We are given the inequality:

$3 |10 - x| - 6 \geq 0$

Step 1: Isolate the absolute value term.

Add $6$ to both sides of the inequality:

$3 |10 - x| \geq 6$

Step 2: Divide by 3 to simplify.

$|10 - x| \geq 2$

Step 3: Solve the absolute value inequality.

The general rule for absolute value inequalities is:

$|A| \geq B \quad \text{means} \quad A \leq -B \quad \text{or} \quad A \geq B$

In our case:

$10 - x \leq -2 \quad \text{or} \quad 10 - x \geq 2$

Step 4: Solve each inequality.

For $10 - x \leq -2$:

$-x \leq -12 \quad \text{(Subtract 10 from both sides)}$ $x \geq 12 \quad \text{(Divide both sides by -1, flip the inequality)}$

For $10 - x \geq 2$:

$-x \geq -8 \quad \text{(Subtract 10 from both sides)}$ $x \leq 8 \quad \text{(Divide both sides by -1, flip the inequality)}$

Step 5: Combine the solutions.

The solution is:

$x \geq 12 \quad \text{or} \quad x \leq 8$

In interval notation, the solution is:

$(-\infty, 8] \cup [12, \infty)$

Would you like more details or have any questions?

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Here are 5 related questions to deepen your understanding:

1. How would the solution change if the inequality was $3 |10 - x| - 6 > 0$?
2. How do you solve an absolute value equation like $|x - 3| = 5$?
3. What happens if the coefficient in front of the absolute value, $3$, is negative?
4. Can you graph the solution to $|10 - x| \geq 2$ on a number line?
5. How would you handle solving $|ax + b| \geq c$ for any constants $a$, $b$, and $c$?

Tip: Always check your final solution by plugging test values from each interval into the original inequality.