$\displaystyle \frac{2}{x+1}=\frac{2x}{2x-3}+3$
after having a few goes at this equation i have become stuck would any be able to show us the best way to approch it?
I'd start with a common denominator.
$\displaystyle \frac{2}{x+1}=\frac{2x}{2x-3}+3$
$\displaystyle \frac{2}{x+1}=\frac{2x}{2x-3}+\frac{3(2x-3)}{2x-3}$
$\displaystyle \frac{2}{x+1}=\frac{2x+3(2x-3)}{2x-3}$
(You could simplify the right further, and indeed it is advisable to do so, but I'll leave you to that.)
Then I'd cross multiply to obtain:
$\displaystyle 2(2x-3)=[2x+3(2x-3)](x+1)$
Do you follow these steps? If yes, then finish; if not, then ask and I'll clarify
Edit: Grammar changes.
Anyone on this site is perfectly willing to do that if we know where your problems are arising. What exactly is it that you've not quite nailed? Do you need a review on fractions? They will come up everywhere - trig, calculus, you just can't get away from them! If you understand common denominators, then it's just algebraic manipulation, and that will come easier with practice.
As for other sites, you could try Wolfram Alpha, although while that tends to give you a fully worked through solution and it's explained, albeit in minimal detail, it sometimes does not presents us with the most rational approach to a question for a human mind, and may give you a false sense of understanding if you come to rely on it.