Chapter 10 - Section 10.5 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 810: 12

$(x,y)=${$(-1,3),(3,-5)$}

Re-write the second equation as: $y=-2x+1$ First equation yields: $x^2+(-2x+1)=4$ or, $x^2-2x-3=0$ or,$(x-3)(x+1)=0$ or, $x=${$-1,3$} From second equation when $x=-1$, we have $y=3$ From second equation when $x=3$, we have $y=-5$ Thus, solution set is $(x,y)=${$(-1,3),(3,-5)$}