Multiplying and Dividing Rational Expressions 7.3Adding and Subtracting Rational Expressions with the Same Denominator 7.4Adding and Subtracting Rational Expressions with Different Denominators 7.5Complex Rational Expressions 7.6Solving Equations Containing Rational Expressions 7.7Applications with Rational Expressions 7

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Simplifying Rational Expressions Evaluate rational expressions. 2.Find numbers that cause a rational expression to be undefined. 3.Simplify rational expressions containing only monomials. 4.Simplify rational expressions containing multiterm polynomials.

Slide 7- 4 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 1 Evaluate rational expressions.

Slide 7- 5 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Rational expression: An expression that can be written in the form, where P and Q are polynomials and Q 0. Some rational expressions are

Slide 7- 6 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Evaluate the expression when a. x = 2 b. x = –1 Solution a. b. which is undefined

Slide 7- 7 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 2 Find numbers that cause a rational expression to be undefined.

Slide 7- 8 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Finding Values That Make a Rational Expression Undefined To determine the value(s) that make a rational expression undefined: 1. Write an equation that has the denominator set equal to zero. 2. Solve the equation.

Slide 7- 9 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Find every value for the variable that makes the expression undefined. Solution The original expression is undefined if y is replaced by 0, –4, or –1. Set the denominator equal to 0. Factor out the monomial GCF, y. Use the zero factor theorem.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 3 Simplify rational expressions containing only monomials.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Simplifying Rational Expressions To simplify a rational expression to lowest terms: 1. Write the numerator and denominator in factored form. 2. Divide out all the common factors in the numerator and denominator. 3.Multiply the remaining factors in the numerator and the remaining factors in the denominator.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Simplify. a. b. Solution a. b. Write the numerator and denominator in factored form, then eliminate the common factors. Multiply the remaining factors.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 4 Simplify rational expressions containing multiterm polynomials.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Simplify. Solution Write the numerator and denominator in factored form, then divide out the common factor, which is x + 2.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Simplify. Solution Factor out the GCF. Factor the polynomial factors. Divide out common factors.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Simplify. a) 5 b) c) d)

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Simplify. a) 5 b) c) d)

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Simplify. a) –10x b) c) d)

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Simplify. a) –10x b) c) d)

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Multiplying and Dividing Rational Expressions Multiply rational expressions. 2.Divide rational expressions. 3.Convert units of measurement using dimensional analysis.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 1 Multiply rational expressions.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Multiplying Rational Expressions To multiply rational expressions: 1. Write each numerator and denominator in factored form. 2.Divide out any numerator factor with any matching denominator factor. 3. Multiply numerator by numerator and denominator by denominator. 4. Simplify as needed.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Multiply. Solution

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Multiply. Solution Write numerators and denominators in factored form. Multiply the remaining numerator factors and denominator factors.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 2 Divide rational expressions.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Dividing Rational Expressions To divide rational expressions: 1. Write an equivalent multiplication statement with the reciprocal of the divisor. 2.Write each numerator and denominator in factored form. (Steps 1 and 2 are interchangeable.) 3. Divide out any numerator factor with any matching denominator factor. 4. Multiply numerator by numerator and denominator by denominator. 5.Simplify as needed.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Divide. Solution Write an equivalent multiplication statement. Divide out common factors, and multiply remaining factors.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 3 Convert units of measurement using dimensional analysis.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Using Dimensional Analysis to Convert Between Units of Measurement To convert units using dimensional analysis, multiply the given measurement by conversion factors so that the undesired units divide out, leaving the desired units.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Convert 17 yards to inches. Solution 17 yards

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Multiply. a) b) c) d)

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Multiply. a) b) c) d)

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Divide. a) b) c) d)

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Divide. a) b) c) d)

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Adding and Subtracting Rational Expressions with the Same Denominator Add or subtract rational expressions with the same denominator.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 1 Add or subtract rational expressions with the same denominator.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Adding or Subtracting Rational Expressions (Same Denominator) To add or subtract rational expressions that have the same denominator: 1. Add or subtract the numerators and keep the same denominator. 2. Simplify to lowest terms (remember to write the numerators and denominators in factored form in order to simplify).

