advertisement

§ 6.6 Rational Equations Solving a Rational Equation A rational equation, also called a fractional equation, is an equation containing one or more rational expressions. The following is an example of a rational equation: 7x 4 9 4 . 5x 5 x Do you see that there is a variable in the denominator? This is a characteristic of many rational equations. P 441 Blitzer, Intermediate Algebra, 5e – Slide #2 Section 6.6 Solving a Rational Equation When adding or subtracting rational expressions, we find the LCD and convert fractions to equivalent fractions that have the common denominator. See Study Tip on page 443 for a comparison of the two procedures. By contrast, when we solve rational equations, the LCD is used as a multiplier that clears an equation of fractions. Thus in an equation – you can say…. “I really don’t like these fractions” and then just proceed to clear them out! Blitzer, Intermediate Algebra, 5e – Slide #3 Section 6.6 P 442 Solving a Rational Equation EXAMPLE Solve: 7 x 4 9 4 . 5x 5 x 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. 7x 4 9 4 5x 5 x 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, 5e – Slide #4 Section 6.6 P 442 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. 7x 4 9 4 , x0 5x 5 x 5x 7x 4 9 4 5 x 5x 5 x 5x 7 x 4 5x 9 5x 4 1 5x 1 5 1 x Blitzer, Intermediate Algebra, 5e – Slide #5 Section 6.6 This is the given equation. Multiply both sides by 5x, the LCD. Use the distributive property. Solving a Rational Equation CONTINUED 7x 4 9x 5 4 Divide out common factors in the multiplications. 7 x 4 9 x 20 Multiply. 4 2 x 20 Subtract. 16 2 x Add. 8 x Divide. The proposed solution, 8, is not part of the restriction x 0 . It should check in the original equation. Blitzer, Intermediate Algebra, 5e – Slide #6 Section 6.6 Solving a Rational Equation CONTINUED Check 8: 7x 4 9 4 5x 5 x 7 8 4 ? 9 4 58 5 8 56 4 ? 9 4 40 5 8 52 ? 9 4 40 5 8 52 ? 9 4 40 40 40 5 8 9 4 52 ? 40 40 5 8 52 ? 8 9 5 4 52 ? 72 20 52 52 true This true statement verifies that the solution is 8 and the solution set is {8}. Blitzer, Intermediate Algebra, 5e – Slide #7 Section 6.6 Solving a Rational Equation Check Point 1 Solve: x 6 x 24 2 2x 5x SOLUTION The denominators are 2x, 5x, and 1. The least common denominator is 10x and x cannot equal zero. x 6 x 24 2, x 0 2x 5x 5 2 10 x x 6 10 x x 24 10 x 2 1 2x 1 5x Divide out common factors in the multiplications. 5 x 30 2 x 48 20 x Multiply. Blitzer, Intermediate Algebra, 5e – Slide #8 Section 6.6 P 442 Solving a Rational Equation CONTINUED 5 x 30 2 x 48 20 x 7 x 78 20 x 78 13 x Combine like terms. Subtract. Divide by 13. 6x The proposed solution, 6, is not part of the restriction the original equation. Blitzer, Intermediate Algebra, 5e – Slide #9 Section 6.6 x 0 . It should check in 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. P 442 Blitzer, Intermediate Algebra, 5e – Slide #10 Section 6.6 Solving a Rational Equation EXAMPLE Solve: 3 x 1 6 x 5 . x4 2x 7 SOLUTION 1) List restrictions on the variable. 3x 1 6 x 5 x 4 2x 7 This denominator would equal zero if x = 4. This denominator would equal zero if x = 3.5. The restrictions are x 4 and x 3.5. Blitzer, Intermediate Algebra, 5e – Slide #11 Section 6.6 P 443 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). 3x 1 6 x 5 , x 4, x 3.5 x 4 2x 7 This is the given equation. x 42 x 7 3x 1 x 42 x 7 6 x 5 Multiply both sides by x4 2x 73x 1 x 46x 5 2 x 7 the LCD. Simplify. Blitzer, Intermediate Algebra, 5e – Slide #12 Section 6.6 Solving a Rational Equation CONTINUED 3) Solve the resulting equation. 