$\begin{array}{l} \left\{ \begin{array}{l} x - 5y = - 1\left( 1 \right)\\ {x^2} + {y^2} - 3xy + x + y = 10\left( 2 \right) \end{array} \right.\\ \left( 1 \right) \to \left( 2 \right):{\left( {5y - 1} \right)^2} + {y^2} - 3\left( {5y - 1} \right)y + 5y - 1 + y = 10\\ \Leftrightarrow 25{y^2} - 10y + 1 + {y^2} - 15{y^2} + 3y + 5y - 1 + y - 10 = 0\\ \Leftrightarrow 11{y^2} - y - 10 = 0\\ \Leftrightarrow \left( {y - 1} \right)\left( {11y + 10} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} y = 1 \Rightarrow x = 5y - 1 = 4\\ y = - \dfrac{{10}}{{11}} \Rightarrow x = \dfrac{{ - 61}}{{11}} \end{array} \right.\\ \Rightarrow \left( {x;y} \right) = \left( {4;1} \right),\left( { - \dfrac{{61}}{{11}}; - \dfrac{{10}}{{11}}} \right) \end{array}$