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Add. Solution Since the rational expressions have the same denominator, we add numerators and keep the same denominator. Factor. Divide out the common factor, 2.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Subtract. Solution Divide out the common factor, x – 4. Note: The numerator can be factored, so we may be able to simplify.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Add. Solution Combine like terms in the numerator. Factor the numerator and the denominator. Divide out the common factor, b.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Add. Solution Combine like terms in the numerator. Factor the numerator and the denominator.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Divide out the common factors, 2 and x + 2.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Subtract. Solution Note: To write an equivalent addition, change the operation symbol from a minus sign to a plus sign and change all the signs in the subtrahend (second) polynomial.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Add. a) b) c) d)

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Add. a) b) c) d)

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Subtract. a) b) c) d)

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Subtract. a) b) c) d)

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Adding and Subtracting Rational Expressions with Different Denominators Find the LCD of two or more rational expressions. 2.Given two rational expressions, write equivalent rational expressions with their LCD. 3.Add or subtract rational expressions with different denominators.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 1 Find the LCD of two or more rational expressions.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Remember that when adding or subtracting fractions with different denominators, we must first find a common denominator. It is helpful to use the least common denominator (LCD), which is the smallest number that is evenly divisible by all the denominators.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Finding the LCD To find the LCD of two or more rational expressions: 1. Factor each denominator. 2. For each unique factor, compare the number of times it appears in each factorization. Write a product that includes each unique factor the greatest number of times it appears in the denominator factorizations.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Find the LCD. Solution We first factor the denominators 12y 2 and 8y 3 by writing their prime factorizations. The unique factors are 2, 3, and y. To generate the LCD, include 2, 3, and y the greatest number of times each appears in any of the factorizations.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Note: We can compare exponents in the prime factorizations to create the LCD. If two factorizations have the same prime factors, we write that prime factor in the LCD with the greater of the two exponents. The greatest number of times that 2 appears is three times (in 2 3 y 3 ). The greatest number of times that 3 appears is once (in y 2 ). The greatest number of times that y appears is three times (in 2 3 y 3 ).

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Find the LCD. Solution Factor the denominators x 2 – 25 and 2x – 10. The unique factors are 2, (x + 5), and (x – 5). The greatest number of times that 2 appears is once. The greatest number of times that (x + 5) appears is once. The greatest number of times that (x – 5) appears is once.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 2 Given two rational expressions, write equivalent rational expressions with their LCD.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Write as equivalent rational expressions with their LCD. Solution In the previous example, we found the LCD to be. Write the denominator in factored form. Note: Another way to determine the appropriate factor is to think of it as the factor in the LCD that is missing from the original denominator. Multiply the numerator and the denominator by the same factor, 2, to get the LCD, 2(x + 5)(x – 5).

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Write the denominator in factored form. Multiply the numerator and the denominator by the same factor, (x + 5), to get the LCD, 2(x + 5)(x – 5).

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 3 Add or subtract rational expressions with different denominators.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Adding or Subtracting Rational Expressions with Different Denominators To add or subtract rational expressions with different denominators: 1. Find the LCD. 2. Write each rational expression as an equivalent expression with the LCD. 3. Add or subtract the numerators and keep the LCD. 4. Simplify.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Add. Solution The LCD is 24x 2. Write equivalent rational expressions with the LCD, 24x 2. Add numerators. Note: Remember that to add polynomials, we combine like terms.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Add. Solution Since x – 6 and 6 – x are additive inverses, we obtain the LCD by multiplying the numerator and denominator of one of the rational expressions by –1. We chose the second rational expression.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Add. a) b) c) d)

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Add. a) b) c) d)

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Subtract. a) b) c) d)

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Subtract. a) b) c) d)

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Complex Rational Expressions Simplify complex rational expressions.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 1 Simplify complex rational expressions.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Complex rational expression: A rational expression that contains rational expressions in the numerator or denominator. Examples:

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Simplifying Complex Rational Expressions To simplify a complex rational expression, use one of the following methods: Method 1 1. Simplify the numerator and denominator if needed. 2. Rewrite as a horizontal division problem. Method 2 1. Multiply the numerator and denominator of the complex rational expression by their LCD. 2. Simplify.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example—Method 1 Simplify. Solution Write the numerator fractions as equivalent fractions with their LCD, 12, and write the denominator fractions with their LCD, 24.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued 1 2

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example—Method 2 Simplify. Solution Multiply the numerator and denominator by

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Simplify. Solution Write the numerator and denominator as equivalent fractions. (Method 1)

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Simplify. Solution Multiply the numerator and denominator by the LCD of all the rational expressions. (Method 2)

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Simplify. a) b) c) d)

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Simplify. a) b) c) d)

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Simplify. a) b) c) a + 1 d) a – 1

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Simplify. a) b) c) a + 1 d) a – 1

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Equations Containing Rational Expressions Solve equations containing rational expressions.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 1 Solve equations containing rational expressions.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve. Solution Multiply both sides by 24. Distribute and divide out common factors. Subtract 6x from both sides. Divide both sides by  2. Simplify both sides

