You have two inequalities to solve:

1. \(-24 < -3x - 3\)
2. \(-39 > -3x - 3\)

Let's solve each one step-by-step:

### Solving the first inequality: 
\(-24 < -3x - 3\)

1. Add 3 to both sides:
   \[
   -24 + 3 < -3x
   \]
   \[
   -21 < -3x
   \]
   
2. Divide both sides by \(-3\) (remember to flip the inequality sign when dividing by a negative number):
   \[
   \frac{-21}{-3} > \frac{-3x}{-3}
   \]
   \[
   7 > x
   \]
   or
   \[
   x < 7
   \]

### Solving the second inequality:
\(-39 > -3x - 3\)

1. Add 3 to both sides:
   \[
   -39 + 3 > -3x
   \]
   \[
   -36 > -3x
   \]

2. Divide both sides by \(-3\) (again, flip the inequality sign):
   \[
   \frac{-36}{-3} < \frac{-3x}{-3}
   \]
   \[
   12 < x
   \]
   or
   \[
   x > 12
   \]

### Final Solution:
Combining both results, we have:
\[
12 < x < 7
\]

This is not possible, as there is no value of \(x\) that satisfies both inequalities at the same time.

Do you want further clarification or more details?

### 5 Related Questions:
1. How do you handle compound inequalities involving "and" vs. "or"?
2. What happens when two inequalities have no solution, like in this case?
3. Can you graph the solution to inequalities on a number line?
4. What are some real-world applications of solving inequalities?
5. How do you handle inequalities with fractions?

### Tip:
Always remember to flip the inequality sign when multiplying or dividing both sides of an inequality by a negative number!

Learn how to solve compound inequalities such as -24 < -3x - 3 and -39 > -3x - 3 using key algebraic techniques. This detailed solution walks through step-by-step processes, including flipping the inequality sign when dividing by negative numbers. Ideal for students in Grades 8-10.

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