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 $A \leq -B$.