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: 41

Answer

$\frac{3(3x+13)}{(x+4)(x+5)}; x \ne -4,-5;$

Work Step by Step

$\frac{3}{x+4} + \frac{6}{x+5} = \frac{3(x+5)+ 6(x+4)}{(x+4)(x+5)} ; x \ne -4,-5;$ $= \frac{3x+15+ 6x+24}{(x+4)(x+5)} ; x \ne -4,-5;$ $= \frac{9x+39}{(x+4)(x+5)} ; x \ne -4,-5;$ $=\frac{3(3x+13)}{(x+4)(x+5)}; x \ne -4,-5$
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