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Copyright © Cengage Learning. All rights reserved. Rational Expressions and Equations; Ratio and Proportion 6.

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Presentation on theme: "Copyright © Cengage Learning. All rights reserved. Rational Expressions and Equations; Ratio and Proportion 6."— Presentation transcript:

1 Copyright © Cengage Learning. All rights reserved. Rational Expressions and Equations; Ratio and Proportion 6

2 Copyright © Cengage Learning. All rights reserved. Section 6.5 Solving Equations That Contain Rational Expressions

3 3 Objectives Solve an equation that contains one or more rational expressions. Identify extraneous solutions. 1 1 2 2

4 4 Solve an equation that contains one or more rational expressions 1.

5 5 Solve an equation that contains one or more rational expressions To solve equations containing rational expressions, we use the same process we did with equations containing fractions. To clear the fractions, we multiply both sides of the equation by the LCD of the rational expressions that appear in the equation. To review this process, we will solve an equation containing only numerical denominators. Multiply both sides of the equation by 6, the LCD, to clear fractions.

6 6 Solve an equation that contains one or more rational expressions We then use the distributive property to remove parentheses, simplify, and solve the resulting equation for x. 2x + 6 = x x + 6 = 0 x = – 6 Subtract x from both sides. Subtract 6 from both sides.

7 7 Solve an equation that contains one or more rational expressions Check: –2 + 1 ≟ –1 –1 = –1 Because – 6 satisfies the original equation, it is the solution. Simplify. Substitute –6 for x.

8 8 Example Solve: Solution: Note that x = 0 is a restricted value since it creates division by 0. To clear the equation of rational expressions, we multiply both sides of the equation by the LCD of, 1, and, which is x. Multiply both sides by x, the LCD.

9 9 Example – Solution 4 + x = 6 x = 2 Check: cont’d Subtract 4 from both sides. Simplify. Substitute 2 for x. Use the distributive property.

10 10 Example – Solution 2 + 1 ≟ 3 3 = 3 Because 2 satisfies the original equation, it is the solution. Simplify. cont’d

11 11 Identify extraneous solutions 2.

12 12 Identify extraneous solutions If we multiply both sides of an equation by an expression that involves a variable, as we did in Example 1, we must check the apparent solutions. The next example shows why.

13 13 Example Solve: Solution: Find the restricted value of the denominators. In this example x  1. To clear the equation of rational expressions, we multiply both sides by x – 1, the LCD of the fractions contained in the equation. Multiply both sides by (x – 1), the LCD.

14 14 Example – Solution Because both sides were multiplied by an expression containing a variable, we must check to see if the apparent solution is a value that must be excluded. If we replace x with 1 in the original equation, both denominators will become 0. Therefore, 1 is not a solution. Simplify. Subtract 3 from both sides. cont’d

15 15 Example – Solution Such false solutions are often called extraneous solutions. Because 1 does not satisfy the original equation, there is no solution. The solution set of the equation is ∅. cont’d

16 16 Identify extraneous solutions Solving Equations Containing Rational Expressions 1. Find any restrictions on the variable. Remember that the denominator of a fraction cannot be 0. 2. Multiply both sides of the equation by the LCD of the rational expressions appearing in the equation to clear the equation of fractions. 3. Solve the resulting equation.

17 17 Identify extraneous solutions 4. If an apparent solution of an equation is a restricted value, that value must be excluded. Check all solutions for extraneous roots.

18 18 Identify extraneous solutions The formula is used in electronics to calculate parallel resistances. Solve the formula for r, we eliminate the denominators by multiplying both sides by the LCD, which is r r 1 r 2. Multiply both sides by r r 1 r 2 the LCD.

19 19 Identify extraneous solutions Use the distributive property to remove parentheses. Factor out r, the GCF. Divide both sides by r 2 + r 1. or


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