Answer
$$\left\{ {\left( { - 2,2} \right),\left( {1,1} \right)} \right\}$$
Work Step by Step
$$\eqalign{
& {x^2} + y = 2\,\,\,\left( {\bf{1}} \right) \cr
& x - y = 0\,\,\,\,\,\left( {\bf{2}} \right) \cr
& \cr
& {\text{Add both equations to eliminate }}y \cr
& {x^2}\,\, + 0x + y = 2 \cr
& \underline {\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\, - y = 0} \cr
& {x^2} + x\,\,\,\,\,\,\,\,\,\,\,\,\, = 2 \cr
& \cr
& {\text{Solve the quadratic equation }}{x^2} + x = 2 \cr
& {x^2} + x - 2 = 0 \cr
& \left( {x + 2} \right)\left( {x - 1} \right) = 0 \cr
& {x_1} = - 2,\,\,\,{x_2} = 1 \cr
& \cr
& {\text{From the equation }}\left( {\bf{2}} \right){\text{ we have that}} \cr
& x - y = 0 \cr
& x = y \cr
& \cr
& {\text{Let }}{x_1} = - 2 \cr
& x = y \cr
& {y_1} = 2 \cr
& {\text{The first solution is }}\left( { - 2, - 2} \right) \cr
& and \cr
& {\text{Let }}{x_2} = 1 \cr
& x = y \cr
& {y_2} = 1 \cr
& {\text{The second solution is }}\left( {1,1} \right) \cr
& \cr
& {\text{Therefore, the solution set of the system is}} \cr
& \left\{ {\left( { - 2,2} \right),\left( {1,1} \right)} \right\} \cr} $$
