Answer
$$\left\{ {\left( { - \frac{5}{2},\frac{1}{4}} \right),\left( { - 4,1} \right)} \right\}$$
Work Step by Step
$$\eqalign{
& y = {x^2} + 6x + 9\,\,\,\,\,\left( {\bf{1}} \right) \cr
& x + 2y = - 2\,\,\,\,\,\,\,\,\,\,\,\,\left( {\bf{2}} \right) \cr
& \cr
& {\text{Substitute }}{x^2} + 6x + 9{\text{ for }}y{\text{ into the equation }}\left( {\bf{2}} \right) \cr
& x + 2\left( {{x^2} + 6x + 9} \right) = - 2 \cr
& x + 2{x^2} + 12x + 18 = - 2 \cr
& {\text{Solve for }}x \cr
& 2{x^2} + 13x + 20 = 0 \cr
& \left( {2x + 5} \right)\left( {x + 4} \right) = 0 \cr
& {x_1} = - \frac{5}{2},\,\,\,\,{x_2} = - 4 \cr
& \cr
& {\text{Substitute }}{x_1} = - \frac{5}{2}{\text{ into the equation }}\left( {\bf{1}} \right){\text{ to find }}\left( {{x_1},{y_1}} \right) \cr
& y = {\left( { - \frac{5}{2}} \right)^2} + 6\left( { - \frac{5}{2}} \right) + 9 \cr
& y = \frac{1}{4} \cr
& {\text{The first solution is }}\left( { - \frac{5}{2},\frac{1}{4}} \right) \cr
& \cr
& {\text{Substitute }}{x_2} = - 4{\text{ into the equation }}\left( {\bf{1}} \right){\text{ to find }}\left( {{x_2},{y_2}} \right) \cr
& y = {\left( { - 4} \right)^2} + 6\left( { - 4} \right) + 9 \cr
& y = 1 \cr
& {\text{The second solution is }}\left( { - 4,1} \right) \cr
& \cr
& {\text{Therefore, the solution set of the system is}} \cr
& \left\{ {\left( { - \frac{5}{2},\frac{1}{4}} \right),\left( { - 4,1} \right)} \right\} \cr} $$
