# How do you solve -abs(5/3y+4)>2/5?

Apr 22, 2015

The answer is x in (-oo;-2 16/25) uu (-2 4/25;+oo)

First you have to simplify the absolute value. To do this you use the identity

$| x | > a \iff \left(x < - a \vee x > a\right)$
So you get a set of linear inequalities

$\frac{5}{3} y + 4 > \frac{2}{5} \vee \frac{5}{3} y + 4 < - \frac{2}{5}$

To simplify the inequalities you can multiply both sides by 15 to get rid of the fractions.