Answer
$\dfrac{3(x^2-3)}{2x-1}$
Work Step by Step
Since the denominators are the same, we have
$$
\frac{3x^2}{2x-1}-\frac{9}{2x-1}=\frac{3x^2-9}{2x-1}=\frac{3(x^2-3)}{2x-1}
$$
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
by Sullivan III, Michael
Published by
Pearson
ISBN 10:
0-32193-104-1
ISBN 13:
978-0-32193-104-7
Appendix A - Review - A.6 Rational Expressions - A.6 Assess Your Understanding - Page A53: 38
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