Precalculus: Mathematics for Calculus, 7th Edition

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