Chapter 10 - Section 10.5 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 810: 5

$(x,y)=${$(-3,11),(4,-10)$}

From the first and second equations, we have $x^2-4x-10=-x^2-2x+14$ or, $x^2-x-12=0$ or, $(x-4)(x+3)=0$ or, $x=${$-3,4$} From first equation when $x=-3$, we have $y=11$ From first equation when $x=4$, we have $y=-10$ Thus, solution set is $(x,y)=${$(-3,11),(4,-10)$}