Question 1089671
Neither of those answers is correct.
From eq. 2,
{{{x=y+2}}}
Substitute into eq. 1,
{{{(y+2)^2+y^2=3}}}
{{{y^2+4y+4+y^2=3}}}
{{{2y^2+4y+1=0}}}
{{{y^2+2y+1/2=0}}}
Complete the square,
{{{(y^2+2y+1)+1/2=1}}}
{{{(y+1)^2=1/2}}}
{{{y+1=0 +- sqrt(1/2)}}}
{{{y=-1 +- sqrt(2)/2}}}

Then use eq. 2 to solve for x.
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*[illustration 347.JPG].
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