# Chapter 11 - Rational Expressions and Functions - 11-5 Solving Rational Equations - Practice and Problem-Solving Exercises - Page 696: 38

The equation has two solutions, 0 and 2.

#### Work Step by Step

$\frac{u+1}{u+2}=\frac{-1}{u-3}+\frac{u-1}{u^2-u-6}$ $\frac{u+1}{u+2}=\frac{-1}{u-3}+\frac{u-1}{(u+2)(u-3)}$ $(u+2)(u-3)\frac{u+1}{u+2}=(u+2)(u-3)(\frac{-1}{u-3}+\frac{u-1}{(u+2)(u-3)})$ $(u+1)(u-3)=-1(u+2)+u-1$ $u^2-2u-3=-u-2+u-1$ $u^2-2u-3=-3$ $u^2-2u=0$ $u(u-2)=0$ $u=0$ or $u-2=0$ $u=0$ or $u=2$ Check: $\frac{0+1}{0+2}=1/2; \frac{-1}{0-3}+\frac{0-1}{0-0-6}=1/3+1/6=1/2$ $\frac{2+1}{2+2}=3/4; \frac{-1}{2-3}+\frac{2-1}{4-2-6}=1+(-1/4)=3/4$ The equation has two solutions, 0 and 2.

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