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: 42



Work Step by Step

$x^2+3x-10=x^2+5x-2x-10=x(x+5)-2(x+5)=(x+5)(x-2)$, so LCD$=(x+5)(x-2)$, thus $\frac{x^2-39}{x^2+3x-10}-\frac{x-7}{x-2}=\frac{x^2-39}{(x+5)(x-2)}-\frac{(x-7)(x+5)}{(x+5)(x-2)}=\frac{x^2-39-(x^2-2x-35)}{(x+5)(x-2)}=\frac{2x-4}{(x+5)(x-2)}=\frac{2(x-2)}{(x+5)(x-2)}=\frac{2}{x+5}$
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