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: 41


$=\displaystyle \frac{3(3x+13)}{(x+4)(x+5)}, \quad x\neq-4,-5$

Work Step by Step

The denominators are different. Find a common denominator. LCD=$(x+4)(x+5)$ $\displaystyle \frac{3}{x+4}+\frac{6}{x+5}$=$\displaystyle \frac{3(x+5)}{(x+4)(x+5)}+\frac{6(x+4)}{(x+4)(x+5)}$ $=\displaystyle \frac{3x+15+6x+24}{(x+4)(x+5)}$ $=\displaystyle \frac{9x+39}{(x+4)(x+5)}$ $=\displaystyle \frac{3(3x+13)}{(x+4)(x+5)}$ ... exclude values that yield 0 in the denominator: $x\neq-4,-5$ ... and, cancel common factors (here, there are none) $=\displaystyle \frac{3(3x+13)}{(x+4)(x+5)}, \quad x\neq-4,-5$
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