Chapter 9 Quadratic Relations and Conic Sections - 9.7 Solve Quadratic Systems - 9.7 Exercises - Skill Practice - Page 661: 21

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Given: $6x^2+8y^2-5x+y=23\\-x+y=-1$ The second equation becomes: $y=-x+1$ Substitute $y$ into the second equation: $6x^2+8(-x+1)^2-5x+(x-1)=23\\6x^2+8x^2-5x-16x+8+x-1=0\\14x^2-20x-16=0\\7x^2-10x-8=0$ Solving this we get $x_1=2\\x_2=\frac{-4}{7}$ Find y: $y_1=x_1-1=1\\y_2=x_2-1=\frac{-11}{7}$