Solve the following inequality *algebraically*.

vertical bar, x, plus, 6, vertical bar, is less than or equal to, 8

∣*x*+6∣≤8

To solve the inequality \( |x + 6| \leq 8 \), we need to break it down into two separate cases, since absolute value inequalities involve two scenarios:

### Step 1: Break into two cases

For \( |x + 6| \leq 8 \), the expression inside the absolute value, \( x + 6 \), can either be between \(-8\) and \(8\).

Thus, we can write the inequality as:
\[
-8 \leq x + 6 \leq 8
\]

### Step 2: Solve the compound inequality

Now, solve for \( x \) by subtracting 6 from all parts of the inequality:
\[
-8 - 6 \leq x + 6 - 6 \leq 8 - 6
\]
\[
-14 \leq x \leq 2
\]

### Final Answer:

The solution to the inequality \( |x + 6| \leq 8 \) is:
\[
x \in [-14, 2]
\]

This means that \( x \) must lie within the interval \([-14, 2]\), inclusive of both endpoints.

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Would you like further details or clarification on any of the steps?

Here are some related questions for further practice:
1. Solve \( |2x - 3| \leq 7 \).
2. Solve \( |x - 5| > 10 \) algebraically.
3. Find the solution set for \( |3x + 1| \geq 4 \).
4. Solve \( |x + 9| \leq 5 \).
5. Determine when \( |x - 4| = 6 \).

**Tip**: When solving absolute value inequalities, always remember that \( |A| \leq B \) can be rewritten as \( -B \leq A \leq B \), and \( |A| \geq B \) splits into two cases: \( A \geq B \) or \( A \leq -B \).

Solve the following inequality algebraically: ∣x + 6∣ ≤ 8

Learn how to solve the absolute value inequality ∣x + 6∣ ≤ 8 with a detailed step-by-step solution. This problem covers key algebraic concepts, including properties of absolute values and solving compound inequalities. Suitable for students in Grades 7-9.

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