Solution set!

$\begin{aligned} \frac {2x+1}{3x-2} & > 2 \end{aligned}$

$\implies \begin{cases} 2x+1 \ {\color{#3D99F6}>} \ 2(3x-2) & \text{for }3x - 2 \ {\color{#3D99F6}>} \ 0 & \implies x \ {\color{#3D99F6}>} \ \dfrac 23 & ...(1) \\ 2x+1 \ {\color{#D61F06}<} \ 2(3x-2) & \text{for }3x - 2 \ {\color{#D61F06}<} \ 0 & \implies x \ {\color{#D61F06}<} \ \dfrac 23 & ... (2) \end{cases}$

$\begin{aligned} (1): \quad 2x + 1 & > 6x -4 & \text{for } x > \frac 23 \\ 4x & < 5 \\ \implies \frac 23 < x & < \frac 54 & \color{#3D99F6} \text{A solution.} \end{aligned}$

$\begin{aligned} (2): \quad 2x + 1 & < 6x -4 & \text{for } x < \frac 23 \\ 4x & > 5 \\ \implies \frac 23 > x & > \frac 54 & \color{#D61F06} \text{No solution.} \end{aligned}$

Therefore, $\dfrac 23 < x < \dfrac 54$ $\implies ab = \dfrac 23 \times \dfrac 54 = \dfrac 56$ $\implies m+n = 5+6 = \boxed{11}$

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

$\frac{2x+1}{3x-2}-2>0$

$\frac{2x+1-6x+4}{3x-2}>0$

$\frac{5-4x}{3x-2}>0$

$\frac{4x-5}{3x-2}<0$

So $(a,b)$ is $(\frac{2}{3},\frac{5}{4})$ $ab=\frac{5}{6}$

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

Let's evaluate the following two cases:

$\text {Case 1: } 3x - 2 > 0 \iff x > \frac {2}{3}$

$2x + 1 > 2(3x -2)$

$2x + 1 > 6x - 4$

$5 > 4x$

$\frac {5}{4} > x$

Hence, our solution in this case:

$\frac {2}{3} < x < \frac {5}{4}$

$\text {Case 2: } 3x - 2 < 0 \iff x < \frac {2}{3}$

Multiplying by a negative number changes the direction of the inequality to the opposite:

$2x + 1 < 2(3x -2)$

$2x + 1 < 6x - 4$

$5 < 4x$

$\frac {5}{4} < x$

$\text {Since } x < \frac {2}{3} < \frac {5}{4} \text { , therefore we don't have a solution in this case.}$

This means, that our only solution is:

$x \in ( \frac {2}{3} , \frac {5}{4} )$

$a = \frac {2}{3}$

$b = \frac {5}{4}$

$ab = \frac {2}{3} × \frac {5}{4} = \frac {10}{12} = \frac {5}{6}$

Hence m = 5, n = 6 and our solution should be:

$m + n = 5 + 6 = \boxed {11}$