Giải Hệ Phương Trình: $\left \{ {{x-5y=-1} \atop {x^{2}} +y^{2}-3xy+x+y=10} \right.$

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