Question 1085706
I think this is what you mean.
{{{(y+2)/4+ (y-3)/8= 1/y}}}
Multiply both sides by {{{8y}}}.
{{{2y(y+2)+y(y-3)=8}}}
{{{2y^2+4y+y^2-3y=8}}}
{{{3y^2+y-8=0}}}
{{{3(y^2+(1/3)y)-8=0}}}
{{{3(y^2+(1/3)y+(1/6)^2)-8=3(1/6)^2}}}
{{{3(y+1/6)^2=3/36+8}}}
{{{3(y+1/6)^2=1/12+96/12}}}
{{{3(y+1/6)^2=97/12}}}
{{{(y+1/6)^2=97/36}}}
{{{y+1/6=0 +- sqrt(97)/6}}}
{{{y=-1/6+- sqrt(97)/6}}}