§ 6.6 Rational Equations

Blitzer, Intermediate Algebra, 5e – Slide #2 Section 6.6 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: 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 #3 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. 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! See Study Tip on page 443 for a comparison of the two procedures. P 442

Blitzer, Intermediate Algebra, 5e – Slide #4 Section 6.6 Solving a Rational EquationEXAMPLE 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. Therefore, we see that x cannot equal zero. P 442

Blitzer, Intermediate Algebra, 5e – Slide #5 Section 6.6 Solving a Rational Equation 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. CONTINUED This is the given equation. Multiply both sides by 5x, the LCD. Use the distributive property.

Blitzer, Intermediate Algebra, 5e – Slide #6 Section 6.6 Solving a Rational EquationCONTINUED 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, 5e – Slide #7 Section 6.6 Solving a Rational EquationCONTINUED Check 8: This true statement verifies that the solution is 8 and the solution set is {8}. ? ? ? ? ? ? ? true

Blitzer, Intermediate Algebra, 5e – Slide #8 Section 6.6 Solving a Rational Equation Check Point 1 Solve: SOLUTION P 442 The denominators are 2x, 5x, and 1. The least common denominator is 10x and x cannot equal zero. Divide out common factors in the multiplications. Multiply.

Blitzer, Intermediate Algebra, 5e – Slide #9 Section 6.6 Solving a Rational EquationCONTINUED Subtract. Divide by 13. The proposed solution, 6, is not part of the restriction. It should check in the original equation. Combine like terms.

Blitzer, Intermediate Algebra, 5e – Slide #10 Section 6.6 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 #11 Section 6.6 Solving a Rational EquationEXAMPLE 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 P 443

Blitzer, Intermediate Algebra, 5e – Slide #12 Section 6.6 Solving a Rational Equation 2) Multiply both sides by the LCD. The denominators are x – 4 and 2x – 7. Thus, the LCD is (x – 4)(2x - 7). CONTINUED This is the given equation. Multiply both sides by the LCD. Simplify.

Blitzer, Intermediate Algebra, 5e – Slide #13 Section 6.6 Solving a Rational Equation 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. CONTINUED This is the equation cleared of fractions. Use FOIL on each side. Subtract from both sides. Subtract 19x from both sides. 3) Solve the resulting equation.

Solving a Rational Equation Blitzer, Intermediate Algebra, 5e – Slide #14 Section 6.6 Do Check Point 2 on page 443 The LCD is (x + 1)(x + 6). The restrictions are Do Check Point 4 on page 445 The LCD is 2x. The restrictions are

Blitzer, Intermediate Algebra, 5e – Slide #15 Section 6.6 Solving a Rational EquationEXAMPLE 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. The restrictions are This denominator would equal zero if x = 2.

Blitzer, Intermediate Algebra, 5e – Slide #16 Section 6.6 Solving a Rational Equation 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). CONTINUED This is the given equation. Multiply both sides by the LCD. Use the distributive property.

Blitzer, Intermediate Algebra, 5e – Slide #17 Section 6.6 Solving a Rational Equation 3) Solve the resulting equation. CONTINUED Simplify. 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, 5e – Slide #18 Section 6.6 Solving a Rational Equation 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. CONTINUED

Solving a Rational Equation Blitzer, Intermediate Algebra, 5e – Slide #19 Section 6.6 Do Check Point 5 on page 446 The LCD is (x - 3)(x - 4). The restrictions are

DONE

Solving a Rational EquationEXAMPLE 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, 5e – Slide #21 Section 6.6

Solving a Rational Equation Blitzer, Intermediate Algebra, 5e – Slide #22 Section 6.6 CONTINUED

Solving a Rational Equation Blitzer, Intermediate Algebra, 5e – Slide #23 Section 6.6 CONTINUED SOLUTION How many learning trials are necessary for 0.5 of the responses to be correct? Identify your solution as a point on the graph. 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.

Solving a Rational Equation Blitzer, Intermediate Algebra, 5e – Slide #24 Section 6.6 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, 5e – Slide #25 Section 6.6 Solving a Rational Equation 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. Important to Remember: