Download presentation

Presentation is loading. Please wait.

1
§ 6.6 Rational Equations

2
**Solving a Rational Equation**

EXAMPLE Solve: SOLUTION Notice that the variable x appears in two of the denominators. We must avoid any values of the variable that make a denominator zero. This denominator would equal zero if x = 0. This denominator would equal zero if x = 0. Therefore, we see that x cannot equal zero. Blitzer, Intermediate Algebra, 4e – Slide #87

3
**Blitzer, Intermediate Algebra, 4e – Slide #88**

Solving a Rational Equation CONTINUED The denominators are 5x, 5, and x. The least common denominator is 5x. We begin by multiplying both sides of the equation by 5x. We will also write the restriction that x cannot equal zero to the right of the equation. This is the given equation. Multiply both sides by 5x, the LCD. Use the distributive property. Blitzer, Intermediate Algebra, 4e – Slide #88

4
**Blitzer, Intermediate Algebra, 4e – Slide #89**

Solving a Rational Equation CONTINUED Divide out common factors in the multiplications. Multiply. Subtract. Add. Divide. The proposed solution, 8, is not part of the restriction It should check in the original equation. Blitzer, Intermediate Algebra, 4e – Slide #89

5
**Blitzer, Intermediate Algebra, 4e – Slide #90**

Solving a Rational Equation CONTINUED Check 8: ? ? ? ? ? true ? This true statement verifies that the solution is 8 and the solution set is {8}. ? Blitzer, Intermediate Algebra, 4e – Slide #90

6
**Solving Rational Equations**

Solving a Rational Equation Solving Rational Equations 1) List restrictions on the variable. Avoid any values of the variable that make a denominator zero. 2) Clear the equation of fractions by multiplying both sides by the LCD of all rational expressions in the equation. 3) Solve the resulting equation. 4) Reject any proposed solution that is in the list of restrictions on the variable. Check other proposed solutions in the original equation. Blitzer, Intermediate Algebra, 4e – Slide #91

7
**Solving a Rational Equation**

EXAMPLE Solve: SOLUTION 1) List restrictions on the variable. This denominator would equal zero if x = 4. This denominator would equal zero if x = 3.5. The restrictions are Blitzer, Intermediate Algebra, 4e – Slide #92

8
**Blitzer, Intermediate Algebra, 4e – Slide #93**

Solving a Rational Equation CONTINUED 2) Multiply both sides by the LCD. The denominators are x – 4 and 2x – 7. Thus, the LCD is (x – 4)(2x + 7). This is the given equation. Multiply both sides by the LCD. Simplify. Blitzer, Intermediate Algebra, 4e – Slide #93

9
**Blitzer, Intermediate Algebra, 4e – Slide #94**

Solving a Rational Equation CONTINUED 3) Solve the resulting equation. This is the equation cleared of fractions. Use FOIL on each side. Subtract from both sides. Subtract 19x from both sides. 4) Check the proposed solution in the original equation. Notice, there is no proposed solution. And of course, -7 = -20 is not a true statement. Therefore, there is no solution to the original rational equation. We say the solution set is , the empty set. Blitzer, Intermediate Algebra, 4e – Slide #94

10
**Solving a Rational Equation**

EXAMPLE Solve: SOLUTION 1) List restrictions on the variable. By factoring denominators, it makes it easier to see values that make the denominators zero. This denominator is zero if x = -4 or x = 2. This denominator would equal zero if x = -4. This denominator would equal zero if x = 2. The restrictions are Blitzer, Intermediate Algebra, 4e – Slide #95

11
**Blitzer, Intermediate Algebra, 4e – Slide #96**

Solving a Rational Equation CONTINUED 2) Multiply both sides by the LCD. The factors of the LCD are x + 4 and x – 2. Thus, the LCD is (x + 4)(x - 2). This is the given equation. Multiply both sides by the LCD. Use the distributive property. Blitzer, Intermediate Algebra, 4e – Slide #96

12
**Blitzer, Intermediate Algebra, 4e – Slide #97**

Solving a Rational Equation CONTINUED Simplify. 3) Solve the resulting equation. This is the equation with cleared fractions. Use the distributive property. Combine like terms. Subtract x from both sides. Add 5 to both sides. Divide both sides by 3. Blitzer, Intermediate Algebra, 4e – Slide #97

13
**Blitzer, Intermediate Algebra, 4e – Slide #98**

Solving a Rational Equation CONTINUED 4) Check the proposed solutions in the original equation. The proposed solution, 3, is not part of the restriction that Substitute 3 for x, in the given (original) equation. The resulting true statement verifies that 3 is a solution and that {3} is the solution set. Blitzer, Intermediate Algebra, 4e – Slide #98

14
**Blitzer, Intermediate Algebra, 4e – Slide #99**

Solving a Rational Equation EXAMPLE Rational functions can be used to model learning. Many of these functions model the proportion of correct responses as a function of the number of trials of a particular task. One such model, called a learning curve, is where f (x) is the proportion of correct responses after x trials. If f (x) = 0, there are no correct responses. If f (x) = 1, all responses are correct. The graph of the rational function is shown on the next page. Use the function to solve the following problem. Blitzer, Intermediate Algebra, 4e – Slide #99

15
**Blitzer, Intermediate Algebra, 4e – Slide #100**

Solving a Rational Equation CONTINUED Blitzer, Intermediate Algebra, 4e – Slide #100

16
**Blitzer, Intermediate Algebra, 4e – Slide #101**

Solving a Rational Equation CONTINUED How many learning trials are necessary for 0.5 of the responses to be correct? Identify your solution as a point on the graph. SOLUTION Substitute 0.5, the proportion of correct responses, for f (x) and solve the resulting rational equation for x. The LCD is 0.9x Multiply both sides by the LCD. Simplify. Blitzer, Intermediate Algebra, 4e – Slide #101

17
**Blitzer, Intermediate Algebra, 4e – Slide #102**

Solving a Rational Equation CONTINUED Use the distributive property on the left side. Subtract 0.9x from both sides. Subtract 0.05 from both sides. Divide both sides by The number of learning trials necessary for 0.5 of the responses to be correct is 1. The solution is identified as a point on the graph at the beginning of the problem. Blitzer, Intermediate Algebra, 4e – Slide #102

Similar presentations

OK

Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 8.4 - 1.

Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 8.4 - 1.

© 2018 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google