# How do you solve abs(4x-5)>16 ?

Jul 7, 2015

$\left\mid 4 x - 5 \right\mid > 16$ when $x < - \frac{11}{4}$ and $x > \frac{21}{4}$ #### Explanation:

The "absolute value" function is defined as :

When $x \in \mathbb{R} ,$

|x|=x,ifx≥0
|x|=−x,ifx<0

And $4 x - 5 \ge 0$ when $x \ge \frac{5}{4}$

Then, $\left\mid 4 x - 5 \right\mid = 4 x - 5 , \mathmr{if} x \ge \frac{5}{4}$
And, $\left\mid 4 x - 5 \right\mid = - \left(4 x - 5\right) = - 4 x + 5 , \mathmr{if} x < \frac{5}{4}$

To resolve the inequality, we have to separate them in two parts :
when $x \ge \frac{5}{4}$ and when $x < \frac{5}{4}$

If $x \ge \frac{5}{4}$ :
$\left\mid 4 x - 5 \right\mid > 16 \iff 4 x - 5 > 16 \iff 4 x > 21 \iff \textcolor{red}{x > \frac{21}{4}}$


If $x < \frac{5}{4}$ :
$\left\mid 4 x - 5 \right\mid > 16 \iff - 4 x + 5 > 16 \iff - 4 x > 11 \iff - x > \frac{11}{4}$

Recall : when we multiply or divide each side of an inequality by a negative number, we have to inverse the sign of the inequality

Ex : If $- a > 0$ then $a \textcolor{red}{<} 0$ (multiplied by $- 1$)


Then $\left\mid 4 x - 5 \right\mid > 16 \iff \textcolor{red}{x < - \frac{11}{4}}$


Therefore, $\left\mid 4 x - 5 \right\mid > 16$ when $x < - \frac{11}{4}$ and $x > \frac{21}{4}$


Other notation :
$\left\mid 4 x - 5 \right\mid > 16$ if x in ]-oo;-11/4[uu]21/4;+oo[