# How do you solve abs(3x + 2) - 1 >=10?

##### 1 Answer
Jul 11, 2016

$x \le - \frac{13}{3}$
$x \ge 3$

#### Explanation:

$I 3 x + 2 I \ge 11$
Separate solving into 2 parts:
a.$\left(3 x + 2\right) \ge 11$
3x >= 9@ x >= 3 b. -(3x + 2) >= 11  - 3x - 2 >= 11  - 3x >= 13 x <= -13/3 Note. The 2 end points (-13/3)# and (3) are included in the solution set.
Graph:

=============== -13/3 ---------0---------- 3 =============