Chapter 9 Quadratic Relations and Conic Sections - 9.7 Solve Quadratic Systems - 9.7 Exercises - Skill Practice - Page 661: 11

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Given: $-2x+y-8=0\\x^2+4y^2-40=0$ From the first equation we have $y=2x+8$ Substitute into the second equation: $x^2+4(2x+8)^2-40=0\\x^2+16x^2+128x+256-40=0\\17x^2+128x+216=0$ Solving this we get $x_1=-2.55\\x_2=-4.98$ Find y: $y_1=2x_1+8=2.89\\y_2=2x_2+8=-1.96$