Answer
$$\left\{ {\left( { - 2, - 2} \right),\left( { - 2,2} \right),\left( {2, - 2} \right),\left( {2,2} \right)} \right\}$$
Work Step by Step
$$\eqalign{
& {x^2} + {y^2} = 8\,\,\,\,\,\,\left( {\bf{1}} \right) \cr
& \,{x^2} - {y^2} = 0\,\,\,\,\,\,\left( {\bf{2}} \right) \cr
& {\text{Add both equations to eliminate }}{y^2} \cr
& {x^2} + {y^2} = 8 \cr
& \underline {{x^2} - {y^2} = 0} \cr
& 2{x^2}\,\,\,\,\,\,\,\,\,\, = 8 \cr
& \cr
& {\text{Solve the quadratic equation 2}}{x^2} = 8 \cr
& {x^2} = 4 \cr
& {x_1} = - 2,\,\,\,{x_2} = - 2 \cr
& \cr
& {\text{From the equation }}\left( {\bf{1}} \right){\text{ we have that}} \cr
& {x^2} + {y^2} = 8\, \cr
& {y^2} = 8 - {x^2} \cr
& \cr
& {\text{Substitute }}{x_1} = - 2{\text{ into the equation }}{y^2} = 8 - {x^2}{\text{ }} \cr
& {y^2} = 8 - {\left( { - 2} \right)^2} \cr
& {y^2} = 4 \cr
& y = \pm 2 \cr
& {\text{The first and second solutions are }}\left( { - 2, - 2} \right){\text{ and }}\left( { - 2,2} \right) \cr
& \cr
& {\text{Substitute }}{x_2} = 2{\text{ into the equation }}{y^2} = 8 - {x^2}{\text{ }} \cr
& {y^2} = 8 - {\left( 2 \right)^2} \cr
& {y^2} = 4 \cr
& y = \pm 2 \cr
& {\text{The third and fourth solutions are }}\left( {2, - 2} \right){\text{ and }}\left( {2,2} \right) \cr
& \cr
& {\text{Therefore, the solution set of the system is}} \cr
& \left\{ {\left( { - 2, - 2} \right),\left( { - 2,2} \right),\left( {2, - 2} \right),\left( {2,2} \right)} \right\} \cr} $$
