Question

1. solve and graph: 3|2x - 1|-2≤10

Explanation:

Step1: Isolate the absolute - value expression

Add 2 to both sides of the inequality $3|2x - 1|-2\leq10$.
$3|2x - 1|\leq10 + 2$, so $3|2x - 1|\leq12$. Then divide both sides by 3: $|2x - 1|\leq4$.

Step2: Split the absolute - value inequality into two cases

Case 1: $2x-1\geq0$ (i.e., $2x - 1\leq4$ when $2x\geq1$ or $x\geq\frac{1}{2}$).
Solve $2x - 1\leq4$. Add 1 to both sides: $2x\leq4 + 1$, so $2x\leq5$. Then divide by 2: $x\leq\frac{5}{2}$.
Case 2: $2x-1\lt0$ (i.e., $-(2x - 1)\leq4$ when $2x\lt1$ or $x\lt\frac{1}{2}$).
Expand $-(2x - 1)\leq4$ to $-2x + 1\leq4$. Subtract 1 from both sides: $-2x\leq4 - 1$, so $-2x\leq3$. Divide both sides by - 2 and reverse the inequality sign: $x\geq-\frac{3}{2}$.

Answer:

The solution of the inequality is $-\frac{3}{2}\leq x\leq\frac{5}{2}$. To graph it, draw a number line, mark the points $x =-\frac{3}{2}$ and $x=\frac{5}{2}$ with closed circles (because the inequality includes equality), and shade the region between these two points.