Answer
$$\left\{ {\left( {2,1} \right),\left( {\frac{1}{3},\frac{4}{9}} \right)} \right\}$$
Work Step by Step
$$\eqalign{
& y = {x^2} - 2x + 1\,\,\,\,\,\left( {\bf{1}} \right) \cr
& x - 3y = - 1\,\,\,\,\,\,\,\,\,\,\,\,\left( {\bf{2}} \right) \cr
& \cr
& {\text{Substitute }}{x^2} - 2x + 1{\text{ for }}y{\text{ into the equation }}\left( {\bf{2}} \right) \cr
& x - 3\left( {{x^2} - 2x + 1} \right) = - 1 \cr
& x - 3{x^2} + 6x - 3 = - 1 \cr
& {\text{Solve for }}x \cr
& - 3{x^2} + 7x - 2 = 0 \cr
& 3{x^2} - 7x + 2 = 0 \cr
& \left( {x - 2} \right)\left( {3x - 1} \right) = 0 \cr
& {x_1} = 2,\,\,\,\,{x_2} = \frac{1}{3} \cr
& \cr
& {\text{Substitute }}{x_1} = 2{\text{ into the equation }}\left( {\bf{1}} \right){\text{ to find }}\left( {{x_1},{y_1}} \right) \cr
& y = {\left( 2 \right)^2} - 2\left( 2 \right) + 1 \cr
& y = 1 \cr
& {\text{The first solution is }}\left( {2,1} \right) \cr
& \cr
& {\text{Substitute }}{x_2} = \frac{1}{3}{\text{ into the equation }}\left( {\bf{1}} \right){\text{ to find }}\left( {{x_2},{y_2}} \right) \cr
& y = {\left( {\frac{1}{3}} \right)^2} - 2\left( {\frac{1}{3}} \right) + 1 \cr
& y = \frac{4}{9} \cr
& {\text{The second solution is }}\left( {\frac{1}{3},\frac{4}{9}} \right) \cr
& \cr
& {\text{Therefore, the solution set of the system is}} \cr
& \left\{ {\left( {2,1} \right),\left( {\frac{1}{3},\frac{4}{9}} \right)} \right\} \cr} $$
