# How do you solve and graph m-2< -8 or m/8>1?

Jun 29, 2018

$m < - 6 \mathmr{and} m > 8$ or in interval notation $m \in \left(- \infty , - 6\right) \cup \left(8 , + \infty\right)$

#### Explanation:

$m - 2 < - 8 \mathmr{and} \frac{m}{8} > 1$

First consider: $m - 2 < - 8$

$\to m < - 8 + 2$

$m < - 6$

Next consider: $\frac{m}{8} > 1$

Multiply both sides by 8.

$\to m > 8$

Hence our compound inequality simplifies to: $m < - 6 \mathmr{and} m > 8$
or in interval notation $m \in \left(- \infty , - 6\right) \cup \left(8 , + \infty\right)$

We can represent this graphically on a 2D plane where $m$ is the horizontal axis, as below.

graph{(x+6)(-x+8)<0 [-14.24, 14.23, -7.12, 7.11]}

So, $m$ may take all values on the horizontal axis in the shaded areas.