Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 8 Rational Functions - 8.6 Solve Rational Equations - 8.6 Exercises - Skill Practice - Page 593: 28



Work Step by Step

We use cross multiplication to solve the equation. Doing this, we find: $$2\left(x^2-2x-3\right)=\left(x-3\right)\cdot \:1 \\ x=3,\:x=-\frac{1}{2}$$ Since 3 is an extraneous solution, $-1/2$ is the actual solution.
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