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

$(x,y)=${$(-12,1),(-2,6)$}

Re-write the second equation as: $x=2y-14$ First equation yields: $(2y-14)y=-12$ or, $y^2-7y+6=0$ or, $(y-6)(y-1)=0$ or, $y=${$1,6$} From second equation when $y=1$, we have $x=-12$ From second equation when $y=6$, we have $x=-2$ Thus, solution set is $(x,y)=${$(-12,1),(-2,6)$}