Here is one more that isnotright I think but I can't figure anything else?

$\displaystyle \frac {x+3}{2x-7}-\frac {2x-1}{x-3}=0$

I have to solve this and show validation.

Here is what I did (or tried to do):

$\displaystyle \frac {(x+3)}{(2x-7)}=\frac {(x+3)}{(2x+1)}$

$\displaystyle (x+3)(2x+1)=(2x-7)(x+3)$

$\displaystyle 2x^2+x+6x+3=2x^2+6x-7x-21$

$\displaystyle 2x^2+x+6x-2x^2-6x+7x=-21-3$

$\displaystyle 8x=-24$

$\displaystyle x=-3$

Validation:

$\displaystyle \frac {-3+3}{2(-3)-7} = \frac {-3+3}{2(-3)+1}$

$\displaystyle \frac {0}{-13}=\frac {0}{-5}$

$\displaystyle 0=0$