Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.4 - Rational Expressions - 1.4 Exercises: 53

Answer

$\frac{x-2}{(x+3)(x-3)}$

Work Step by Step

$\frac{1}{x+3}+\frac{1}{x^{2}-9}$ Factor $x^{2}-9$ and replace in the denominator: $=\frac{1}{x+3}+\frac{1}{(x+3)(x-3)}$ Find the lowest common denominator (i.e. $(x+3)(x-3)$) and adjust the fractions accordingly: $=\frac{1\times (x-3)}{(x+3)\times (x-3)}+\frac{1}{(x+3)(x-3)}$ $=\frac{x-3}{(x+3)(x-3)}+\frac{1}{(x+3)(x-3)}$ Combine the fractions: $=\frac{x-3+1}{(x+3)(x-3)}$ Simplify: $=\frac{x-2}{(x+3)(x-3)}$
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