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 6.3 - Complex Rational Expressions - Exercise Set - Page 435: 10

Answer

$\frac{1}{x+3}$.

Work Step by Step

The given expression is $=\frac{\frac{1}{x+2}}{1+\frac{1}{x+2}}$ Multiply the numerator and the denominator by $(x+2)$. $=\frac{(x+2)}{(x+2)}\cdot \frac{\frac{1}{x+2}}{1+\frac{1}{x+2}}$ Use the distributive property. $=\frac{(x+2)\cdot\frac{1}{x+2}}{(x+2)\cdot (1)+(x+2)\cdot \frac{1}{x+2}}$ Simplify. $=\frac{1}{x+2+1}$ $=\frac{1}{x+3}$.
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