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: 42

Answer

$\frac{2(5x-14)}{(x-2)(x-3)}; x \ne 2,3;$

Work Step by Step

$\frac{8}{(x-2)} + \frac{2}{(x-3)} = \frac{8(x-3)+2 (x-2)}{(x-2)(x-3)} ;x \ne 2,3;$ $= \frac{8x-24+2x-4}{(x-2)(x-3)} ;x \ne 2,3;$ $= \frac{10x-28}{(x-2)(x-3)} ;x \ne 2,3;$ $=\frac{2(5x-14)}{(x-2)(x-3)}; x \ne 2,3;$
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