Answer
$$\left\{ {\left( {4, - \frac{5}{2}} \right),\left( {5, - 2} \right)} \right\}$$
Work Step by Step
$$\eqalign{
& xy = - 10\,\,\,\,\,\,\,\,\left( {\bf{1}} \right) \cr
& x + 2y = 1\,\,\,\,\,\,\left( {\bf{2}} \right) \cr
& \cr
& {\text{Solve the equation }}\left( {\bf{1}} \right){\text{ for }}y \cr
& y = - \frac{{10}}{x} \cr
& {\text{Substitute }} - \frac{{10}}{x}{\text{ for }}y{\text{ into the equation }}\left( {\bf{2}} \right) \cr
& x + 2\left( { - \frac{{10}}{x}} \right) = 1\, \cr
& x - \frac{{20}}{x} = 1\, \cr
& {x^2} - 20 = x\, \cr
& {x^2} - x - 20 = 0\, \cr
& {\text{Solve for }}x \cr
& \left( {x - 5} \right)\left( {x - 4} \right) = 0 \cr
& {x_1} = 4,\,\,\,\,{x_2} = 5 \cr
& \cr
& {\text{Substitute }}{x_1} = 4{\text{ into the equation }}y = - \frac{{10}}{x}{\text{ to find }}\left( {{x_1},{y_1}} \right) \cr
& y = - \frac{{10}}{4} \cr
& y = - \frac{5}{2} \cr
& {\text{The first solution is }}\left( {4, - \frac{5}{2}} \right) \cr
& \cr
& {\text{Substitute }}{x_2} = 5{\text{ into the equation }}y = - \frac{{10}}{x}{\text{ to find }}\left( {{x_2},{y_2}} \right) \cr
& y = - \frac{{10}}{5} \cr
& y = - 2 \cr
& {\text{The second solution is }}\left( {5, - 2} \right) \cr
& \cr
& {\text{Therefore, the solution set of the system is}} \cr
& \left\{ {\left( {4, - \frac{5}{2}} \right),\left( {5, - 2} \right)} \right\} \cr} $$
