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The inequality given is a compound inequality:

\[
3x + 1 > 2x - 5 > x - 7
\]

We will break this into two parts:

1. \( 3x + 1 > 2x - 5 \)
2. \( 2x - 5 > x - 7 \)

### Solving part 1:
\[
3x + 1 > 2x - 5
\]
Subtract \( 2x \) from both sides:
\[
x + 1 > -5
\]
Now subtract 1 from both sides:
\[
x > -6
\]

### Solving part 2:
\[
2x - 5 > x - 7
\]
Subtract \( x \) from both sides:
\[
x - 5 > -7
\]
Now add 5 to both sides:
\[
x > -2
\]

### Conclusion:

The solution is the intersection of the two conditions: \( x > -6 \) and \( x > -2 \). The stricter condition is \( x > -2 \), so the solution to the compound inequality is:

\[
x > -2
\]

Would you like further details on any of these steps or have additional questions?

Here are five follow-up questions:
1. How do you solve a compound inequality with "less than" instead of "greater than"?
2. Can you visualize the solution on a number line?
3. How does the solution change if we have \( \geq \) instead of \( > \)?
4. How can compound inequalities be applied in real-life situations?
5. What happens if you reverse the inequality symbols?

**Tip:** Always solve each part of a compound inequality separately before combining results.

Solve the compound inequality: 3x + 1 > 2x - 5 > x - 7.

Learn how to solve the compound inequality 3x + 1 > 2x - 5 > x - 7 with a detailed step-by-step solution. This problem covers key algebraic concepts like inequalities and compound inequalities. Suitable for students in Grades 8-10.

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