ok so i have never been given a quesiton like this and im trying to figure it out for the first time. here it is
Solve |2x-3/4x+1| = > 2
one way to go, is to square both sides. you get
$\displaystyle \frac {(2x - 3)^2}{(4x + 1)^2} \ge 2$
now solve for $\displaystyle x$. be sure to check if there are any extraneous solutions.
another approach is to account for the inside of the absolute values to be positive and negative. that is, solve the following two inequalities separately:
$\displaystyle \frac {2x - 3}{4x + 1} \ge 2$ and $\displaystyle - \frac {2x - 3}{4x + 1} \ge 2$