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

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...
