Chapter 10 - Section 10.5 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 810: 40

{$(0,-3),(2,1)$}

Re-write the first equation as: $y=2x-3$ First equation yields: $(x-1)^2+(2x-3+1)^2=5$ or, $5x^2-10x=0$ or,$5x(x-2)=0$ or, $x=${$0,2$} From second equation $2x-y=3$ when $x=0$, we have $y=-3$ From second equation $2x-y=3$ when $x=2$, we have $y=1$ Thus, solution set is {$(0,-3),(2,1)$}