Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Section 2.7 - Polynomial and Rational Inequalities - Exercise Set - Page 413: 67

Answer

$(-3,2)$

Work Step by Step

Step 1. Rewrite the inequality as $\frac{3}{x+3}-\frac{3}{x-2}\gt0$, $\frac{3x-6-3x-9}{(x+3)(x-2)}\gt0$ and $\frac{-15}{(x+3)(x-2)}\gt0$, Step 2. Reorder the inequality; we have $\frac{15}{(x+3)(x-2)}\lt0$ and the boundary points are $x=-3, 2$ Step 3. Using test points to examine signs of the left side across the boundary points, we have $...(+)...(-3)...(-)...(2)...(+)...$ thus the solutions are $-3\lt x\lt 2$; that is, $(-3,2)$ Step 4. We can express the solutions on a real number line as shown in the figure.
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