We are given the following compound inequality:

$-24 \leq -3x - 3 \quad \text{or} \quad -39 > -3x - 3$

Step 1: Solve each inequality separately.

For the first inequality:

$-24 \leq -3x - 3$

1. Add 3 to both sides: $-24 + 3 \leq -3x$ $-21 \leq -3x$
2. Divide both sides by -3, and remember to reverse the inequality when dividing by a negative number: $\frac{-21}{-3} \geq \frac{-3x}{-3}$ $7 \geq x$ This simplifies to: $x \leq 7$

For the second inequality:

$-39 > -3x - 3$

1. Add 3 to both sides: $-39 + 3 > -3x$ $-36 > -3x$
2. Divide both sides by -3, and reverse the inequality: $\frac{-36}{-3} < \frac{-3x}{-3}$ $12 < x$ This simplifies to: $x > 12$

Step 2: Combine the results.

The solution to the compound inequality is: $x \leq 7 \quad \text{or} \quad x > 12$

This means the solution set is all values of $x$ such that:

• $x \leq 7$, or
• $x > 12$.

Final answer:

$x \leq 7 \quad \text{or} \quad x > 12$

Do you want any more details or explanations? Here are some related questions you might want to explore:

1. How do you solve compound inequalities involving "and" conditions?
2. Can you solve an inequality involving absolute values?
3. What are the key differences between solving linear equations and inequalities?
4. How do you graph the solution of a compound inequality on a number line?
5. How do you approach inequalities with fractions or decimals?

Tip: Always remember to reverse the inequality symbol when multiplying or dividing by a negative number!