Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 7 - Rational Functions - 7.5 Solving Rational Equations - 7.5 Exercises - Page 600: 39

Answer

Our solution is $x = 1$. To check our answer, we plug the solution into the original equation: $\frac{4}{1 + 5} = \frac{6}{1 + 8}$ Simplify the denominator: $\frac{4}{6} = \frac{6}{9}$ Simplify the fractions: $\frac{2}{3} = \frac{2}{3}$ The two sides are equal to one another. The solution is correct.

Work Step by Step

The first thing we want to do is get rid of the fractions. We do this by first finding the least common denominator (LCD) for all the terms. We find the LCD by taking the highest power of each factor in the denominators of the fractions: LCD = $(x + 5)(x + 8)$ Now, we will multiply each numerator by the factor its denominator is missing from the LCD: $4(x + 8) = 6(x + 5)$ Use the distributive property on both sides of the equation: $4x + 32 = 6x + 30$ Collect constants on the right side of the equation: $4x = 6x - 2$ Collect variables on the left side of the equation: $-2x = -2$ Divide each side of the equation by $-2$: $x = 1$ Our solution is $x = 1$.
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