Answer
$$\left\{ {\left( { - 7,\frac{{24}}{7}} \right),\left( {3,2} \right)} \right\}$$
Work Step by Step
$$\eqalign{
& {x^2} + 2xy = 15 + 2x\,\,\,\left( {\bf{1}} \right) \cr
& xy - 3x + 3 = 0\,\,\,\,\,\,\,\,\,\,\left( {\bf{2}} \right) \cr
& {\text{Solve the equation }}\left( {\bf{2}} \right){\text{ for }}y \cr
& x\left( {y - 3} \right) = - 3 \cr
& y - 3 = - \frac{3}{x} \cr
& y = 3 - \frac{3}{x} \cr
& {\text{Substitute }}3 - \frac{3}{x}{\text{ for }}y{\text{ into the equation }}\left( {\bf{1}} \right) \cr
& {x^2} + 2x\left( {3 - \frac{3}{x}} \right) = 15 + 2x \cr
& {x^2} + 6x - 6 = 15 + 2x \cr
& {x^2} + 4x - 21 = {\text{0}} \cr
& {\text{Solve for }}x \cr
& \left( {x + 7} \right)\left( {x - 3} \right) = \cr
& {x_1} = - 7,\,\,\,\,{x_2} = 3 \cr
& \cr
& {\text{Substitute }}{x_1} = - 7{\text{ into the equation }}y = 3 - \frac{3}{x}{\text{ to find }}\left( {{x_1},{y_1}} \right) \cr
& y = 3 - \frac{3}{{ - 7}} \cr
& y = \frac{{24}}{7} \cr
& {\text{The first solution is }}\left( { - 7,\frac{{24}}{7}} \right) \cr
& \cr
& {\text{Substitute }}{x_2} = 3{\text{ into the equation }}y = 3 - \frac{3}{x}{\text{ to find }}\left( {{x_2},{y_2}} \right) \cr
& y = 3 - \frac{3}{3} \cr
& y = 2 \cr
& {\text{The second solution is }}\left( {3,2} \right) \cr
& \cr
& {\text{Therefore, the solution set of the system is}} \cr
& \left\{ {\left( { - 7,\frac{{24}}{7}} \right),\left( {3,2} \right)} \right\} \cr} $$
