Answer
$$\left\{ {\left( {\sqrt 3 ,2i} \right),\left( { - \sqrt 3 ,2i} \right),\left( {\sqrt 3 , - 2i} \right),\left( { - \sqrt 3 , - 2i} \right)} \right\}$$
Work Step by Step
$$\eqalign{
& 2{x^2} - 3{y^2} = 18\,\,\,\left( {\bf{1}} \right) \cr
& 2{x^2} - 2{y^2} = 14\,\,\,\left( {\bf{2}} \right) \cr
& \cr
& {\text{Multiply the equation }}\left( {\bf{1}} \right){\text{ by }} - 1{\text{ and add both equations to}} \cr
& {\text{eliminate }}{x^2} \cr
& - 2{x^2} + 3{y^2} = - 18 \cr
& \,\,\,\underline {2{x^2} - 2{y^2} = 14} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{y^2} = - 4 \cr
& {\text{Solve the quadratic equation }}{y^2} = - 4 \cr
& {y^2} = - 4 \cr
& {y_1} = 2i,\,\,\,{x_2} = - 2i \cr
& \cr
& {\text{From the equation }}\left( {\bf{2}} \right){\text{ we have that}} \cr
& 2{x^2} - 2{y^2} = 14 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{x^2} = \frac{{14 + 2{y^2}}}{2} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{x^2} = 7 + {y^2} \cr
& \cr
& {\text{Substitute }}{y_1} = 2i{\text{ into the equation }}{x^2} = 7 + {y^2} \cr
& {x^2} = 7 + {\left( {2i} \right)^2} \cr
& {x^2} = 3 \cr
& x = \pm \sqrt 3 \cr
& {\text{The first and second solutions are }}\left( {\sqrt 3 ,2i} \right){\text{ and }}\left( { - \sqrt 3 ,2i} \right) \cr
& \cr
& {\text{Substitute }}{y_2} = - 2i{\text{ into the equation }}{x^2} = 7 + {y^2} \cr
& {x^2} = 7 + {\left( {2i} \right)^2} \cr
& {x^2} = 3 \cr
& x = \pm \sqrt 3 \cr
& {\text{The first and second solutions are }}\left( {\sqrt 3 , - 2i} \right){\text{ and }}\left( { - \sqrt 3 , - 2i} \right) \cr
& \cr
& {\text{Therefore, the solution set of the system is}} \cr
& \left\{ {\left( {\sqrt 3 ,2i} \right),\left( { - \sqrt 3 ,2i} \right),\left( {\sqrt 3 , - 2i} \right),\left( { - \sqrt 3 , - 2i} \right)} \right\} \cr} $$
