Answer
$(0,2),(\frac{4}{3},\frac{2}{3})$
Work Step by Step
Substituting the second equation into the first one we get: $6x^2+3(-x+2)^2=12\\6x^2+3(x^2-4x+4)=12\\9x^2-12x+12=12\\3x(3x-4)=0$
Thus $x=0$ or $x=\frac{4}{3}$.
If $x=0$, then $y=2$, and if $x=\frac{4}{3}$, then $y=\frac{2}{3}$. Thus the solutions are: $(0,2),(\frac{4}{3},\frac{2}{3})$
