College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.6 - Page 86: 49

Answer

$\frac{13}{6(x+2)}; x \ne -2;$

Work Step by Step

The given expression is $\frac{3}{2x+4} + \frac{2}{3x+6}$ Factorizing the denominator, $\frac{3}{2(x+2)} + \frac{2}{3(x+2)} ; $ Taking LCD, the denominator becomes $6(x+2)$ and the expression becomes $\frac{3(3)}{6(x+2)} + \frac{2(2)}{6(x+2)} = \frac{9}{6(x+2)} + \frac{4} {6(x+2)}$ $ = \frac{13}{6(x+2)} ; x \ne -2;$
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