Chapter 10 - Section 10.5 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 810: 6

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

From the first and second equations, we have $x^2+4x+5=x^2+2x-1$ or, $2x=-6$ or, $x=-3$ From first equation when $x=-3$, we have $y=(-3)^2+4(-3)+5=9-12+5=2$ Thus, solution set is $(x,y)=${$(-3,2)$}