We are given the compound inequality:

\[
-x + 3 > 2 \quad \text{or} \quad -6 > -x + 3
\]

### Step 1: Solve each inequality separately.

#### Inequality 1:
\[
-x + 3 > 2
\]

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

Multiply both sides by \(-1\) (and reverse the inequality):
\[
x < 1
\]

#### Inequality 2:
\[
-6 > -x + 3
\]

Subtract 3 from both sides:
\[
-6 - 3 > -x
\]
\[
-9 > -x
\]

Multiply both sides by \(-1\) (and reverse the inequality):
\[
9 < x \quad \text{or} \quad x > 9
\]

### Step 2: Combine the results.

The solution is:
\[
x < 1 \quad \text{or} \quad x > 9
\]

This means that \(x\) is either less than 1 or greater than 9. These are two separate regions on the number line.

### Final Answer:
\[
x \in (-\infty, 1) \cup (9, \infty)
\]

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Here are five more questions to consider:
1. How would the solution change if the inequalities were connected by "and" instead of "or"?
2. How do you approach inequalities that include variables on both sides?
3. How would you represent the solution \(x < 

Learn how to solve the compound inequality -x + 3 > 2 or -6 > -x + 3. This problem focuses on algebraic manipulation and the reversal of inequalities when multiplying by negative numbers. Suitable for students in Grades 8-10.

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