solve the following absolute value inequality. 3|x - 7| ≤ 27 x ≤ ? x ≥

Explanation:

Step1: Isolate the absolute - value expression

Divide both sides of the inequality $3|x - 7|\leq27$ by 3. We get $|x - 7|\leq9$. [|x - 7|\leq9]

Step2: Consider the two cases for the absolute - value inequality

Case 1: When $x−7\geq0$ (i.e., $x\geq7$), the inequality becomes $x - 7\leq9$. Add 7 to both sides: $x\leq9 + 7$, so $x\leq16$. Case 2: When $x−7<0$ (i.e., $x<7$), the inequality becomes $-(x - 7)\leq9$. Expand to get $-x + 7\leq9$. Subtract 7 from both sides: $-x\leq9 - 7$, so $-x\leq2$. Multiply both sides by - 1 and reverse the inequality sign, we get $x\geq - 2$.

Answer:

$x\leq16$ $x\geq - 2$