Chapter 7 - Section 7.4 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 850: 3

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

Here, we have $y=x^2-4x+4$ Now, $x+(x^2-4x+4)=2$ This gives: $x^2-3x+2=0$ This yields : $(x-2)(x-1)=0$ when $x=2$ then we have $y=0$ and when $x=1$ then we have $y=(1)^2-4(1)+4=1$ Hence, our answers are: $(2,0)$ or $(1,1)$