Question 884130
What you expressed is not clear.   A guess,  {{{(y+2)/(y-1)-(4-y)/2=7/3}}}.


Multiply both sides by the simplest common denominator, {{{2*3*(y-1)}}}.

{{{6(y+2)-(4-y)3*(y-1)=7(2)(y-1)}}}
{{{6y+12-3(4-y)(y-1)=14y-14}}}
{{{12-3(4-y)(y-1)=-14}}}
{{{26+3(y-4)(y-1)=0}}}
{{{26+3(y^2-5y+4)=0}}}
{{{3y^2-15y+12+26=0}}}
{{{highlight_green(3y^2-15y+38=0)}}}
You can finish the rest from here.  General solution for quadratic formula would be the way to do.  (Maybe check for factorability.)