Solve  |x+8| < 16
  Then graph the solution set on a separate sheet of paper.

To solve the inequality \( |x+8| < 16 \), we follow these steps:

### Step 1: Definition of Absolute Value Inequality
The absolute value inequality \( |x+8| < 16 \) can be rewritten as a compound inequality:
\[
-16 < x + 8 < 16
\]
This comes from the property of absolute value inequalities that:
\[
|A| < B \implies -B < A < B
\]
where \( A = x+8 \) and \( B = 16 \).

### Step 2: Solve the Compound Inequality
Now, solve the inequality:
\[
-16 < x + 8 < 16
\]

Subtract 8 from all parts of the inequality:
\[
-16 - 8 < x + 8 - 8 < 16 - 8
\]
\[
-24 < x < 8
\]

### Step 3: Solution Set
The solution to the inequality is:
\[
x \in (-24, 8)
\]
This means that \( x \) is any real number between -24 and 8, not including -24 or 8.

### Step 4: Graph the Solution Set
To graph the solution set, you can draw a number line and mark the interval from \( -24 \) to \( 8 \), excluding the endpoints (use open circles at \( -24 \) and \( 8 \) to indicate they are not included in the solution).

Would you like a visual graph or further details on any part of this solution?

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Here are five related questions to explore further:
1. How would the solution change if the inequality were \( |x+8| \leq 16 \)?
2. What happens if we have \( |x+8| > 16 \) instead?
3. Can you explain why the absolute value inequality turns into a compound inequality?
4. How would you solve \( |x+8| < -5 \)? What does the solution mean?
5. What if the inequality were \( |2x+3| < 10 \)?

**Tip:** When solving absolute value inequalities, always check whether the inequality is less than (compound inequality) or greater than (or) to determine the correct method of solution.

Solve |x+8| < 16. Then graph the solution set on a separate sheet of paper.

Discover how to solve the absolute value inequality |x+8| < 16 and graph the solution set. This guide explains how to transform the absolute value into a compound inequality and provides a detailed step-by-step solution. Suitable for students in Grades 9-10.

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