Chapter 9 Quadratic Relations and Conic Sections - 9.7 Solve Quadratic Systems - Guided Practice for Examples 3 and 4 - Page 661: 7

$(0,1),(0,-1),(-0.5,\sqrt{0.75}),(-0.5,\sqrt{0.75})$

Adding the first and twice the second equation we get: $x+2+2x^2-2=0\\2x^2+x=0\\x(2x+1)=0$ Thus $x=0$ or $x=-0.5$. When $x=0$ then $y=\pm\sqrt{1-0^2}=\pm1$ When $x=-0.5$ then $y=\pm\sqrt{1-(-0.5)^2}=\pm\sqrt{0.75}$ Thus the solutions are: $(0,1),(0,-1),(-0.5,\sqrt{0.75}),(-0.5,\sqrt{0.75})$