# How do you solve abs(20+1/2x)>6?

May 30, 2018

See a solution process below:

#### Explanation:

The absolute value function takes any term and transforms it to its non-negative form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent.

$- 6 > 20 + \frac{1}{2} x > 6$

$- 6 - \textcolor{red}{20} > 20 - \textcolor{red}{20} + \frac{1}{2} x > 6 - \textcolor{red}{20}$

$- 26 > 0 + \frac{1}{2} x > - 14$

$\textcolor{red}{2} \times - 26 > \textcolor{red}{2} \times \frac{1}{2} x > \textcolor{red}{2} \times - 14$

$- 52 > \frac{\textcolor{red}{2}}{2} x > - 28$

$- 52 > 1 x > - 28$

$- 52 > x > - 28$

Or

$x < - 52$; $x > - 28$

Or, in interval notation:

$\left(- \infty , - 52\right)$; $\left(- 28 , + \infty\right)$

May 30, 2018

$x < - 52$ or $x > - 28$

#### Explanation:

One way to approach the evaluation of an absolute value inequality like $\left\mid 20 + \frac{1}{2} x \right\mid > 6$ is to first temporarilly replace $\left(20 + \frac{1}{2} x\right)$ with some other variable.
For example if we let $z = \left(20 + \frac{1}{2} x\right)$ then we have the simpler inequality:
$\textcolor{w h i t e}{\text{XXX}} \left\mid z \right\mid > 6$
Presented on the number line this would look like:

That is
$\textcolor{w h i t e}{\text{XXXxxxxxx")z < -6color(white)("xxxx")"or"color(white)("xxxxxxxxx}} z > + 6$
then since we know $z = 20 + \frac{1}{2} x$
these become
$\textcolor{w h i t e}{\text{XXX")20+1/2x < -6color(white)("xxxx")"or"color(white)("xxxx}} 20 + \frac{1}{2} x > + 6$

$\textcolor{w h i t e}{\text{XXX20+")1/2x < -26color(white)("xxxx")"or"color(white)("xxxx20+}} \frac{1}{2} x > - 14$

$\textcolor{w h i t e}{\text{XXX"20+2)x < -52color(white)("xxxx"20+2)"or"color(white)("xxxx}} x > - 28$