Chapter 9 Quadratic Relations and Conic Sections - 9.7 Solve Quadratic Systems - 9.7 Exercises - Skill Practice - Page 662: 34

Option D

Given: $x^2+y^2+6x=0\\y^2+x-6=0$ The second equation becomes: $$y^2=6-x$$ Substitute $x$: $x^2+(6-x)+6x=0\\x^2=5+6=0\\(x+2)(x+3)=0\\x=-2\lor x=-3$ Find y: $y^2+(-2)-6=0\\y^2=8\\y=\pm\sqrt 8=\pm 2\sqrt 2$ and $y^2+(-3)-6=0\\y^2=9\\y=\pm 3$ The solutions are $(-2,-2\sqrt 2);(-2,2\sqrt 2);(-3,-3);(-3,3)$ Hence, the correct answer is D