The problem presented is an absolute value inequality:

$\frac{|x + 6|}{8} \leq 1$

To solve this, follow these steps:

Step 1: Eliminate the denominator.

Multiply both sides of the inequality by 8 to get rid of the denominator:

$|x + 6| \leq 8$

Step 2: Split into two inequalities.

Since we are dealing with an absolute value, we split this into two separate inequalities:

$-8 \leq x + 6 \leq 8$

Step 3: Solve for $x$.

Now, solve for $x$ by subtracting 6 from all parts of the inequality:

$-8 - 6 \leq x \leq 8 - 6$

$-14 \leq x \leq 2$

Final Answer:

The solution is:

$x \geq -14 \quad \text{and} \quad x \leq 2$

So, the solution in interval notation would be:

$x \in [-14, 2]$

The boxes in the image would be filled with:

• $x \leq 2$ for the first blank.
• $x \geq -14$ for the second blank.

Would you like more details or have any further questions?

Here are 5 related questions for practice:

1. How do you solve an absolute value inequality like $|2x - 3| \leq 5$?
2. What happens if the inequality is strict, for example $\frac{|x + 3|}{2} < 4$?
3. How do you solve absolute value inequalities when the inequality symbol is flipped, such as $\frac{|x - 4|}{5} > 3$?
4. What is the graphical interpretation of the solution to an absolute value inequality?
5. Can you explain how to solve an absolute value inequality with two absolute values, such as $|x - 1| + |x + 2| \leq 7$?

Tip: When solving absolute value inequalities, always split the inequality into two cases and remove the absolute value bars!