Answer
$$\left\{ {\left( {4, - 1} \right),\left( {4,1} \right),\left( { - 4, - 1} \right),\left( { - 4,1} \right)} \right\}$$
Work Step by Step
$$\eqalign{
& {x^2} + {y^2} = 17\,\,\,\,\,\,\left( {\bf{1}} \right) \cr
& 2{x^2} - {y^2} = 31\,\,\,\,\left( {\bf{2}} \right) \cr
& {\text{Add both equations to eliminate }}{y^2} \cr
& {x^2} + {y^2} = 17 \cr
& \underline {2{x^2} - {y^2} = 31} \cr
& 3{x^2}\,\,\,\,\,\,\,\,\,\, = 48 \cr
& \cr
& {\text{Solve the quadratic equation 3}}{x^2} = 48 \cr
& {x^2} = 16 \cr
& {x_1} = - 4,\,\,\,{x_2} = 4 \cr
& \cr
& {\text{From the equation }}\left( {\bf{1}} \right){\text{ we have that}} \cr
& {x^2} + {y^2} = 17\,\,\, \cr
& \,\,\,\,\,\,\,\,\,\,{y^2} = 17\, - {x^2} \cr
& \cr
& {\text{Substitute }}{x_1} = - 4{\text{ into the equation }}{y^2} = 17\, - {x^2} \cr
& {y^2} = 17\, - {\left( { - 4} \right)^2} \cr
& {y^2} = 1 \cr
& y = \pm 1 \cr
& {\text{The first and second solutions are }}\left( { - 4, - 1} \right){\text{ and }}\left( { - 4,1} \right) \cr
& \cr
& {\text{Substitute }}{x_2} = 4{\text{ into the equation }}{y^2} = 17\, - {x^2} \cr
& {y^2} = 17\, - {\left( 4 \right)^2} \cr
& {y^2} = 1 \cr
& y = \pm 1 \cr
& {\text{The first and second solutions are }}\left( {4, - 1} \right){\text{ and }}\left( {4,1} \right) \cr
& \cr
& {\text{Therefore, the solution set of the system is}} \cr
& \left\{ {\left( {4, - 1} \right),\left( {4,1} \right),\left( { - 4, - 1} \right),\left( { - 4,1} \right)} \right\} \cr} $$