2x 73x 1 x 46x 5 This is the equation cleared of fractions. 6 x 2 19x 7 6 x 2 19x 20 19 x 7 19 x 20 Use FOIL on each side. Subtract 6x 2 from both sides. 7 20 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, 5e – Slide #13 Section 6.6 Solving a Rational Equation Do Check Point 2 on page 443 x3 x2 . x 1 x 6 x 4 The LCD is (x + 1)(x + 6). The restrictions are x 1 and x 6. Do Check Point 4 on page 445 x 12 5 2 x x 4,6 The LCD is 2x. The restrictions are x 0 Blitzer, Intermediate Algebra, 5e – Slide #14 Section 6.6 Solving a Rational Equation EXAMPLE Solve: 2x 1 2 1 . 2 x 2x 8 x 4 x 2 SOLUTION 1) List restrictions on the variable. By factoring denominators, it makes it easier to see values that make the denominators zero. 2x 1 2 1 x 4x 2 x 4 x 2 This denominator is zero if x = -4 or x = 2. This denominator would equal zero if x = -4. The restrictions are x 4 and x 2. Blitzer, Intermediate Algebra, 5e – Slide #15 Section 6.6 This denominator would equal zero if x = 2. 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). 2x 1 2 1 , x 4, x 2 x 4x 2 x 4 x 2 This is the given equation. 2x 1 2 1 x 4x 2 x 4x 2 x2 x 4x 2 x 4 x 4x 2 Multiply both sides by the LCD. 2x 1 2 1 x 4x 2 x 4x 2 x 4x 2 x4 x2 Use the distributive property. Blitzer, Intermediate Algebra, 5e – Slide #16 Section 6.6 Solving a Rational Equation CONTINUED 2 x 1 2x 2 x 4 Simplify. 3) Solve the resulting equation. 2 x 1 2x 2 x 4 2x 1 2x 4 x 4 4x 5 x 4 3x 5 4 3x 9 x3 Blitzer, Intermediate Algebra, 5e – Slide #17 Section 6.6 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. 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 x 4 and x 2. 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, 5e – Slide #18 Section 6.6 Solving a Rational Equation Do Check Point 5 on page 446 3 5 x 2 20 2 x 3 x 4 x 7 x 12 The LCD is (x - 3)(x - 4). The restrictions are x 4 and x 3 x 1,7 Blitzer, Intermediate Algebra, 5e – Slide #19 Section 6.6 DONE 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 0 .9 x 0 .4 f x 0 . 9 x 0. 1 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, 5e – Slide #21 Section 6.6 Solving a Rational Equation CONTINUED Proportion of Correct Responses A Learning Curve 1.2 1.1 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 f x 0 1 2 3 4 5 6 7 8 0 .9 x 0 .4 0 . 9 x 0. 1 9 10 11 12 13 14 15 Number of Learning Trials Blitzer, Intermediate Algebra, 5e – Slide #22 Section 6.6 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. 0 .9 x 0 .4 0 .5 0 .9 x 0 .1 0.9 x 0.4 0.9 x 0.1 0.5 0.9 x 0.1 0.9 x 0.1 0.50.9 x 0.1 0.9 x 0.4 Blitzer, Intermediate Algebra, 5e – Slide #23 Section 6.6 The LCD is 0.9x + 0.1. Multiply both sides by the LCD. Simplify. Solving a Rational Equation CONTINUED 0.45 x 0.05 0.9 x 0.4 0.45 x 0.05 0.4 0.45 x 0.45 x 1 Use the distributive property on the left side. Subtract 0.9x from both sides. Subtract 0.05 from both sides. Divide both sides by -0.45. 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, 5e – Slide #24 Section 6.6 Solving a Rational Equation Important to Remember: A common error when solving rational equations is to forget to list the restrictions for the variable in the very beginning and then to reject any of those values in the end as solutions. Don’t forget to say in the beginning what x can’t be! And don’t forget to throw out a solution if it’s something x “can’t be”…. Another common error is to check the proposed solutions in one of the later equations. You must check your proposed solutions in the original equation. Blitzer, Intermediate Algebra, 5e – Slide #25 Section 6.6