Question 687818
x^2+xy+y^2=12
x+y=2 
solve the following system
x^2+xy+y^2=12
add xy to both sides
x^2+2xy+y^2=xy+12
(x+y)^2=xy+12
From 2nd equation:
(x+y)^2=4=xy+12
xy=-8
y=-8/x
x+y=2
x-8/x=2
x^2-8=2x
x^2-2x-8=0
(x-4)(x+2)=0
x=4
y=-2
or
x=-2
y=4