# How do you solve -6x+14<-28 or 9x+15<-12?

Jun 20, 2017

$\left(- \infty , - 3\right) \cup \left(7 , \infty\right)$

#### Explanation:

Solve each inequality separately, then combine them with the "or" operator, and then simplify if possible.

$- 6 x + 14 < - 28$

$- 6 x < - 42$

$6 x > 42$

$x > 7$

Now, we solve the second inequality:

$9 x + 15 < - 12$

$9 x < - 27$

$x < - 3$

Combining the two, we get:

$x < - 3 \mathmr{and} x > 7$

This cannot be simplified, since the two solution regions do not overlap. The more formal way to write this (with interval notation) is:

$\left(- \infty , - 3\right) \cup \left(7 , \infty\right)$

On a number line, the solution set looks like this: