Chapter 10 - Section 10.5 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 810: 3

$(x,y)=${$(1,1),(2,0)$}

Re-write the first equation as: $y=-x+2$ The second equation yields: $-x+2=x^2-4x+4$ or, $x^2-3x+2=0$ or, $(x-2)(x-1)=0$ or, $x=${$1,2$} From first equation when $x=1$, we have $y=1$ From first equation when $x=2$, we have $y=0$ Thus, solution set is $(x,y)=${$(1,1),(2,0)$}