Question

solve the absolute value inequality. other than ∅, use interval notation to express the solution set and graph the solution set on a number line. 31) |3(x + 1)+9|≤15

Explanation:

Step1: Simplify the absolute - value inequality

First, simplify the expression inside the absolute - value. $|3(x + 1)+9|\leq15$ becomes $|3x+3 + 9|\leq15$, then $|3x + 12|\leq15$.

Step2: Rewrite the absolute - value inequality as a compound inequality

If $|u|\leq a$ ($a\geq0$), then $-a\leq u\leq a$. So, $-15\leq3x + 12\leq15$.

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

Subtract 12 from all parts: $-15-12\leq3x+12 - 12\leq15 - 12$. This gives $-27\leq3x\leq3$.

Step4: Solve for x

Divide all parts by 3: $\frac{-27}{3}\leq\frac{3x}{3}\leq\frac{3}{3}$, which simplifies to $-9\leq x\leq1$.

Answer:

The solution set in interval notation is $[-9,1]$. On the number line, we mark closed circles at $x=-9$ and $x = 1$ and shade the region between them.