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Extraneous Solution: An apparent solution that does not solve its equation. Solving Equations Containing Rational Expressions To solve an equation that contains rational expressions: 1. Eliminate the rational expressions by multiplying both sides of the equation by their LCD. 2. Solve the equation using the methods we learned in Chapters 2 (linear equations) and 6 (quadratic equations). 3. Check your solution(s) in the original equation. Discard any extraneous solutions.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve. Solution If x = 6, then the equation is undefined, so the solution cannot be 6. Divide out common factors. Distribute 2 to clear the parentheses. Subtract 2x on both sides. Divide both sides by 4.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve. Solution In the original equation, when x = -1/3 it is undefined, so the only solution is x = ½.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve. Solution

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued This solves the original equation

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve. a)  2 b) 5 c) 7 d) 10

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve. a)  2 b) 5 c) 7 d) 10

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve. a)  2 b) 1 c)  2 and  5 d) 3 and 4

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve. a)  2 b) 1 c)  2 and  5 d) 3 and 4

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Applications with Rational Expressions Use tables to solve problems with two unknowns involving rational expressions. 2.Solve problems involving direct variation. 3.Solve problems involving inverse variation.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 1 Use tables to solve problems with two unknowns involving rational expressions.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Tables are helpful in problems involving two or more people working together to complete a task. For each person involved, the rate of work, time at work, and amount of the task completed are related as follows: Since the people are working together, the sum of their individual amounts of the task completed equals the whole task. Person’s rate of work Time at work Amount of the task completed by that person  = Amount completed by one person Amount completed by the other person Whole task  =

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Louis and Rebecca own a painting business. Louis can paint an average size room in 5 hours. Rebecca can paint the same room in 3 hours. How long would it take them to paint the same room working together? Understand Louis paints at a rate of 1 room in 5 hours, or 1/5 of a room per hour. Rebecca paints at a rate of 1 room in 3 hours, or 1/3 of a room per hour.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Plan and Execute Louis’ amount completed + Rebecca’s amount completed = 1 room CategoryRate of Work Time at Work Amount of Task Completed Louist t Rebeccatt

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Answer Working together, it takes Louis and Rebecca or 1 7/8 hours to paint an average size room. Check

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 2 Solve problems involving direct variation.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Direct variation: Two variables y and x are in direct variation if y = kx, where k is a constant. In words, direct variation is written as “y varies directly as x” or “y is directly proportional to x” and these phrases translate to y = kx.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Suppose y varies directly as x. When y = 12, x = 6. Find y when x = 9. Solution Replace y with 12 and x with 6, then solve for k. y = kx 12 = k  6 2 = k Replace k with 2 in y = kx so that we have y = 2x. y = 2x y = 2(9) = 18

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Nigel’s paycheck varies directly as the number of hours worked. For 15 hours of work, the pay is $ Find the pay for 27 hours of work. Understand Translating “paycheck varies directly as the number of hours worked,” we write p = kh, where p represents the paycheck and h represents the hours. Plan Use p = kh and replace p with and h with 15, in order to find the value of k.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Execute = k  = k Replace k with p = kh p = 12.55h When hours = 27: p = 12.55(27) p = Answer Working 27 hours will earn a paycheck of $

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective 3 Solve problems involving inverse variation.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Inverse variation: Two variables, y and x, are in inverse variation if where k is a constant. In words, inverse variation is written as “y varies inversely as x” or “y is inversely proportional to x, and these phrases translate to.”

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example If the temperature is constant, the pressure of a gas in a container varies inversely as the volume of the container. If the pressure is 15 pounds per square foot in a container with 4 cubic feet, what is the pressure in a container with 1.5 cubic feet? Understand Because the pressure and volume vary inversely, we can write where P represents the pressure and V represents the volume.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Plan Use pressure is 15 when volume is 4 to determine the value of k. Execute Answer The pressure is 40 pounds per square foot when the volume is 1.5 cubic feet. Replacing k with 60, solve for P when V is 1.5.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Danielle can paint a 4800-square foot house in 8 days, and Paige can do the same job in only 6 days. How long will it take them to paint a square foot house if they work together? a) 14 days b) 7 days c) d)

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Danielle can paint a 4800-square foot house in 8 days, and Paige can do the same job in only 6 days. How long will it take them to paint a square foot house if they work together? a) 14 days b) 7 days c) d)

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The distance a car travels varies directly with the amount of gas it carries. On the highway, a van travels 112 miles using 7 gallons of gas. How many gallons are required to travel 544 miles? a) 15 gallons b) 16 gallons c) 32 gallons d) 34 gallons

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The distance a car travels varies directly with the amount of gas it carries. On the highway, a van travels 112 miles using 7 gallons of gas. How many gallons are required to travel 544 miles? a) 15 gallons b) 16 gallons c) 32 gallons d) 34 gallons

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley If y varies inversely as x and y = 8 when x = 10, what is y when x is 16? a) 2 b) 5 c) 10 d) 16

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley If y varies inversely as x and y = 8 when x = 10, what is y when x is 16? a) 2 b) 5 c) 10 d) 16