Giải thích các bước giải:

 Ta có:

$2x^2-3xy+3x-6y=18$
$\to 2x^2+3x-18=3xy+6y$

$\to 2x^2+3x-18=3y(x+2)$

$\to 2x^2+4x-x-2-16=3y(x+2)$

$\to 2x(x+2)-(x+2)-16=3y(x+2)$

Vì $3y(x+2)\quad\vdots\quad x+2$

$\to 16\quad\vdots\quad x+2$

$\to x+2\in U(16)$

$\to x+2\in\{-16, -8, -4, -2, -1, 1, 2, 4, 8, 16\}$

$\to x\in\{-18, -10, -6, -4, -3, -1, 0, 2, 6, 14\}$

$\to y\in\{-\dfrac{195}{16}, -\dfrac{161}{24}, -\dfrac{15}4, -\dfrac{11}6, 0, -\dfrac{10}3, -\dfrac32, \dfrac5{12}, \dfrac{27}8, \dfrac{425}{48}\}$

Do $x,y\in Z$

$\to x=-3, y=0$