Question 435357
{{{2x^2-xy+y^2=8}}}

{{{xy=4}}}....=>...{{{x=4/y}}}.substitute in {{{2x^2-xy+y^2=8}}}


{{{2(4/y)^2-(4/y)y+y^2=8}}}....solve for {{{y}}}

{{{2(16/y^2)-(4y/y)+y^2=8}}}

{{{2(16/y^2)-(4cross(y)/cross(y))+y^2=8}}}

{{{2(16/y^2)-4+y^2=8}}}...both sides divide by {{{2}}}

{{{2(16/y^2)/2-4/2+y^2/2=8/2}}}

{{{16/y^2- 2 +y^2/2=4}}}

{{{16/y^2+y^2/2=4+2}}}

{{{16/y^2+y^2/2=6}}}...both sides divide by {{{2y^2}}}

{{{16*2y^2/y^2+ y^2*2y^2/2=6*2y^2}}}

{{{32cross(y^2)/cross(y^2)+ y^2*cross(2)y^2/cross(2)= 12y^2}}}

{{{32 + y^4= 12y^2}}}

{{{y^4 -12y^2+32=0}}}

{{{y^4 -4y^2-8y^2+32=0}}}

{{{(y^4 -4y^2)- (8y^2-32)=0}}}

{{{y^2(y^2 -4)- 8(y^2-4)=0}}}

{{{(y^2- 8)(y^2-4)=0}}}

if
{{{y^2- 8=0}}}...=>

{{{y^2= 8}}}

{{{y}}}=+-{{{sqrt(8)}}}

{{{y=2.8}}}, or {{{y= -2.8}}}


now find {{{x}}}


{{{x=4/2.8}}}

{{{x=1.43}}}, or {{{x= -1.43}}}


check:

{{{xy=4}}}

{{{1.43*2.8=4}}}

{{{4.004=4}}}

{{{4=4}}}