We are tasked with solving the inequality:

$\left| \frac{2x - 6}{3} \right| \geq \frac{5}{6}$

Step 1: Split the absolute value inequality

The absolute value inequality $|A| \geq B$ means: $A \geq B \quad \text{or} \quad A \leq -B$ So, we can split the inequality as:

$\frac{2x - 6}{3} \geq \frac{5}{6} \quad \text{or} \quad \frac{2x - 6}{3} \leq -\frac{5}{6}$

Step 2: Solve each inequality

Inequality 1:

$\frac{2x - 6}{3} \geq \frac{5}{6}$ Multiply both sides by 6 to eliminate the fractions: $6 \times \frac{2x - 6}{3} \geq 6 \times \frac{5}{6}$ This simplifies to: $2(2x - 6) \geq 5$ Distribute the 2: $4x - 12 \geq 5$ Add 12 to both sides: $4x \geq 17$ Divide by 4: $x \geq \frac{17}{4}$

Inequality 2:

$\frac{2x - 6}{3} \leq -\frac{5}{6}$ Multiply both sides by 6: $6 \times \frac{2x - 6}{3} \leq 6 \times -\frac{5}{6}$ This simplifies to: $2(2x - 6) \leq -5$ Distribute the 2: $4x - 12 \leq -5$ Add 12 to both sides: $4x \leq 7$ Divide by 4: $x \leq \frac{7}{4}$

Step 3: Combine the results

From the two inequalities, we have: $x \geq \frac{17}{4} \quad \text{or} \quad x \leq \frac{7}{4}$

Thus, the solution is: $x \in \left( -\infty, \frac{7}{4} \right] \cup \left[ \frac{17}{4}, \infty \right)$

Would you like further details on this? Here are 5 related questions:

1. What happens if the inequality was $<$ instead of $\geq$?
2. How do we solve an absolute value equation instead of an inequality?
3. How does multiplying or dividing by a negative number affect inequalities?
4. What if the inequality involved a different fraction inside the absolute value?
5. Can absolute value inequalities have no solutions?

Tip: When working with absolute value inequalities, always consider the two cases (positive and negative) carefully.