Let's solve the system of inequalities step by step:

### Inequality 1:  
\[
-3x + 1 < 10
\]

1. Subtract 1 from both sides:
   \[
   -3x < 9
   \]

2. Divide both sides by -3. Remember, dividing by a negative number flips the inequality:
   \[
   x > -3
   \]

### Inequality 2:  
\[
-11 < -3x + 1
\]

1. Subtract 1 from both sides:
   \[
   -12 < -3x
   \]

2. Divide both sides by -3, flipping the inequality:
   \[
   4 > x \quad \text{or} \quad x < 4
   \]

### Combined Solution:
We have two inequalities:
\[
x > -3 \quad \text{and} \quad x < 4
\]

So, the solution is:
\[
-3 < x < 4
\]

This means \( x \) is between -3 and 4.

Would you like more details or have any questions about this solution?

Here are 5 related questions for practice:

1. Solve \( 2x + 5 > 9 \) and \( x + 3 < 7 \).
2. Solve \( -4x - 2 \leq 10 \) and \( -8 > -4x + 4 \).
3. What is the solution to \( -5x + 6 > 1 \) and \( 3x - 4 \leq 5 \)?
4. How do you graph the solution \( -2 < x \leq 5 \)?
5. Solve the inequality \( 7x - 4 \geq 3x + 12 \).

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

Learn how to solve the system of linear inequalities -3x + 1 < 10 and -11 < -3x + 1 step-by-step. This problem involves solving linear inequalities and understanding the rule for reversing the inequality sign when dividing by a negative number. Suitable for students in Grades 7-9.

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