Chapter 10 - Section 10.5 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 810: 33

{$(4,1),(2,2)$}

Re-write the second equation as: $x=-2y+6$ First equation yields: $(-2y+6)^2+4y^2=20$ or, $y^2-3y+2=0$ or,$(y-2)(y-1)=0$ or, $y=${$1,2$} From second equation when $y=1$, we have $x=4$ From second equation when $y=2$, we have $x=2$ Thus, solution set is: {$(4,1),(2,2)$}