# How do you graph and solve  3|x-1|+2>=8?

Dec 15, 2015

$\left(- \infty , - 1\right] \text{ U } \left[3 , \infty\right)$

#### Explanation:

$| x | \ge k \implies \text{ " x>= k " " or " " } x \le - k$

$| x | \le k \implies \text{ } k \le x \le k$

Given:

$3 | x - 1 | + 2 \ge 8$

Isolate the inequality, so absolute value can be by itself.

Step 1: Subtract 2 to both side

$3 | x - 1 | + \cancel{2 \textcolor{red}{- 2}} \ge 8 \textcolor{red}{- 2}$

$3 | x - 1 | \ge 6$

Step 2 : Divide by 3 to both side

$\frac{3 | x - 1 |}{\textcolor{b l u e}{3}} \ge \frac{6}{\textcolor{b l u e}{3}}$

$| x - 1 | \ge 2$

Step 3 : Undo the absolute value like the information mention at the beginning

$\implies x - 1 \ge 2 \text{ " " " or " " " } x - 1 \le - 2$

Step 4 : Solve for $x$

$x \ge 3 \text{ " " or " " } x \le - 1$

Step 5: Draw the number line and determine the interval from Step 4.

Solution: $\left(- \infty , - 1\right] \text{ U } \left[3 , \infty\right)$