Answer
$(0,1),(-\frac{2}{3},-\frac{1}{3})$
Work Step by Step
1. Use the 1st equation in the 2nd to get $2x^2+(2x+1)^2=1$ or $6x^2+4x=0$, thus $x=-\frac{2}{3},0$
2. For $x=-\frac{2}{3}$, use the 1st equation again to get $y=2(-\frac{2}{3})+1=-\frac{1}{3}$
3. For $x=0$, use the 1st equation again to get $y=1$
4. Thus the solutions $(0,1),(-\frac{2}{3},-\frac{1}{3})$
