\[ \Leftrightarrow 3x(x + y) - 17 = 9x - 2(x + y)\]

\[ \Leftrightarrow (x + y)\,(3x + 2) = 17 + 9x\]

\[ \Leftrightarrow (x + y)\,(3x + 2) = 3(3x + 2) + 11\]

\[ \Leftrightarrow (3x + 2)\,(x + y - 3) = 11\]

\[ \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}3x + 2 = 11\\x + y - 3 = 1\end{array} \right.\\\left\{ \begin{array}{l}3x + 2 = - 1\\x + y - 3 = - 11\end{array} \right.\\\left\{ \begin{array}{l}3x + 2 = - 11\\x + y - 3 = - 1\end{array} \right.\\\left\{ \begin{array}{l}3x + 2 = 1\\x + y - 3 = 11\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x = 3\\y = 1\end{array} \right.\\\left\{ \begin{array}{l}x = - 1\\y = - 7\end{array} \right.\end{array} \right.\]

Vậy (x; y) = {(3; 1); (–1; –7)}