Answer
$$\left\{ {\left( { - 3,4} \right),\left( {1,12} \right)} \right\}$$
Work Step by Step
$$\eqalign{
& y = 2x + 10\,\,\,\,\,\left( {\bf{1}} \right) \cr
& {x^2} + y = 13\,\,\,\,\,\,\left( {\bf{2}} \right) \cr
& {\text{Substitute }}2x + 10{\text{ for }}y{\text{ into the equation }}\left( {\bf{2}} \right) \cr
& {x^2} + 2x + 10 = 13\, \cr
& {\text{Solve for }}x \cr
& {x^2} + 2x - 3 = 0 \cr
& \left( {x + 3} \right)\left( {x - 1} \right) = 0 \cr
& {x_1} = - 3,\,\,\,\,{x_2} = 1 \cr
& \cr
& {\text{Substitute }}{x_1} = - 3{\text{ into the equation }}y = 2x + 10{\text{ to find }}\left( {{x_1},{y_1}} \right) \cr
& y = 2\left( { - 3} \right) + 10 \cr
& y = 4 \cr
& {\text{The first solution is }}\left( { - 3,4} \right) \cr
& \cr
& {\text{Substitute }}{x_2} = 1{\text{ into the equation }}y = 2x + 10{\text{ to find }}\left( {{x_2},{y_2}} \right) \cr
& y = 2\left( 1 \right) + 10 \cr
& y = 12 \cr
& {\text{The second solution is }}\left( {1,12} \right) \cr
& \cr
& {\text{Therefore, the solution set of the system is}} \cr
& \left\{ {\left( { - 3,4} \right),\left( {1,12} \right)} \right\} \cr} $$
