Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.8 - Inequalities - 1.8 Exercises - Page 89: 36


$\left[\frac{11}{12}, \frac{13}{6}\right]$ Refer to the attached image below for the graph.

Work Step by Step

For easier work, get rid of the fractions by multiplying the LCD (which is 20) to each part to have: $\\20(-\frac{1}{2}) \le 20 \cdot \frac{4-3x}{5} \le 20(\frac{1}{4}) \\-10 \le 4(4-3x) \le 5 \\-10 \le 16-12x \le 5$ Subtract 16 to each part to have: $\\-10-16 \le 16-12x-16 \le 5-16 \\-26 \le -12x \le -11$ Divide $-12$ to each part. Note that dividing a negative number to the inequality will flip the inequality symbols. Thus, $\\\frac{-26}{-12} \ge \frac{-12x}{-12} \ge \frac{-11}{-12} \\\frac{13}{6} \ge x \ge \frac{11}{12} \\\frac{11}{12} \le x \le \frac{13}{6}$ Thus, the solution is $\left[\frac{11}{12}, \frac{13}{6}\right]$.
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