Answer
See below
Work Step by Step
Obtain: $$\frac{1}{x-3}+3=\frac{2x}{x+3}\\\frac{3x-8}{x-3}=\frac{2x}{x+3}\\(3x-8)(x+3)=2x(x-3)\\3x^2+x-24=2x^2-6x\\x^2+7x-24=0$$
The solutions are $x=\frac{1}{2}(-7\pm \sqrt 145)$
Algebra 2 (1st Edition)
by Larson, Ron; Boswell, Laurie; Kanold, Timothy D.; Stiff, Lee
Published by
McDougal Littell
ISBN 10:
0618595414
ISBN 13:
978-0-61859-541-9
Chapter 10 Counting Methods and Probability - 10.2 Use Combinations and the Binomial Theorem - 10.2 Exercises - Mixed Review - Page 697: 63
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