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 695: 21


The equation has one solution, -1.

Work Step by Step

$\frac{a}{a+3}=\frac{2a}{a-3}-1$ $\frac{a}{a+3}-\frac{2a}{a-3}+1=0$ $\frac{a(a-3)}{(a+3)(a-3)}-\frac{2a(a+3)}{(a+3)(a-3)}+\frac{(a+3)(a-3)}{(a+3)(a-3)}=0$ $\frac{a^2-3a-2a^2-6a+a^2-9}{(a+3)(a-3)}=0$ $(a^2+a^2-2a^2)-(3a+6a)-9=0*(a+3)(a-3)$ $-9a-9=0$ $-9a=9$ $a=-1$ Check: $-1/(-1+3)=-1/2;(2\times-1)/(-1-3)-1=1/2-1=-1/2$ The equation has one solution, -1.
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