Chapter 10 - Section 10.5 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 810: 1

$(x,y)=${$(-3,5),(2,0)$}

Re-write the first equation as: $y=-x+2$ The second equation yields: $-x+2=x^2-4$ or, $x^2+x-6=0$ or, $(x+3)(x-2)=0$ or, $x=${$-3,2$} From first equation when $x=-3$, we have $y=5$ From first equation when $x=2$, we have $y=0$ Thus, solution set is $(x,y)=${$(-3,5),(2,0)$}