Lesson 11-5 Warm-Up

“Solving Rational Equations” (11-5) How do you solve rational equation containing one or more rational expressions? To solve a rational equation with one or more rational expressions (i.e. unlike denominators containing variables), multiply both sides of the equation by the LCD to eliminate the rational expressions. Example:

Solve + = . Check the solution. Solving Rational Equations LESSON 11-5 Additional Examples Solve + = . Check the solution. 3 8 4 x 14 2x 3 8 4 x 14 2x + = The denominators are 8, x, and 2x. The LCD is 8x. 8x = 8x 3 8 4 x + 14 2x Multiply each side by 8x. This means you multiply every term in the equation by 8x. 1 1 4 8x = 8x 3 8 4 x + 14 2x 8x Use the Distributive Property. 1 1 1 No rational expressions (fractions), so now you can solve. 3x + 32 = 56 -32 -32 Subtract 32 from each side. 3x = 24 Simplify. + 0 3 3 Divide each side by 3. x = 8 Simplify

Check: + Substitute 8 for x. = (continued) 3 8 4 (8) 14 2(8) 7 8 14 16 Solving Rational Equations LESSON 11-5 Additional Examples (continued) Check: 3 8 4 (8) 14 2(8) + Substitute 8 for x. 7 8 14 16 7 8 =

Solve = – 1. Check the solution. Solving Rational Equations LESSON 11-5 Additional Examples 6 x2 5 x Solve = – 1. Check the solution. 5 x – 1 6 x2 = Multiply each side by the LCD, x2. This Means multiply every term in the equation by x2. 5 x – x2 (1) 6 x2 = Use the Distributive Property. 1 x 1 1 6 = 5x – x2 Simplify. x2 – 5x + 6 = 0 Collect like terms on one side. (x – 3)(x – 2) = 0 Factor the quadratic expression. x – 3 = 0 or x – 2 = 0 Use the Zero-Product Property. + 3 + 3 + 2 + 2 Isolate the x’s. x = 3 or x = 2 Solve.

– 1 – 1 Check: = = (continued) 6 (3)2 5 (3) 6 (2)2 5 (2) 2 3 3 2 Solving Rational Equations LESSON 11-5 Additional Examples (continued) 6 (3)2 5 (3) – 1 6 (2)2 5 (2) – 1 Check: 2 3 = 3 2 =

Person Work Rate Time Worked Part of (part of job/min.) (min) Job Done Solving Rational Equations LESSON 11-5 Additional Examples Renee can mow the lawn in 20 minutes. Joanne can do the same job in 30 minutes. How long will it take them if they work together? Define: Let n = the time to complete the job if they work together (in minutes). Person Work Rate Time Worked Part of (part of job/min.) (min) Job Done Renee n Joanne n 1 20 30 n Words: Renee’s part done + Joanne’s part done = complete job.

Multiply each side by the LCD, 60. Solving Rational Equations LESSON 11-5 Additional Examples (continued) n 20 30 + Translate: = 1 60 = 60(1) Multiply each side by the LCD, 60. 3n + 2n = 60 Use the Distributive Property. 5n = 60 Combine like terms. n = 12 Divide both sides by 5. It will take two of them 12 minutes to mow the lawn working together. 1 20 Check: Renee will do • = of the job, and Joanne will do of the job. Together, they will do = 1, or the whole job. 12 3 5 2 30 • = +

“Solving Rational Equations” (11-5) How can you solve a rational equation in the form of a proportion (equal fractions)? To solve a rational equation in the form of a proportion, or equal fractions, simply cross multiply as you would a proportion with rational numbers. Example: Given Note: The process of cross multiplying may give an extraneous solution, or a solution that solves the new equation, but not the original one. Therefore, always check your solutions, because they may not make not work in the original equation.

Solve = . Check the solution. Solving Rational Equations LESSON 11-5 Additional Examples 4 x2 1 x + 8 Solve = . Check the solution. 4 x2 1 x + 8 = 4(x + 8) = x2(1) Write cross products. 4x + 32 = x2 Use the Distributive Property. x2 – 4x – 32 = 0 Collect all of the terms on one side of the equal sign. (x – 8)(x + 4) = 0 Factor the quadratic expression. x – 8 = 0  or  x + 4 = 0 Use the Zero-Product Property. x = 8  or x = –4 Solve.

Check: = Substitute 8 and -4 for x. = = (continued) 4 x2 1 x + 8 4 Solving Rational Equations LESSON 11-5 Additional Examples (continued) Check: 4 x2 1 x + 8 = 4 (8)2 1 (8) + 8 4 (–4)2 1 (–4) + 8 Substitute 8 and -4 for x. = 4 64 1 16 4 16 1 =

(x + 3) (x – 1) = 4(x – 1) Write the cross products. Solving Rational Equations LESSON 11-5 Additional Examples x + 3 x – 1 4 x – 1 Solve = . x + 3 x – 1 4 = (x + 3) (x – 1) = 4(x – 1) Write the cross products. x2 – x + 3x – 3 = 4x – 4 Use the Distributive Property to multiply x2 + 2x – 3 = 4x – 4 Combine like terms. x2 – 2x + 1 = 0 Collect all of the terms on one side of the equal sign. (x – 1) (x – 1) = 0 Factor. x – 1 = 0 Use the Zero-Product Property. x = 1 Simplify.

Undefined! There is no division by 0. Solving Rational Equations LESSON 11-5 Additional Examples (continued) Check: 1 + 3 (1) – 1 4 Substitute 1 for x. 4 = Undefined! There is no division by 0. The equation has no solution because 1 makes a denominator equal 0.