I'm trying to take the inverse of the function $\displaystyle \frac{2x-3}{x+4}$

I think I've got it following this procedure

$\displaystyle x=\frac{2y-3}{y+4}$

$\displaystyle x(y+4) = 2y-3$

$\displaystyle xy+4x = 2y-3$

$\displaystyle xy-2y = -3-4x$

$\displaystyle y(x-2) = -3-4x$

[tex]y = \frac{-3-4x}{x-2}[tex]

$\displaystyle y = -\frac{4x+3}{x-2}$

I worked that a few times and am pretty confident in my answer. The proof is in plugging that back into the original equation and seeing if I get x, and this is where I'm fumbling about

$\displaystyle \frac{2(-\frac{4x+3}{x-2})-3}{-(\frac{4x+3}{x-2})+4}$

$\displaystyle \frac{-(\frac{6+8x-3x+6}{x-2})}{-(\frac{3+4x+4x-8}{x-2})}$

$\displaystyle \frac{-(\frac{5x+12}{x-2})}{-(\frac{8x-5}{x-2})}$

$\displaystyle -\frac{5x+12}{x-2} * -\frac{x-2}{8x-5}$

$\displaystyle \frac{5x+12}{8x+5}$

which, unless I'm blind, does not equal x. I hope the above isn't too confusing.

Assuming my initial answer is correct, what am I doing wrong when I'm trying to verify it?