Tìm số nguyên x, y biết xy – x + 2y = 3.

Ta có :

xy – x + 2y = 3

⇔ xy – x + 2y – 3 = 0

⇔ xy – x + 2y – 3 + 1 = 1

⇔ x(y - 1) + 2(y - 1) = 1

⇔ (y - 1)(x + 2) = 1

Suy ra: \(\left[ \begin{array}{l}\left\{ \begin{array}{l}y - 1 = 1\\x + 2 = 1\end{array} \right.\\\left\{ \begin{array}{l}y - 1 = - 1\\x + 2 = - 1\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}y = 2\\x = - 1\end{array} \right.\\\left\{ \begin{array}{l}y = 0\\x = - 3\end{array} \right.\end{array} \right.\)

Vậy (x;y) ∈ {(-1; 2), (-3; 0)}.