We are tasked with solving the compound inequality:

\[
-11 \leq -32x + 1 \leq -2
\]

### Step 1: Break down the compound inequality into two parts:
1. \(-11 \leq -32x + 1\)
2. \(-32x + 1 \leq -2\)

We'll solve both inequalities separately.

### Step 2: Solve the first inequality \( -11 \leq -32x + 1 \)

1. Subtract 1 from both sides:
   \[
   -11 - 1 \leq -32x
   \]
   \[
   -12 \leq -32x
   \]

2. Divide by \(-32\) and reverse the inequality (since dividing by a negative reverses the inequality):
   \[
   \frac{-12}{-32} \geq x
   \]
   \[
   \frac{3}{8} \geq x
   \]
   So, \(x \leq \frac{3}{8}\).

### Step 3: Solve the second inequality \( -32x + 1 \leq -2 \)

1. Subtract 1 from both sides:
   \[
   -32x \leq -2 - 1
   \]
   \[
   -32x \leq -3
   \]

2. Divide by \(-32\) and reverse the inequality:
   \[
   x \geq \frac{-3}{-32}
   \]
   \[
   x \geq \frac{3}{32}
   \]

### Step 4: Combine the results:
We have:
\[
\frac{3}{32} \leq x \leq \frac{3}{8}
\]

### Final Answer:
The solution to the compound inequality is:

\[
\frac{3}{32} \leq x \leq \frac{3}{8}
\]

Would you like further details or clarifications?

Here are some related questions:
1. How do you solve inequalities with fractions?
2. What happens when multiplying or dividing inequalities by negative numbers?
3. How do you represent solutions to inequalities on a number line?
4. How do compound inequalities differ from simple inequalities?
5. Can inequalities have no solutions or all real numbers as solutions?

**Tip**: When solving inequalities, always remember to flip the inequality sign when multiplying or dividing by a negative number!

Learn how to solve the compound inequality −11 ≤ −32x + 1 ≤ −2 step-by-step. This problem involves reversing inequalities when dividing by negative numbers and working with fractions, making it suitable for students in Grades 8-10.

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