Question

help me please
solve the system
x^2+y^2=3
x-y=2

Answer 1

x=2+y
(y+2)^2+y^2=3
y^2+4y+4+y^2=3
2y^2+4y+1=0
D=8
y1,2=(-4+-sqrt(8))/4=(-2+-sqrt(2))/2=-1+-0,5sqrt(2)
x1=2-1+0,5sqrt(2)=1+0,5sqrt(2)
x2=2-1-0,5sqrt(2)=1-0,5sqrt(2)