Chapter 10 - Section 10.5 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 810: 16

{$(-6,3),(-2,-1)$}

Re-write the first equation as: $y=-x-3$ First equation yields: $x^2+2(-x-3)^2=12(-x-3)+18$ or, $x^2+8x+12=0$ or,$(x+6)(x+2)=0$ or, $x=${$-6,-2$} From first equation when $x=-6$, we have $y=3$ From first equation when $x=-2$, we have $y=-1$ Thus, solution set is {$(-6,3),(-2,-1)$}