Solving Rational Equations

are rational expressions. For example, A rational expression is a fraction with polynomials for the numerator and denominator. are rational expressions. For example, If x is replaced by a number making the denominator of a rational expression zero, the value of the rational expression is undefined. Example: Evaluate for x = –3, 0, and 1. x  3 undefined undefined 9 undefined 1 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Rational Expression

A rational equation is an equation between rational expressions. For example, and are rational equations. To solve a rational equation: 1. Find the LCM of the denominators. 2. Clear denominators by multiplying both sides of the equation by the LCM. 3. Solve the resulting polynomial equation. 4. Check the solutions. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Rational Equation

Examples: 1. Solve: . LCM = x – 3. 1 = x + 1 x = 0 (0) LCM = x(x – 1). Find the LCM. 1 = x + 1 Multiply by LCM = (x – 3). x = 0 Solve for x. (0) Check. Substitute 0. Simplify. True. 2. Solve: . LCM = x(x – 1). Find the LCM. Multiply by LCM. x – 1 = 2x Simplify. x = –1 Solve. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Examples: Solve

In this case, the value is not a solution of the rational equation. After clearing denominators, a solution of the polynomial equation may make a denominator of the rational equation zero. In this case, the value is not a solution of the rational equation. It is critical to check all solutions. Example: Solve: . Since x2 – 1 = (x – 1)(x + 1), LCM = (x – 1)(x + 1). 3x + 1 = x – 1 2x = – 2  x = – 1 Check. Since – 1 makes both denominators zero, the rational equation has no solutions. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Solve

Example: Solve: . x2 – 8x + 15 = (x – 3)(x – 5) x(x – 5) = – 6 Factor. The LCM is (x – 3)(x – 5). Original Equation. x(x – 5) = – 6 Polynomial Equation. x2 – 5x + 6 = 0 Simplify. (x – 2)(x – 3) = 0 Factor. Check. x = 2 is a solution. x = 2 or x = 3 Check. x = 3 is not a solution since both sides would be undefined. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Solve

Example: Using Work Formula To solve problems involving work, use the formula, part of work completed = rate of work time worked. Example: If it takes 5 hours to paint a room, what part of the work is completed after 3 hours? If one room can be painted in 5 hours then the rate of work is (rooms/hour). The time worked is 3 hours. Therefore, part of work completed = rate of work time worked part of work completed . Three-fifths of the work is completed after three hours. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Using Work Formula

Let t be the time it takes them to paint the room together. Example: If a painter can paint a room in 4 hours and her assistant can paint the room in 6 hours, how many hours will it take them to paint the room working together? Let t be the time it takes them to paint the room together. painter assistant rate of work time worked part of work completed t t LCM = 12. Multiply by 12. Simplify. Working together they will paint the room in 2.4 hours. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Word Problem

Examples: Using Motion Formulas To solve problems involving motion, use the formulas, distance = rate  time and time = . Examples: 1. If a car travels at 60 miles per hour for 3 hours, what distance has it traveled? Since rate = 60 (mi/h) and time = 3 h, then distance = rate time = 60 3 = 180. The car travels 180 miles. 2. How long does it take an airplane to travel 1200 miles flying at a speed of 250 miles per hour? Since distance = 1200 (mi) and rate = 250 (mi/h), time = = = 4.8. It takes 4.8 hours for the plane make its trip. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Examples: Using Motion Formulas

Let r be the rate of travel (speed) in miles per hour. Example: A traveling salesman drives from home to a client’s store 150 miles away. On the return trip he drives 10 miles per hour slower and adds one-half hour in driving time. At what speed was the salesperson driving on the way to the client’s store? Let r be the rate of travel (speed) in miles per hour. Trip to client Trip home distance rate time 150 r 150 r – 10 LCM = 2r (r – 10). 300r – 300(r – 10) = r(r – 10) Multiply by LCM. Example continued Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Word Problem

The return trip took one-half hour longer. Example continued 300r – 300r + 3000 = r2 – 10r 0 = r2 – 10r – 3000 0 = (r – 60)(r + 50) r = 60 or – 50 (–50 is irrelevant.) The salesman drove from home to the client’s store at 60 miles per hour. Check: At 60 mph the time taken to drive the 150 miles from the salesman’s home to the clients store is = 2.5 h. At 50 mph (ten miles per hour slower) the time taken to make the return trip of 150 miles is = 3 h. The return trip took one-half hour longer. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example Continued