One more formulae manipulation question Just making sure I am right...
Make c the subject of the formula:
$ y = \frac{2a+b}{3c -d}$
so $ 3c -d = \frac{2a+b}{y}$
so $ 3c = \frac{2a+b}{y} +d$
so $ c =\frac{6a+3b}{y} + \frac{d}{3}$
If I am not right please say why thanks
 A: You are correct up to this point:$$3c-d=\frac{2a+b}{y}$$you then add $d$ to both sides to get:$$3c=\frac{2a+b}{y}+d$$Finally you divide both sides by $3$ to get:$$c=\frac{1}{3}\left(\frac{2a+b}{y}+d\right)=\frac{1}{3}\times\frac{2a+b}{y}+\frac{1}{3}\times d$$Hopefully you can complete from here...
A: Yes, this is correct: $$3c = \frac{2a+b}{y} +d$$
But...Your last line is incorrect; you should get $$c=\frac{2a+b}{3y} + \frac d3$$ since we need to divide both sides by 3 in order to isolate $c$.
Remark Do be careful. The above only holds if $2a \neq -b$, else $y = 0$ and we cannot divide by zero. So I'd simply add the provision: $2a \neq -b$ with the equation for $c$.
A: As Mufasa pointed out, you are correct up to the point where you added $b$ by accident instead of $d$. There is actually a simpler way, I think, to get what you want (solving for $c$, that is):
\begin{align}
y = \frac{2a+b}{3c-d} &\Longleftrightarrow y(3c-d)=2a+b\\[1em]
                      &\Longleftrightarrow 3yc-yd = 2a+b\\[1em]
                      &\Longleftrightarrow c = \frac{2a+b+yd}{3y}.
\end{align}
Does that solve it in the way you desired? 
