# How do you solve 3abs(x-10)+2>=8?

Jun 22, 2015

$x \le 8$ or $x \ge 12$

#### Explanation:

Given $3 \left(x - 10\right) + 2 \ge 8$

We can subtract $2$ from both sides without effecting the orientation of the inequality:
$\textcolor{w h i t e}{\text{XXXX}}$$3 \left\mid x - 10 \right\mid \ge 6$

Similarly dividing by a value $> 0$ does not effect the orientation of the inequality:
$\textcolor{w h i t e}{\text{XXXX}}$$\left\mid x - 10 \right\mid \ge 2$

Case 1: $x < 10$
$\textcolor{w h i t e}{\text{XXXX}}$$\left(x - 10\right)$ is negative so we have
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$10 - x \ge 2$
$\textcolor{w h i t e}{\text{XXXX}}$Subracting 10 from both sides:
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$- x \ge - 8$
$\textcolor{w h i t e}{\text{XXXX}}$Multiplying by $\left(- 1\right)$ and remembering to reverse the direction of the inequality for a negative multiplier:
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$x \le 8$

Combining the two conditions $x < 10$ and $x \le 8$ for Case 1, we have
$\textcolor{w h i t e}{\text{XXXX}}$$x \le 8$

Case 2: $x \ge 10$
$\textcolor{w h i t e}{\text{XXXX}}$$\left(x - 10\right)$ is positive so we have
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$x - 10 \ge 2$
$\textcolor{w h i t e}{\text{XXXX}}$Adding 10 to both sides:
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$x \ge 12$

Combining the two conditions $x \ge 10$ and $x \ge 12$ for Case 2, we have
$\textcolor{w h i t e}{\text{XXXX}}$$x \ge 12$

Combining Case 1 with Case 2
(Note either Case 1 or Case 2 is possible)
$\textcolor{w h i t e}{\text{XXXX}}$$x \le 8$ or $x \ge 12$