Question 1086028
{{{x=6-y}}}
Substituting,
{{{(6-y)^2+y^2=28}}}
{{{y^2-12y+36+y^2=28}}}
{{{2y^2-12y+8=0}}}
{{{y^2-6y+4=0}}}
{{{y^2-6y+9+4=9}}}
{{{(y-3)^2=5}}}
{{{y-3=0 +- sqrt(5)}}}
{{{y=3 +- sqrt(5)}}}
So then,
{{{x=6-y}}}
Solve for x.
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*[illustration 96.JPG].