Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Appendix A - Review - A.10 Internal Notation; Solving Inequalities - A.10 Assess Your Understanding - Page A87: 87

Answer

$\{x| x\gt3\}$ see graph.

Work Step by Step

1. Solve the left part to get $2x-4\gt0$ or $x\gt2$ 2. For the right part, $\frac{1}{2x-4}-\frac{1}{2}\lt0$ gives $\frac{2-2x+4}{2x-4}\lt0$ or $\frac{2x-6}{2x-4}\gt0$ 3. The $\gt$ sign requires that the solution be outside of the region formed by the boundary points $(x=2,3)$, thus the solution $x\lt2$ or $x\gt3$ 4. Combine the above results to get $\{x| x\gt3\}$ 5. see graph.
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