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: 37



Work Step by Step

$x^2-25=(x+5)(x-5)$, so LCD$=(x+5)(x-5)$, thus $\frac{3x}{x^2-25}-\frac{4}{x+5}=\frac{3x}{(x+5)(x-5)}-\frac{4(x-5)}{(x+5)(x-5)}=\frac{3x-4x+20}{(x+5)(x-5)}=\frac{-x+20}{(x+5)(x-5)}$
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