Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 6 - Section 6.2 - Adding and Subtracting Rational Expressions - Exercise Set - Page 427: 52

Answer

$\frac{(x-5)}{4(x-3)}$.

Work Step by Step

The given expression is $=\frac{x+5}{4x+12}+\frac{x}{9-x^2}$ First denominator $=4x+12$. Factor $=4(x+3)$. $=4(3+x)$ Second denominator $=9-x^2$. Factor $=3^2-x^2$ Use the algebraic identity $a^2-b^2=(a+b)(a-b)$. $=(3+x)(3-x)$ Substitute both factors into the given expression. $=\frac{x+5}{4(3+x)}+\frac{x}{(3+x)(3-x)}$ The LCM of both denominators is $=4(3+x)(3-x)$. $=\frac{x+5}{4(3+x)} \times \frac{3-x}{3-x}+\frac{x}{(3+x)(3-x)}\times \frac{4}{4}$ Simplify. $=\frac{(x+5)(3-x)}{4(3+x)(3-x)}+\frac{4x}{4(3+x)(3-x)}$ $=\frac{(x+5)(3-x)+4x}{4(3+x)(3-x)}$ $=\frac{3x-5x-x^2+15+4x}{4(3+x)(3-x)}$ Simplify. $=\frac{-x^2+2x+15}{4(3+x)(3-x)}$ Factor the numerator as shown below. $=-x^2+2x+15$ $=-x^2+5x-3x+15$ $=-x(x-5)-3(x-5)$ $=(x-5)(-x-3)$ $=-(x-5)(x+3)$ Substitute into the fraction. $=\frac{-(x-5)(x+3)}{4(3+x)(3-x)}$ Cancel common terms. $=\frac{-(x-5)}{4(3-x)}$ Rearrange. $=\frac{(x-5)}{4(x-3)}$.
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