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