To solve the compound inequality 

\[
32 < -5x + 7 \quad \text{or} \quad -5x + 7 < -13,
\]

we need to solve each inequality separately.

### Solving the First Inequality: \(32 < -5x + 7\)

1. Subtract 7 from both sides:

\[
32 - 7 < -5x \implies 25 < -5x.
\]

2. Divide both sides by -5 (remember to reverse the inequality sign):

\[
\frac{25}{-5} > x \implies -5 > x \implies x < -5.
\]

### Solving the Second Inequality: \(-5x + 7 < -13\)

1. Subtract 7 from both sides:

\[
-5x + 7 - 7 < -13 - 7 \implies -5x < -20.
\]

2. Divide both sides by -5 (again, reverse the inequality sign):

\[
\frac{-5x}{-5} > \frac{-20}{-5} \implies x > 4.
\]

### Combining the Solutions

The solution to the compound inequality is:

\[
x < -5 \quad \text{or} \quad x > 4.
\]

### Interval Notation

In interval notation, the solution can be written as:

\[
(-\infty, -5) \cup (4, \infty).
\]

Would you like further details, or do you have any questions?

Here are 5 related questions you might find interesting:

1. How do we solve a compound inequality involving "and" instead of "or"?
2. What happens when both inequalities have overlapping solutions?
3. How do you graph the solution set of a compound inequality on a number line?
4. Can you explain why we reverse the inequality sign when dividing or multiplying by a negative number?
5. How does the solution change if we switch the signs in the original inequalities?

**Tip:** Always remember to reverse the inequality sign when multiplying or dividing by a negative number.

Learn how to solve the compound inequality 32 < −5x + 7 or −5x + 7 < −13. This step-by-step guide covers core algebraic concepts, such as handling inequalities, reversing the inequality sign, and solving compound inequalities. Suitable for students in Grades 7-9.

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