Answer
$$\left\{ {\left( { - 2,0} \right),\left( {1,1} \right)} \right\}$$
Work Step by Step
$$\eqalign{
& {x^2} + 2xy + {y^2} = 4\,\,\,\left( {\bf{1}} \right) \cr
& x - 3y = - 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\bf{2}} \right) \cr
& {\text{Solve the equation }}\left( {\bf{2}} \right){\text{ for }}y \cr
& y = \frac{{x + 2}}{3} \cr
& {\text{Substitute }}\frac{{x + 2}}{3}{\text{ for }}y{\text{ into the equation }}\left( {\bf{1}} \right) \cr
& {x^2} + 2x\left( {\frac{{x + 2}}{3}} \right) + {\left( {\frac{{x + 2}}{3}} \right)^2} = 4 \cr
& {x^2} + \frac{2}{3}{x^2} + \frac{4}{3}x + \frac{{{x^2}}}{9} + \frac{{4x}}{9} + \frac{4}{9} = 4 \cr
& \frac{{16}}{9}{x^2} + \frac{{16}}{9}x + \frac{4}{9} = 4 \cr
& {\text{Solve for }}x \cr
& 16{x^2} + 16x + 4 = 36 \cr
& 4{x^2} + 4x + 1 = 9 \cr
& 4{x^2} + 4x - 8 = 0 \cr
& {x^2} + x - 2 = 0 \cr
& \left( {x + 2} \right)\left( {x - 1} \right) = 0 \cr
& {x_1} = - 2,\,\,\,\,{x_2} = 1 \cr
& \cr
& {\text{Substitute }}{x_1} = - 2{\text{ into the equation }}y = \frac{{x + 2}}{3}{\text{ to find }}\left( {{x_1},{y_1}} \right) \cr
& y = \frac{{ - 2 + 2}}{3} \cr
& y = 0 \cr
& {\text{The first solution is }}\left( { - 2,0} \right) \cr
& \cr
& {\text{Substitute }}{x_2} = 1{\text{ into the equation }}y = \frac{{x + 2}}{3}{\text{ to find }}\left( {{x_2},{y_2}} \right) \cr
& y = \frac{{1 + 2}}{3} \cr
& y = 1 \cr
& {\text{The second solution is }}\left( {1,1} \right) \cr
& \cr
& {\text{Therefore, the solution set of the system is}} \cr
& \left\{ {\left( { - 2,0} \right),\left( {1,1} \right)} \right\} \cr} $$
