Answer
$$\left\{ {\left( { - 3, - 9} \right),\left( {1,7} \right)} \right\}$$
Work Step by Step
$$\eqalign{
& y = 6x + {x^2} \cr
& 4x - y = - 3 \cr
& \cr
& {\text{Substitute }}6x + {x^2}{\text{ for }}y{\text{ into the equation }}\left( {\bf{2}} \right) \cr
& 4x - 6x - {x^2} = - 3 \cr
& 4x - 6x - {x^2} + 3 = 0 \cr
& - {x^2} - 2x + 3 = 0 \cr
& {x^2} + 2x - 3 = 0 \cr
& {\text{Solve for }}x \cr
& \left( {x + 3} \right)\left( {x - 1} \right) = 0 \cr
& {x_1} = - 3,\,\,\,\,{x_2} = 1 \cr
& \cr
& {\text{Substitute }}{x_1} = - 3{\text{ into the equation }}\left( {\bf{1}} \right){\text{ to find }}\left( {{x_1},{y_1}} \right) \cr
& y = 6\left( { - 3} \right) + {\left( { - 3} \right)^2} \cr
& y = - 9 \cr
& {\text{The first solution is }}\left( { - 3, - 9} \right) \cr
& \cr
& {\text{Substitute }}{x_2} = 1{\text{ into the equation }}\left( {\bf{1}} \right){\text{ to find }}\left( {{x_1},{y_1}} \right) \cr
& y = 6\left( 1 \right) + {\left( 1 \right)^2} \cr
& y = 7 \cr
& {\text{The second solution is }}\left( {1,7} \right) \cr
& \cr
& {\text{Therefore, the solution set of the system is}} \cr
& \left\{ {\left( { - 3, - 9} \right),\left( {1,7} \right)} \right\} \cr} $$
