Chapter 10 - Section 10.5 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 810: 8

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

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