Question 930367
From eq. 2,
{{{2x=y-2}}}
{{{(2x)^2=(y-2)^2}}}
{{{4x^2=y^2-4y+4}}}
Substitute into eq. 1,
{{{y^2-4y+4+y^2=4}}}
{{{2y^2-4y=0}}}
{{{y^2-2y=0}}}
{{{y(y-2)=0}}}
Two solutions:
{{{y=0}}}
Then,
{{{2x=0-2}}
{{{x=-1}}}
and
{{{y-2=0}}}
{{{y=2}}}
Then,
{{{2x=2-2}}}
{{{2x=0}}}
{{{x=0}}}
So the solutions are (-1,0) and (0,2)