Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 11 - Rational Expressions and Functions - 11-5 Solving Rational Equations - Practice and Problem-Solving Exercises - Page 697: 47


The solutions are $n=0$ or $n=\frac{1}{2}$.

Work Step by Step

$\frac{n}{n-2}+\frac{n}{n+2}=\frac{n}{n^2-4}$ $\frac{n(n+2)}{(n-2)(n+2)}+\frac{n(n-2)}{(n-2)(n+2)}=\frac{n}{(n-2)(n+2)}$ $\frac{2n^2}{(n-2)(n+2)}=\frac{n}{(n-2)(n+2)}$ $2n^2=n$ $2n^2-n=0$ $n=0$ or $n=\frac{1}{2}$ Check: $\frac{0}{0-2}+\frac{0}{0+2}=\frac{0}{0^2-4}$ $0=0$ $\frac{\frac{1}{2}}{\frac{1}{2}-2}+\frac{\frac{1}{2}}{\frac{1}{2}+2}=\frac{\frac{1}{2}}{(\frac{1}{2})^2-4}$ $\frac{-2}{15}=\frac{-2}{15}$ The solutions are $n=0$ or $n=\frac{1}{2}$.
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