Chapter 4 - Section 4.3 - Equations and Inequalities Involving Absolute Value - Exercise Set - Page 284: 83

$(-\infty,\dfrac{-1}{3}]\cup [3, \infty)$

Given: $|4-3x| \geq 5$ or, $4-3x \geq 5$ or,$-3x \geq 1$ or, $x \leq \dfrac{-1}{3}$ Now, $|4-3x| \geq 5$ or, $4-3x \leq -5$ or,$x \leq -5-43$ or, $x \geq 3$ Hence, $(-\infty,\dfrac{-1}{3}]\cup [3, \infty)$