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