Answer
$$\left\{ {\left( { - 3, - 1} \right),\left( { - 3,1} \right),\left( {3, - 1} \right),\left( {3,1} \right)} \right\}$$
Work Step by Step
$$\eqalign{
& \,\,\,{x^2} + {y^2} = 10\,\,\,\,\left( {\bf{1}} \right) \cr
& 2\,{x^2} - {y^2} = 17\,\,\,\,\left( {\bf{2}} \right) \cr
& {\text{Add both equations to eliminate }}{y^2} \cr
& \,\,\,{x^2} + {y^2} = 10 \cr
& \underline {2\,{x^2} - {y^2} = 17} \cr
& 3{x^2}\,\,\,\,\,\,\,\,\,\, = 27 \cr
& \cr
& {\text{Solve the quadratic equation 3}}{x^2} = 27 \cr
& 3{x^2} = 27 \cr
& {x^2} = 9 \cr
& {x_1} = - 3,\,\,\,{x_2} = 3 \cr
& \cr
& {\text{From the equation }}\left( {\bf{1}} \right){\text{ we have that}} \cr
& {x^2} + {y^2} = 10 \cr
& {y^2} = 10 - {x^2} \cr
& \cr
& {\text{Substitute }}{x_1} = - 3{\text{ into the equation }}{y^2} = 10 - {x^2}{\text{ }} \cr
& {y^2} = 10 - {\left( { - 3} \right)^2} \cr
& {y^2} = 1 \cr
& y = \pm 1 \cr
& {\text{The first and second solutions are }}\left( { - 3, - 1} \right){\text{ and }}\left( { - 3,1} \right) \cr
& \cr
& {\text{Substitute }}{x_2} = 3{\text{ into the equation }}{y^2} = 10 - {x^2}{\text{ }} \cr
& {y^2} = 10 - {\left( 3 \right)^2} \cr
& {y^2} = 1 \cr
& y = \pm 1 \cr
& {\text{The third and fourth solutions are }}\left( {3, - 1} \right){\text{ and }}\left( {3,1} \right) \cr
& \cr
& {\text{Therefore, the solution set of the system is}} \cr
& \left\{ {\left( { - 3, - 1} \right),\left( { - 3,1} \right),\left( {3, - 1} \right),\left( {3,1} \right)} \right\} \cr} $$
