Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 6 - Section 6.2 - Adding and Subtracting Rational Expressions - Exercise Set - Page 427: 38



Work Step by Step

$x^2-16=(x+4)(x-4)$, so LCD$=(x+4)(x-4)$, thus $\frac{8x}{x^2-16}-\frac{5}{x+4}=\frac{8x}{(x+4)(x-4)}-\frac{5(x-4)}{(x+4)(x-4)}=\frac{8x-5x+20}{(x+4)(x-4)}=\frac{3x+20}{(x+4)(x-4)}$
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