Chapter 10 - Section 10.5 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 810: 7

$(x,y)=${$(-3,-4),(4,3)$}

Re-write the second equation as: $x=y+1$ First equation yield: $(y+1)^2+y^2=25$ or, $y^2+y-12=0$ or, $(y+4)(y-3)=0$ or, $y=${$-4,3$} From first equation when $y=-4$, we have $x=-3$ From first equation when $y=3$, we have $x=4$ Thus, solution set is $(x,y)=${$(-3,-4),(4,3)$}