Chapter 9 Quadratic Relations and Conic Sections - 9.7 Solve Quadratic Systems - 9.7 Exercises - Skill Practice - Page 661: 19

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Given: $4x^2+2y^2-x-y=6\\3x-y=2$ The second equation becomes: $y=3x-2$ Substitute $y$ into the second equation: $4x^2+2(3x-2)^2-x-y=6\\4x^2+18x^2-24x+8-4x-4=0\\22x^2-28x+4=0$ Solving this we get $x_1=\frac{7+3\sqrt 3}{11}\\x_2=\frac{7-3\sqrt 3}{11}$ Find y: $y_1=3x_1-2=\frac{-1+9\sqrt 3}{11}\\y_2=3x_2-2=\frac{-1-9\sqrt 3}{11}$