Chapter 10 - Section 10.5 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 810: 38

{$(2,2),(2,4)$}

After adding the two equations, we get $2x^2-8x+8=0$ or, $(x-2)(x-2)=0$; or $x=${$2$} From first equation $x^2-y^2-4x+6y-4=0$ when $x=2$, we have $y^2-6y+8=0$ or, $(y-4)(y-2)=0$ or, $y=${$2,4$} Thus, solution set is: {$(2,2),(2,4)$}