Answer
$$\left\{ {\left( {\frac{{15}}{4}, - 4} \right),\left( { - 3,5} \right)} \right\}$$
Work Step by Step
$$\eqalign{
& \,\,\,\,\,\,\,\,\,\,\,\,xy = - 15\,\,\,\left( {\bf{1}} \right) \cr
& \,4x + 3y = 3\,\,\,\,\,\,\,\,\left( {\bf{2}} \right) \cr
& \cr
& {\text{Solve the equation }}\left( {\bf{1}} \right){\text{ for }}y \cr
& y = - \frac{{15}}{x} \cr
& \cr
& {\text{Substitute }} - \frac{{15}}{x}{\text{ for }}y{\text{ into the equation }}\left( {\bf{2}} \right) \cr
& 4x + 3\left( { - \frac{{15}}{x}} \right) = 3 \cr
& {\text{Solve for }}x \cr
& 4x - \frac{{45}}{x} = 3 \cr
& 4{x^2} - 45 = 3x \cr
& 4{x^2} - 3x - 45 = 0 \cr
& \left( {4x - 15} \right)\left( {x + 3} \right) = 0 \cr
& {x_1} = \frac{{15}}{4},\,\,\,\,{x_2} = - 3 \cr
& \cr
& {\text{Substitute }}{x_1} = \frac{{15}}{4}{\text{ into the equation }}y = - \frac{{15}}{x} \cr
& y = - \frac{{15}}{{15/4}} \cr
& y = - 4 \cr
& {\text{The first solution is }}\left( {\frac{{15}}{4}, - 4} \right) \cr
& \cr
& {\text{Substitute }}{x_2} = - 3{\text{ into the equation }}y = - \frac{{15}}{x} \cr
& y = - \frac{{15}}{{ - 3}} \cr
& y = 5 \cr
& {\text{The second solution is }}\left( { - 3,5} \right) \cr
& \cr
& {\text{Therefore, the solution set of the system is}} \cr
& \left\{ {\left( {\frac{{15}}{4}, - 4} \right),\left( { - 3,5} \right)} \right\} \cr} $$
