$\begin{array}{l} \left\{ \begin{array}{l} {x^2} - xy - 2{y^2} = 0\\ 3x + y = 1 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} {x^2} + xy - 2xy - 2{y^2} = 0\\ 3x + y = 1 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} x\left( {x + y} \right) - 2y\left( {x + y} \right) = 0\\ 3x + y = 1 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} \left( {x - 2y} \right)\left( {x + y} \right) = 0\\ 3x + y = 1 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x + y = 0\\ 3x + y = 1 \end{array} \right.\\ \left\{ \begin{array}{l} x - 2y = 0\\ 3x + y = 1 \end{array} \right. \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x = \dfrac{1}{2}\\ y = - \dfrac{1}{2} \end{array} \right.\\ \left\{ \begin{array}{l} x = \dfrac{2}{7}\\ y = \dfrac{1}{7} \end{array} \right. \end{array} \right.\\ \Rightarrow \left( {x;y} \right) = \left( {\dfrac{1}{2}; - \dfrac{1}{2}} \right),\left( {\dfrac{2}{7};\dfrac{1}{7}} \right) \end{array}$