Solution
= Solution
Use the first of the <simultaneous equations> to put $y=1-x$. Substitution into the second gives
$$
x-x^2+2(1-x)^2=2,\qquad x^2-3x=x(x-3)=0.
$$
Thus $x=0$ or $x=3$. The corresponding values of $y$ give exactly
$$
\boxed{(x,y)=(0,1)\text{ or }(3,-2).}
$$