Answer
$$\left\{ {\left( { - \frac{4}{3}, - 6} \right),\left( { - 4, - 2} \right)} \right\}$$
Work Step by Step
$$\eqalign{
& \,\,\,\,\,\,\,\,\,\,\,\,xy = 8\,\,\,\,\,\,\,\,\,\left( {\bf{1}} \right) \cr
& \,3x + 2y = - 16\,\,\,\left( {\bf{2}} \right) \cr
& \cr
& {\text{Solve the equation }}\left( {\bf{1}} \right){\text{ for }}y \cr
& y = \frac{8}{x} \cr
& \cr
& {\text{Substitute }}\frac{8}{x}{\text{ for }}y{\text{ into the equation }}\left( {\bf{2}} \right) \cr
& 3x + 2\left( {\frac{8}{x}} \right) = - 16 \cr
& {\text{Solve for }}x \cr
& 3x + \frac{{16}}{x} = - 16 \cr
& 3{x^2} + 16 = - 16x \cr
& 3{x^2} + 16x + 16 = 0 \cr
& \left( {3x + 4} \right)\left( {x + 4} \right) = 0 \cr
& {x_1} = - \frac{4}{3},\,\,\,\,{x_2} = - 4 \cr
& \cr
& {\text{Substitute }}{x_1} = - \frac{4}{3}{\text{ into the equation }}y = \frac{8}{x} \cr
& y = \frac{8}{{ - 4/3}} \cr
& y = - 6 \cr
& {\text{The first solution is }}\left( { - \frac{4}{3}, - 6} \right) \cr
& \cr
& {\text{Substitute }}{x_2} = - 4{\text{ into the equation }}y = \frac{8}{x} \cr
& y = \frac{8}{{ - 4}} \cr
& y = - 2 \cr
& {\text{The second solution is }}\left( { - 4, - 2} \right) \cr
& \cr
& {\text{Therefore, the solution set of the system is}} \cr
& \left\{ {\left( { - \frac{4}{3}, - 6} \right),\left( { - 4, - 2} \right)} \right\} \cr} $$
