To solve the compound inequality:

\[
-4 < 3x + 2 < 5
\]

We will solve it in two steps by breaking the inequality into two separate inequalities:

### Step 1: Solve the left inequality \(-4 < 3x + 2\)

1. Subtract 2 from both sides:
   \[
   -4 - 2 < 3x
   \]
   \[
   -6 < 3x
   \]

2. Divide both sides by 3:
   \[
   \frac{-6}{3} < x
   \]
   \[
   -2 < x
   \]
   
So, the first inequality is \(x > -2\).

### Step 2: Solve the right inequality \(3x + 2 < 5\)

1. Subtract 2 from both sides:
   \[
   3x < 5 - 2
   \]
   \[
   3x < 3
   \]

2. Divide both sides by 3:
   \[
   x < 1
   \]

### Final Solution:
Combining the two parts, the solution is:
\[
-2 < x < 1
\]

This means that \(x\) is between \(-2\) and \(1\) (but does not include \(-2\) or \(1\)).

Would you like more details or have any questions?

Here are 5 related questions:
1. How do we solve compound inequalities involving absolute values?
2. What is the solution if the inequality is reversed, such as \(5 < 3x + 2 < 10\)?
3. How do we graph the solution of this inequality on a number line?
4. How does multiplying or dividing by a negative number affect inequalities?
5. Can you solve an inequality where the expression is a quadratic instead of linear?

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

Learn how to solve the compound inequality 4 < 3x + 2 < 5 step by step. This solution breaks down the inequality into two parts, covering essential algebraic concepts. Suitable for students in Grades 7-9.

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