Đáp án: $x=2t+1,y=5t-2, t\in Z, t\ne 0$
Giải thích các bước giải:
ĐKXĐ: $y\ne -2$
Ta có:
$\dfrac{x-1}{y+2}=\dfrac25$
$\to 5(x-1)=2(y+2)$
$\to 5x-5=2y+4$
$\to 5x=2y+9$
$\to 2y+9\quad\vdots\quad 5$
$\to 2y+4+5\quad\vdots\quad 5$
$\to 2y+4\quad\vdots\quad 5$
$\to 2(y+2)\quad\vdots\quad 5$
$\to y+2\quad\vdots\quad 5$
$\to y+2=5t, t\in Z, t\ne 0$ vì $y+2\ne 0$
$\to y=5t-2$
$\to 5x=2(5t-2)+9$
$\to 5x=10t+5$
$\to x=2t+1$
Vậy $x=2t+1,y=5t-2, t\in Z, t\ne 0$
