# How do you solve -3x + 2 >11 or 5(x - 2) <0?

Jul 13, 2017

Solution : $x < - 3$ , in interval notation : $\left(- \infty , - 3\right)$

#### Explanation:

1) $- 3 x + 2 > 11 \mathmr{and} - 3 x > 9 \mathmr{and} - x > 3 \mathmr{and} x < - 3$

2) $5 \left(x - 2\right) < 0 \mathmr{and} x - 2 < 0 \mathmr{and} x < 2$

$x < - 3$ satisfies both inequaity.

Solution : $x < - 3$ , in interval notation : $\left(- \infty , - 3\right)$ [Ans]