We are given the compound inequality:

\[
20 \geq 5x + 10 \quad \text{or} \quad 65 \leq 5x + 10
\]

Let's solve both inequalities separately.

### First inequality: 
\[
20 \geq 5x + 10
\]
1. Subtract 10 from both sides:
   \[
   20 - 10 \geq 5x \quad \Rightarrow \quad 10 \geq 5x
   \]
2. Divide both sides by 5:
   \[
   \frac{10}{5} \geq x \quad \Rightarrow \quad 2 \geq x \quad \text{or} \quad x \leq 2
   \]

### Second inequality: 
\[
65 \leq 5x + 10
\]
1. Subtract 10 from both sides:
   \[
   65 - 10 \leq 5x \quad \Rightarrow \quad 55 \leq 5x
   \]
2. Divide both sides by 5:
   \[
   \frac{55}{5} \leq x \quad \Rightarrow \quad 11 \leq x \quad \text{or} \quad x \geq 11
   \]

### Final solution:
We now combine the two inequalities:

\[
x \leq 2 \quad \text{or} \quad x \geq 11
\]

Thus, the solution to the compound inequality is:
\[
x \in (-\infty, 2] \cup [11, \infty)
\]

Would you like more details on any specific step?

Here are five related questions to explore further:
1. What happens when solving a compound inequality with "and" instead of "or"?
2. How do we graph inequalities on a number line?
3. How do we solve compound inequalities involving fractions?
4. Can inequalities have no solution? If so, how?
5. What happens when you reverse the inequality sign during solving?

**Tip**: Always pay attention to whether the compound inequality uses "and" or "or," as this changes the way you combine solutions.

Learn how to solve the compound inequality 20 ≥ 5x + 10 or 65 ≤ 5x + 10 with a detailed step-by-step breakdown. This problem covers key algebraic concepts, including solving inequalities and combining solutions. Suitable for students in Grades 8-10.

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