`xy - 10 +5x - 3y = 2`
`=> xy + 5x - 10 - 3y = 2`
`=> x(y+5) - 3(y+5) = -3`
`=> y+5(x - 3) = -3`
`=> y +5 ; x - 3 \in Ư(3)={±1 ; ±3}`
Ta có bảng sau :
$\begin{array}{|c|c|} \hline x -3&1&3&-1&-3 \\\hline y+5&-3&-1&3&1 \\\hline x&4&6&2&0 \\\hline y&-8&-6&-2&-4 \\\hline\end{array}$
Vậy `(x ; y) = ( 4 ; -8) ; (6 ; -6 ) ; (2 ; -2) ; (0 ; -4)`