To find values of $x$ that are within 4 units of 12, we need to set up an inequality that captures this condition.

"Within 4 units" means that the distance between $x$ and 12 is less than or equal to 4. Mathematically, the absolute value of the difference between $x$ and 12 must be less than or equal to 4:

$|x - 12| \leq 4$

Now, let's break down the absolute value inequality. The absolute value inequality $|x - 12| \leq 4$ means that $x - 12$ lies between -4 and 4:

$-4 \leq x - 12 \leq 4$

Next, solve for $x$ by adding 12 to all parts of the inequality:

$-4 + 12 \leq x \leq 4 + 12$

$8 \leq x \leq 16$

Thus, the inequality representing all values of $x$ that are within 4 units of 12 is:

$8 \leq x \leq 16$

Would you like further details or have any questions?

Here are some related questions:

1. How would the inequality change if the range were "within 5 units" of 12?
2. What values of $x$ would satisfy $|x - 12| > 4$?
3. What is the geometric interpretation of $|x - 12| \leq 4$?
4. How would the inequality look for values of $x$ within 4 units of a different number, like 20?
5. Can you graph the solution $8 \leq x \leq 16$ on a number line?

Tip: When dealing with absolute value inequalities, think of them as describing a range or distance around a central point.