Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 11 - Rational Expressions and Functions - 11-5 Solving Rational Equations - Practice and Problem-Solving Exercises - Page 683: 21



Work Step by Step

Given: $a(a-3)=2a(a+3)-(a+3)(a-3)$ $a^2-3a=2a^2+6a-[a^2-3a+3a-9]$ $\implies a^2-3a=2a^2+6a-a^2+3a-3a+9$ $\implies a^2-3a=a^2+6a+9$ $\implies a^2+6a+9-a^2+3a=0$ or, $9a+9=0 \implies a=-\frac{9}{9}$ $\implies a=-1$
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