Chapter 10 - Systems of Equations and Inequalities - Section 10.6 Systems of Nonlinear Equations - 10.6 Assess Your Understanding - Page 795: 50

$(0,1),(2,1)$

1. Multiply $x^2$ to the 2nd equation to get $x^3-2x^2+y^2-y=0$ 2. Take the difference of the above and the 1st equation to get $4y-4=0$, thus $y=1$ 3. Use the 1st equation to get $x^3-2x^2+1+3-4=0$ or $x^2(x-2)=0$, thus $x=0, 2$ 4. . Combine the above the results to get the solutions: $(0,1),(2,1)$