Giúp mk với!!! Giải hệ phương trình: x^2-xy-2y^2=0 3x+y=1

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