1. **State the problem:** Solve the inequality $$\frac{|3x - 4|}{2} \geq \left|\frac{5}{12}\right|$$. 2. **Recall the properties:** The absolute value is always non-negative, and $$|a| \geq |b|$$ means $$a \leq -b$$ or $$a \geq b$$ when $$a$$ and $$b$$ are real numbers. 3. **Simplify the right side:** $$\left|\frac{5}{12}\right| = \frac{5}{12}$$ since $$\frac{5}{12} > 0$$. 4. **Multiply both sides by 2 to clear the denominator:** $$\cancel{2} \times \frac{|3x - 4|}{\cancel{2}} \geq 2 \times \frac{5}{12}$$ which simplifies to $$|3x - 4| \geq \frac{10}{12} = \frac{5}{6}$$. 5. **Rewrite the inequality without absolute value:** $$3x - 4 \leq -\frac{5}{6} \quad \text{or} \quad 3x - 4 \geq \frac{5}{6}$$. 6. **Solve each inequality separately:** For $$3x - 4 \leq -\frac{5}{6}$$: $$3x \leq -\frac{5}{6} + 4 = -\frac{5}{6} + \frac{24}{6} = \frac{19}{6}$$ Divide both sides by 3: $$\cancel{3}x \leq \frac{19}{6} \div \cancel{3}$$ $$x \leq \frac{19}{18}$$. For $$3x - 4 \geq \frac{5}{6}$$: $$3x \geq \frac{5}{6} + 4 = \frac{5}{6} + \frac{24}{6} = \frac{29}{6}$$ Divide both sides by 3: $$\cancel{3}x \geq \frac{29}{6} \div \cancel{3}$$ $$x \geq \frac{29}{18}$$. 7. **Final solution:** $$x \leq \frac{19}{18} \quad \text{or} \quad x \geq \frac{29}{18}$$. This means $$x$$ is in the intervals $$(-\infty, \frac{19}{18}] \cup [\frac{29}{18}, \infty)$$.