Answer
$$\left\{ {\left( { - 4,0} \right),\left( {2,12} \right)} \right\}$$
Work Step by Step
$$\eqalign{
& y = {x^2} + 4x\,\,\,\,\,\left( {\bf{1}} \right) \cr
& 2x - y = - 8\,\,\,\,\,\left( {\bf{2}} \right) \cr
& \cr
& {\text{Substitute }}{x^2} + 4x{\text{ for }}y{\text{ into the equation }}\left( {\bf{2}} \right) \cr
& 2x - {x^2} - 4x = - 8 \cr
& - {x^2} - 2x = - 8 \cr
& {x^2} + 2x - 8 = 0 \cr
& {\text{Solve for }}x \cr
& \left( {x + 4} \right)\left( {x - 2} \right) = 0 \cr
& {x_1} = - 4,\,\,\,\,{x_2} = 2 \cr
& \cr
& {\text{Substitute }}{x_1} = - 4{\text{ into the equation }}\left( {\bf{1}} \right){\text{ to find }}\left( {{x_1},{y_1}} \right) \cr
& y = {\left( { - 4} \right)^2} + 4\left( { - 4} \right) \cr
& y = 0 \cr
& {\text{The first solution is }}\left( { - 4,0} \right) \cr
& \cr
& {\text{Substitute }}{x_2} = 2{\text{ into the equation }}\left( {\bf{1}} \right){\text{ to find }}\left( {{x_1},{y_1}} \right) \cr
& y = {\left( 2 \right)^2} + 4\left( 2 \right) \cr
& y = 12 \cr
& {\text{The second solution is }}\left( {2,12} \right) \cr
& \cr
& {\text{Therefore, the solution set of the system is}} \cr
& \left\{ {\left( { - 4,0} \right),\left( {2,12} \right)} \right\} \cr} $$
