Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.6 - Rational Expressions - Exercise Set - Page 85: 47


$=\displaystyle \frac{2x^{2}+50}{(x-5)(x+5)}, \quad x\neq-5,5$

Work Step by Step

The denominators are different. Find a common denominator. LCD=$(x+5)(x- 5)$ $\displaystyle \frac{x+5}{x-5}+\frac{x-5}{x+5}= \displaystyle \frac{(x+5)(x+5)}{(x+5)(x- 5)}+\frac{(x- 5)(x- 5)}{(x+5)(x- 5)}$ $=\displaystyle \frac{x^{2}+10x+25+x^{2}-10x+25}{(x-5)(x+5)}$ $=\displaystyle \frac{2x^{2}+50}{(x-5)(x+5)}$ ... exclude values that yield 0 in the denominator: $x\neq-5,5$ ... here, there are no common factors to cancel. (the numerator = $2(x+25)$ ) $=\displaystyle \frac{2x^{2}+50}{(x-5)(x+5)}, \quad x\neq-5,5$
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