Question 305894
1.{{{12-x^2 -2xy = 0}}}
2.{{{12-y^2 - 2xy = 0}}}
Use eq. 1 to solve for y.
1.{{{12-x^2 -2xy = 0}}}
{{{12-x^2=2xy}}}
{{{y=(12-x^2)/2x}}}
From that, then,
{{{y^2= (12-x^2)^2/(4x^2)}}}
and 
{{{2xy=12-x^2}}}
Substitute into eq. 2,
2.{{{12-y^2 - 2xy = 0}}}
{{{12- (12-x^2)^2/(4x^2) - (12-x^2) = 0}}}
{{{- (12-x^2)^2/(4x^2) +x^2 = 0}}}
{{{ (12-x^2)^2/(4x^2) =x^2}}}
{{{ (12-x^2)^2 =4x^4}}}
{{{ 12-x^2 =0 +- 2x^2}}}
Two solutions:
{{{12-x^2=2x^2}}}
{{{3x^2=12}}}
{{{x^2=4}}}
{{{x=2}}} and {{{x=-2}}}
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{{{12-x^2=-2x^2}}}
{{{-x^2=12}}}
{{{x^2=-12}}}
No real solution. 
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When {{{x=-2}}},
{{{y=(12-x^2)/2x}}}
{{{y=(12-4)/2(-2)}}}
{{{y=-2}}}
When {{{x=2}}},
{{{y=(12-x^2)/2x}}}
{{{y=(12-4)/4}}}
{{{y=2}}}
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(-2,-2) and (2,2) are the solutions.
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You are searching for the intersection point between two curves.
The first curve is 
{{{y=(12-x^2)/2x}}} (in red below)
The second curve needs to be worked out by completing the square for the second equation,
{{{12-y^2 - 2xy = 0}}}
{{{12=y^2+2xy}}}
{{{12+x^2=y^2+2xy+x^2}}}
{{{12+x^2=(y+x)^2}}}
{{{0 +- sqrt(12+x^2)=y+x}}}
{{{y=-x +- sqrt(12+x^2)}}} (in blue and green below)
The curves and the two solution points are shown below.
{{{drawing(300,300,-8,8,-8,8,grid(1),circle(2,2,.2),circle(-2,-2,.2),
graph(300,300,-8,8,-8,8,(12-x^2)/(2x),sqrt(12+x^2)-x,-sqrt(12+x^2)-x))}}}