Chapter 9 Quadratic Relations and Conic Sections - 9.7 Solve Quadratic Systems - 9.7 Exercises - Quiz for Lessons 9.6-9.7 - Page 664: 12

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Given: $y^2-4x^2-4y=0\\2x^2+y^2-8x-4y=-8$ The first equation becomes: $y^2-2y=6x+3$ Substitute into the second equation: $2y^2-4y+x+6=0\\2(y^2-2y)+x+6=0\\2(6x+3)+x+6=0\\13x=-12\\x=-\frac{12}{13}$ Thus $y^2-2y=6(-\frac{12}{13})+3\\y^2-2y+\frac{33}{13}=0$ The last equation doesn't have any real solutions, so this system can't be solved.