\[12 \cdot (x - 1) - 6x = x(x - 1)\]
\[12x - 12 - 6x = x^{2} - x\]
\[x^{2} - x - 6x + 12 = 0\]
\[x^{2} - 7x + 12 = 0\]
\[x_{1} + x_{2} = 7;\ \ \ \ x_{1} \cdot x_{2} = 12\]
\[x_{1} = 3;\ \ \ x_{2} = 4.\]
\[\left\{ \begin{matrix} x = 3 \\ y = 4 \\ \end{matrix} \right.\ \ \ \ \ \ \ \ или\ \ \ \ \ \left\{ \begin{matrix} x = 4 \\ y = 6 \\ \end{matrix} \right.\ \]
\[Ответ:(3;4);\ \ (4;6).\]