Precalculus (6th Edition) Blitzer

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

Chapter 2 - Test - Page 434: 28



Work Step by Step

Step 1. Rewrite the inequality as $\frac{2x+1}{x-3}-3=\frac{2x+1-3x+9}{x-3}=\frac{-x+10}{x-3}\leq0$ or $\frac{x-10}{x-3}\geq0$ Thus the boundary points are $x=3,10$ Step 2. Using test points to examine the signs of the left side across the boundary points, we have $...(+)...(3)...(-)...(10)...(+)...$ Thus the solutions are: $(-\infty,3)\cup[10,\infty)$ Step 3. We can graph the above solution on a real number line as shown.
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