Question 931584
From eq. 2,
{{{x=2-y}}}
Substitute into eq. 1,
{{{(2-y)^2+y^2=100}}}
{{{4-4y+y^2+y^2=100}}}
{{{2y^2-4y+4=100}}}
{{{2y^2-4y-96=0}}}
{{{y^2-2y-48=0}}}
{{{(y-8)(y+6)=0}}}
Two solutions:
{{{y-8=0}}}
{{{y=8}}
Then,
{{{x=2-8}}}
{{{x=-6}}}
and
{{{y+6=0}}}
{{{y=-6}}}
Then,
{{{x=2-(-6)}}}
{{{x=8}}}
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(-6,8) and (8,-6)