Answer
$${\left( { - 3, - 2} \right),\left( { - 3,2} \right),\left( {3, - 2} \right),\left( {3,2} \right)}$$
Work Step by Step
$$\eqalign{
& 2{x^2} + 3{y^2} = 30\,\,\,\left( {\bf{1}} \right) \cr
& {x^2} + {y^2} = 13\,\,\,\,\,\,\,\,\left( {\bf{2}} \right) \cr
& {\text{Multiply the equation }}\left( {\bf{2}} \right){\text{ by }} - 3{\text{ and add both equations to}} \cr
& {\text{eliminate }}{y^2} \cr
& \,\,\,2{x^2} + 3{y^2} = 30 \cr
& \underline { - 3{x^2} - 3{y^2} = - 39} \cr
& - {x^2}\,\,\,\,\,\,\,\,\,\, = - 9 \cr
& \cr
& {\text{Solve the quadratic equation }} - {x^2} = - 9 \cr
& {x^2} = 9 \cr
& {x_1} = - 3,\,\,\,{x_2} = 3 \cr
& \cr
& {\text{From the equation }}\left( {\bf{2}} \right){\text{ we have that}} \cr
& {x^2} + {y^2} = 13 \cr
& \,\,\,\,\,\,\,\,\,\,{y^2} = 13\, - {x^2} \cr
& \cr
& {\text{Substitute }}{x_1} = - 3{\text{ into the equation }}{y^2} = 13\, - {x^2} \cr
& {y^2} = 13\, - {\left( { - 3} \right)^2} \cr
& {y^2} = 4 \cr
& y = \pm 2 \cr
& {\text{The first and second solutions are }}\left( { - 3, - 2} \right){\text{ and }}\left( { - 3,2} \right) \cr
& \cr
& {\text{Substitute }}{x_2} = 3{\text{ into the equation }}{y^2} = 13\, - {x^2} \cr
& {y^2} = 13\, - {\left( 3 \right)^2} \cr
& {y^2} = 4 \cr
& y = \pm 2 \cr
& {\text{The first and second solutions are }}\left( {3, - 2} \right){\text{ and }}\left( {3,2} \right) \cr
& \cr
& {\text{Therefore, the solution set of the system is}} \cr
& \left\{ {\left( { - 3, - 2} \right),\left( { - 3,2} \right),\left( {3, - 2} \right),\left( {3,2} \right)} \right\} \cr} $$
