Question

solve the following inequality algebraically. |x + 3| ≤ 8 answer attempt 1 out of 5

Explanation:

Step1: Apply absolute - value rule

If \(|a|\leq b\) (\(b\geq0\)), then \(-b\leq a\leq b\). Here \(a = x + 3\) and \(b = 8\), so \(-8\leq x+3\leq8\).

Step2: Solve the left - hand side of the compound inequality

Subtract 3 from both sides of \(-8\leq x + 3\): \(-8-3\leq x+3 - 3\), which simplifies to \(-11\leq x\).

Step3: Solve the right - hand side of the compound inequality

Subtract 3 from both sides of \(x + 3\leq8\): \(x+3-3\leq8 - 3\), which simplifies to \(x\leq5\).

Answer:

\([-11,5]\)