Question 1178359
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I can solve it this way (substitution):

{{{y=1/x}}}
{{{x^2+y^2 = 2 }}}

Substitute 1/x for y in the 2nd eqn: 
{{{ x^2 + (1/x)^2 = 2 }}}

Simplifies to:
{{{ x^4 - 2x^2 + 1 = 0 }}}

Factors to:
{{{ (x^2 - 1) = 0 }}}

Solutions: 
{{{ x^2 = 1 }}}  --> x = -1   and/or   x = 1

For x=-1  we get y=-1   and {{{x^2+y^2 = (-1)^2+(-1)^2 = 1+1 = 2 }}}
For x=1  we get y=1     and {{{x^2+y^2 = (1)^2 + (1)^2 = 1+1 = 2 }}}

Both solutions (1,1) and (-1,-1) satisify the equations.