Simple inequality $\frac{4x+1}{2x-3}>2$ 
$$\frac{4x+1}{2x-3}>2$$

I have started with looking at positive/negative situations and there are 2 or that both expressions are positive or both are negative.
If both a positive we can solve for  $$\frac{4x+1}{2x-3}>2$$
 $${4x+1}>2(2x-3)$$
$${4x+1}>4x-6$$
$${0x}>-7$$
Or both are negative and then  $${4x+1}<2(2x-3)$$
And again $$0x<-7$$
So there is not answer?

HINT
Notice that
$$
\frac{4x+1}{2x-3}
 = \frac{4x-6}{2x-3} + \frac{7}{2x-3}
 = 2 + \frac{7}{2x-3}
$$
which is more than $2$ when the fraction is positive. Can you take it from here?
