Question

question
solve the following inequality algebraically.
|x + 3| ≤ 8

Explanation:

Step1: Apply absolute value rule

For $|A| \leq B$, $-B \leq A \leq B$. So:
$-8 \leq x + 3 \leq 8$

Step2: Subtract 3 from all parts

Subtract 3 to isolate $x$:
$-8 - 3 \leq x \leq 8 - 3$

Step3: Simplify all terms

Calculate the bounds:
$-11 \leq x \leq 5$

Answer:

$-11 \leq x \leq 5$