The system of inequalities you provided is:

\[
\begin{aligned}
- x + 3 &< -3 \\
- x + 3 &> -8
\end{aligned}
\]

Let’s solve each inequality step by step:

### Inequality 1: \(- x + 3 < -3\)

1. Subtract 3 from both sides:
   \[
   - x + 3 - 3 < -3 - 3 \implies - x < -6
   \]
   
2. Multiply both sides by \(-1\) (remember to reverse the inequality):
   \[
   x > 6
   \]

### Inequality 2: \(- x + 3 > -8\)

1. Subtract 3 from both sides:
   \[
   - x + 3 - 3 > -8 - 3 \implies - x > -11
   \]

2. Multiply both sides by \(-1\) (again, reverse the inequality):
   \[
   x < 11
   \]

### Solution:
The solution is the set of values where \(x > 6\) and \(x < 11\), which is the interval:

\[
6 < x < 11
\]

Thus, the solution is \(x \in (6, 11)\).

Would you like more details or have any questions?

### Related Questions:
1. How would the solution change if the inequalities used \(\leq\) and \(\geq\)?
2. Can you graph the solution on a number line?
3. What would happen if you combined two inequalities that had no overlapping solutions?
4. How do you solve compound inequalities involving absolute values?
5. How does reversing an inequality change the solution?

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

Explore how to solve the system of linear inequalities -x + 3 < -3 and -x + 3 > -8. This solution walks you through step-by-step, explaining key algebraic concepts such as reversing the inequality when multiplying by a negative number. Suitable for students in Grades 7-9.

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