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: 50

Answer

$\frac{29}{6(x+4)}; x \ne -4;$

Work Step by Step

$\frac{5}{(2x+8)} + \frac{7}{(3x+12)}$ Take out common factor from the denominator, $=\frac{5}{2(x+4)} + \frac{7}{3(x+4)}$ Taking LCD, the denominator becomes $[2 \times 3 \times (x+4) = 6(x+4)]$ and the expression becomes $=\frac{5(3) + 7(2)}{6(x+4)}; x \ne -4;$ $=\frac{15 + 14}{6(x+4)}; x \ne -4;$ $=\frac{29}{6(x+4)}; x \ne -4;$
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