Solving Rational Equations Digital Lesson Solving Rational Equations

Objective and Vocabulary The objective is to be able to solve rational equations. Lowest Common Denominator (LCD) – The smallest denominator which has all of the original denominators as factors.

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: This denominator would equal zero if x = 0. This denominator would equal zero if x = 0. When there is a variable in the denominator, we must avoid any values of the variable that make a denominator zero. When adding or subtracting rational expressions, we find the LCD and convert fractions to equivalent fractions that have the common denominator.

Solving Rational Equations Solving a Rational Equation Solving Rational Equations 1) Find the LCM of the denominators. *Avoid any values of the variable that make a denominator zero. 2) Clear denominators by multiplying both sides of the equation by the LCM. 3) Solve the equation. 4) Check your solution.

Solving a Rational Equation Example 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.

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.

Solving a Rational Equation CONTINUED Check 8: ? ? ? ? ? true ? This true statement verifies that the solution is 8 and the solution set is {8}. ?

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

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.

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.

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

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.

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.

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.

Objective #1: Example

Objective #1: Example

Method 2 The previous example had two or more fractions on one side of the equation. When there is a single fraction on each side of the equation, the equation can be solved as a proportion by cross-multiplying.

Objective #1: Example

Objective #1: Example

Objective #1: Example

Objective #1: Example

Objective #1: Example

Objective #1: Example

Objective #1: Example

Objective #1: Example

Objective #1: Example CONTINUED

Objective #1: Example CONTINUED

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