Answer
$$\left\{ {\left( {6,\frac{1}{{15}}} \right),\left( { - 1, - \frac{2}{5}} \right)} \right\}$$
Work Step by Step
$$\eqalign{
& \,\, - 5xy + 2 = 0\,\,\,\,\left( {\bf{1}} \right) \cr
& \,\,\,\,\,\,x - 15y = 5\,\,\,\left( {\bf{2}} \right) \cr
& \cr
& {\text{Solve the equation }}\left( {\bf{1}} \right){\text{ for }}y \cr
& - 5xy + 2 = 0 \cr
& y = \frac{2}{{5x}} \cr
& \cr
& {\text{Substitute }}y = \frac{2}{{5x}}{\text{ for }}y{\text{ into the equation }}\left( {\bf{2}} \right) \cr
& x - 15\left( {\frac{2}{{5x}}} \right) = 5 \cr
& {\text{Solve for }}x \cr
& x - \frac{6}{x} = 5 \cr
& {x^2} - 6 = 5x \cr
& {x^2} - 5x - 6 = 0 \cr
& \left( {x - 6} \right)\left( {x + 1} \right) = 0 \cr
& {x_1} = 6,\,\,\,\,{x_2} = - 1 \cr
& \cr
& {\text{Substitute }}{x_1} = 6{\text{ into the equation }}y = \frac{2}{{5x}} \cr
& y = \frac{2}{{5\left( 6 \right)}} \cr
& y = \frac{1}{{15}} \cr
& {\text{The first solution is }}\left( {6,\frac{1}{{15}}} \right) \cr
& \cr
& {\text{Substitute }}{x_2} = - 1{\text{ into the equation }}y = \frac{2}{{5x}} \cr
& y = \frac{2}{{5\left( { - 1} \right)}} \cr
& y = - \frac{2}{5} \cr
& {\text{The second solution is }}\left( { - 1, - \frac{2}{5}} \right) \cr
& \cr
& {\text{Therefore, the solution set of the system is}} \cr
& \left\{ {\left( {6,\frac{1}{{15}}} \right),\left( { - 1, - \frac{2}{5}} \right)} \right\} \cr} $$
