We are given the compound inequality:

\[
\leq -3x + 3 \quad \text{or} \quad -24 > -3x + 3
\]

Let's solve each part of the inequality separately.

### First inequality: \(\leq -3x + 3\)

1. Subtract 3 from both sides:
   \[
   -3 \leq -3x
   \]

2. Divide both sides by \(-3\). When dividing by a negative number, the inequality flips:
   \[
   1 \geq x \quad \text{or} \quad x \leq 1
   \]

### Second inequality: \(-24 > -3x + 3\)

1. Subtract 3 from both sides:
   \[
   -24 - 3 > -3x
   \]
   \[
   -27 > -3x
   \]

2. Divide both sides by \(-3\), flipping the inequality:
   \[
   9 < x \quad \text{or} \quad x > 9
   \]

### Final Solution:

The solution to the compound inequality \(\leq -3x + 3\) or \(-24 > -3x + 3\) is the union of both individual solutions:
\[
x \leq 1 \quad \text{or} \quad x > 9
\]

This means \(x\) can either be less than or equal to 1, or greater than 9.

Would you like more details on solving inequalities, or do you have any specific questions?

Here are 5 related questions for further exploration:
1. How do you solve inequalities with both \(>\) and \(\geq\) symbols involved?
2. What are some key steps to remember when dividing or multiplying inequalities by negative numbers?
3. How does solving compound inequalities differ from single inequalities?
4. Can inequalities have more than two separate intervals as solutions?
5. How does graphing the solution of a compound inequality look like?

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

This solution explains how to solve the compound inequality ≤−3x + 3 or −24 > −3x + 3 step-by-step. Learn to handle inequalities involving negative numbers and understand when to flip inequality signs. Ideal for students in Grades 8-10, covering essential algebra concepts.

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