Express x in terms of y for the expression:

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

and hence evaluate:

$\displaystyle \frac {7x-2y}{3x+y}$

So i express x in terms of y in the first expression and get:

$\displaystyle x=\frac {7y}{8}$

And then proceed to evaluate the second part of the question as follows:

$\displaystyle \frac {7\frac{7y}{8}-2y}{3\frac{7y}{8}+y}$

$\displaystyle \frac {\frac {49y}{8}-\frac{2y}{1}}{\frac {21y}{8}+\frac{y}{1}}$

Multiply here by the LCD 8:

$\displaystyle \frac {\frac {49y}{8}-\frac{16y}{8}}{\frac {21y}{8}+\frac{8y}{8}}$

Which gives:

$\displaystyle \frac {\frac {33y}{8}}{\frac {29y}{8}}$

The final answer in the book is:

$\displaystyle \frac {33}{29}$

So do i just cancel the eights and y's in my above answer? Also is my method in evaluating correct? Any help would be great. Thanks!

P.S why does my latex come out so small :?