Tìm x , y thuộc Z `3y . ( x - 5 ) + 7 . ( x - 5 ) = 69`

`3y . (x - 5) + 7 . (x - 5) = 69`

`<=> (x - 5) . (3y + 7) = 69`

`<=> 69 = {( 3 . 23 ),( 23 . 3 ),( (-3) . (-23) ),( (-23) . (-3) ),( 1 . 69 ),( 69 . 1 ),( (-1) . (-69) ),( (-69) . (-1) ):}`

Ta có các trường hợp sau:

`TH_{1} :`

$\left[\begin{matrix} x - 5 = 3\\ 3y + 7 = 23\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = 3 + 5\\ 3y = 23 - 7\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = 8\\ 3y = 16\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = 8\\ y = 16 : 3\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = 8\\ y = \dfrac{16}{3}\end{matrix}\right.$

`TH_{2} :`

$\left[\begin{matrix} x - 5 = 23\\ 3y + 7 = 3\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = 23 + 5\\ 3y = 3 - 7\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = 28\\ 3y = -4\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = 28\\ y = -4 : 3\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = 28\\ y = \dfrac{-4}{3}\end{matrix}\right.$

`TH_{3} :`

$\left[\begin{matrix} x - 5 = -3\\ 3y + 7 = -23\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = -3 + 5\\ 3y = -23 - 7\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = 2\\ 3y = -30\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = 2\\ y = -30 : 3\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = 2\\ y = -10\end{matrix}\right.$

`TH_{4} :`

$\left[\begin{matrix} x - 5 = -23\\ 3y + 7 = -3\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = -23 + 5\\ 3y = -3 - 7\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = -18\\ 3y = -10\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = -18\\ y = -10 : 3\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = 28\\ y = \dfrac{-10}{3}\end{matrix}\right.$

`TH_{5} :`

$\left[\begin{matrix} x - 5 = 1\\ 3y + 7 = 69\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = 1 + 5\\ 3y = 69 - 7\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = 6\\ 3y = 62\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = 6\\ y = 62 : 3\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = 6\\ y = \dfrac{62}{3}\end{matrix}\right.$

`TH_{6} :` 

$\left[\begin{matrix} x - 5 = 69\\ 3y + 7 = 1\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = 69 + 5\\ 3y = 1 - 7\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = 74\\ 3y = -6\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = 74\\ y = -6 : 3\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = 74\\ y = -2\end{matrix}\right.$

`TH_{7} :`

$\left[\begin{matrix} x - 5 = -1\\ 3y + 7 = -69\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = -1 + 5\\ 3y = -69 - 7\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = 4\\ 3y = -76\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = 4\\ y = -76 : 3\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = 4\\ y = \dfrac{-76}{3}\end{matrix}\right.$

`TH_{8} :`

$\left[\begin{matrix} x - 5 = -69\\ 3y + 7 = -1\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = -69 + 5\\ 3y = -1 - 7\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = -64\\ 3y = -8\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = -64\\ y = -8 : 3\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = -64\\ y = \dfrac{-8}{3}\end{matrix}\right.$

Vậy: các cặp `xy` thỏa mãn là: `(8; 16/3) ; (28; -4/3) ; (2; -10) ; (28; -10/3) ; (6; 62/3) ; (74; -2) ; (4; -76/3) ; (-64; -8/3)`