(3y-1)-1/2(2y+6/8-y-5/3)
Your title said to 'find x' and your expression is in 'y'.
I'm having trouble untangling your expression. Is it one of these:
(1) $\displaystyle 3y-1-\frac{1}{2}\left(\frac{2y+6}{8-y}-\frac{5}{3}\right)$
(2) $\displaystyle 3y-1-\frac{1}{2}\left(2y+\frac{6}{8}-y-\frac{5}{3}\right)$
(3) $\displaystyle 3y-1-\frac{1}{2}\left(\frac{2y+6}{8}-\frac{y-5}{3}\right)$
Or something else entirely?
(3) $\displaystyle 3y-1-\frac{1}{2}\left(\frac{2y+6}{8}-\frac{y-5}{3}\right)$
Simplify what's in parenthesis. The common denominator will be 24.
$\displaystyle 3y-1-\frac{1}{2}\left(\frac{3(2y+6)}{24}-\frac{8(y-5)}{24}\right)$
$\displaystyle 3y-1-\frac{1}{2}\left(\frac{6y+18}{24}-\frac{8y-40}{24}\right)$
$\displaystyle 3y-1-\frac{1}{2}\left(\frac{6y+18-8y+40}{24}\right)$
$\displaystyle 3y-1-\frac{1}{2}\left(\frac{-2y+58}{24}\right)$
$\displaystyle 3y-1+\frac{2y-58}{48}$
Put everything over a denominator of 48.
$\displaystyle \frac{144y-48+2y-58}{48}$
$\displaystyle \frac{146y-106}{48}$
$\displaystyle \boxed{\frac{73y}{24}-\frac{53}{24}}$
And by the way....Go Federer!