Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 6 - Section 6.2 - Adding and Subtracting Rational Expressions - Exercise Set - Page 354: 32

Answer

$\frac{7x+5}{(x-5)(x+5)}$

Work Step by Step

$\frac{(x-1)}{(x-5)}$-$\frac{(x+2)}{(x+5)}$= $\frac{(x-1)(x+5)}{(x-5)(x+5)}$-$\frac{(x+2)(x-5)}{(x-5)(x+5)}$= $\frac{x^2+4x-5)-(x^2-3x-10)}{(x-5)(x+5)}$= $\frac{x^2+4x-5-x^2+3x+10}{(x-5)(x+5)}$= $\frac{7x+5}{(x-5)(x+5)}$
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