Question

solve for s.
2|s + 3| ≥ 2
write a compound inequality like 1 < x < 3 or like x < 1 or x > 3. use integers, proper fractions, or improper fractions in simplest form.

Explanation:

Step1: Isolate the absolute - value.

Divide both sides of $2|s + 3|\geq2$ by 2: $|s + 3|\geq1$.

Step2: Consider two cases.

Case 1: $s + 3\geq1$. Subtract 3 from both sides: $s\geq1 - 3$, so $s\geq - 2$.
Case 2: $s + 3\leq - 1$. Subtract 3 from both sides: $s\leq - 1-3$, so $s\leq - 4$.

Answer:

$s \leq - 4$ or $s\geq - 2$