Let's solve the inequality:

Given inequality:
\[
3(x + 5) < 2(x + 2) + 8
\]

### Step 1: Expand both sides

On the left-hand side:
\[
3(x + 5) = 3x + 15
\]

On the right-hand side:
\[
2(x + 2) + 8 = 2x + 4 + 8 = 2x + 12
\]

So the inequality becomes:
\[
3x + 15 < 2x + 12
\]

### Step 2: Move the variables to one side

Subtract \( 2x \) from both sides:
\[
3x - 2x + 15 < 12
\]
This simplifies to:
\[
x + 15 < 12
\]

### Step 3: Solve for \( x \)

Subtract 15 from both sides:
\[
x < 12 - 15
\]
\[
x < -3
\]

### Final solution:
\[
x < -3
\]

Let me know if you need more details or have any questions!

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Here are 5 related questions for further practice:
1. Solve \( 5(x + 2) \geq 3(x + 4) + 7 \).
2. Solve \( 4x - 7 < 2x + 5 \).
3. Solve the compound inequality \( 2x + 1 < 5x + 7 < 8x + 12 \).
4. Solve \( 7(x - 1) > 4x + 5 \).
5. Solve \( 6x + 9 \leq 2(x + 4) + 13 \).

**Tip:** Always simplify both sides of the inequality before moving variables and constants around.

Learn how to solve the linear inequality 3(x + 5) < 2(x + 2) + 8 through detailed step-by-step expansion and simplification. This problem introduces key algebraic concepts such as the distributive property and linear inequality manipulation. Suitable for students in Grades 7-9.

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