# Thread: Absolute Values Question?

1. ## Absolute Values Question?

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

2. Originally Posted by jamman790
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

$\frac {(2x - 3)^2}{(4x + 1)^2} \ge 2$

now solve for $x$. 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:

$\frac {2x - 3}{4x + 1} \ge 2$ and $- \frac {2x - 3}{4x + 1} \ge 2$

3. It says...show all solutions?...but what does that mean?

4. Originally Posted by Jhevon
one way to go, is to square both sides. you get

$\frac {(2x - 3)^2}{(4x + 1)^2} \ge 2$

now solve for $x$. 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:

$\frac {2x - 3}{4x + 1} \ge 2$ and $- \frac {2x - 3}{4x + 1} \ge 2$
I believe you'll find if you square both sides then the inequation becomes

$\frac{(2x - 3)^2}{(4x + 1)^2}\geq 4$.

5. Originally Posted by Prove It
I believe you'll find if you square both sides then the inequation becomes

$\frac{(2x - 3)^2}{(4x + 1)^2}\geq 4$.
but of course. silly mistake on my part