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?
 A: 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?
A: With these kind of inequalities I believe it is better to bring everything to left side and make everything into one fraction:
$$\frac{4x+1}{2x-3}-\frac{2(2x-3)}{2x-3}>0$$
Simplifying to: $\frac{7}{2x-3}>0$. Since the numerator is positive, focus on denominator.
A: Let's use equivalent but simpler inequalities
$$
\frac{4x+1}{2x-3}>2 \quad \Leftrightarrow \quad
\frac{4x+1}{2x-3}-2>0 \quad \Leftrightarrow \quad
\frac{7}{2x-3}>0
$$
Therefore we must have 
$$
2x-3>0 \quad \Leftrightarrow \quad x>{3\over2}.
$$
A: You made a mistake moving from
$$\frac{4x+1}{2x-3}>2$$
to
$$4x+1 > 2(2x-3).$$
The reason this is potentially wrong is that when you multiply both sides of an inequality by some number $c$, then the inequality flips if $c < 0$. In this case, it depends on whether $2x-3$ is positive or negative.
So, we should have
$$4x+1 > 2(2x-3),\quad \text{if }2x-3 > 0,\\
4x+1 < 2(2x-3),\quad \text{if }2x-3 < 0.$$
In the first case, we simplify (as you had done) to get $4x + 1 > 4x - 6$, so $0x > -7$. This is better written as
$$0 > -7,$$
which is always true. (This was a second mistake in your reasoning: you stopped at $0x > -7$ without making the right conclusion.) Keep in mind, however, that we are assuming for this first case that $2x-3 > 0$, i.e., $x > \frac32$.
In the second case, when $x < \frac32$, we similarly simplify to get
$$0 < -7,$$
which, of course, is never true. We therefore reject $x < \frac32$. In conclusion, the initial inequality holds for $x \in (\frac32,\infty)$.
Alternatively, as others have pointed out, you can move everything to one side of the equation first, then solve. This has the advantage of not having to deal with two cases.
A: For $x > \frac{3}{2}$, we rearrange to $$4x+1 > 4x -6 \iff 1 > -6$$ and so it holds for all $x > \frac{3}{2}$.
If $x < \frac{3}{2}$, we rearrange to  $$4x+1 < 4x-6 \iff 1 < -6$$ and hence it does not hold for $x < \frac{3}{2}$.
