Question

rezolvati in Z ecuatia 2x-5y+3xy-2=0

Answer 1

Răspuns:

$(x,y)\in\{(2,-2),(3,-1),(1,0)\}$

Explicație pas cu pas:

$x=\dfrac{2+5y}{2+3y},\:y\ne -2/3\:(y\in\Bbb{Z})$.

$x\in\Bbb{Z}, y\in\Bbb{Z}\implies (2+3y)\mid (2+5y)$.

$2+3y\mid 2+5y\implies 2+3y\mid 3(2+5y)=6+15y\\2+3y\mid 2+3y\implies 2+3y\mid 5(2+3y)=10+15y$.

Scazand, obtinem: $2+3y\mid 10+15y-6-15y=4\iff 2+3y\mid 4$.

In concluzie, $2+3y\in\mathcal{D}_4=\{-4,-2,-1,1,2,4\}\iff y\in\{-2, -4/3, -1, -1/3, 0, 2/3\}$.

Dar $y\in\Bbb{Z}\implies y\in\{-2,-1,0\}$.

Cazul 1. $y=-2$. Ecuatia devine: $8-4x=0\implies x=2\in\Bbb{Z}$. O solutie este, deci, $(x_1, y_1)=(2,-2)$.

Cazul 2. $y=-1$. Ecuatia devine: $3-x=0\implies x=3\in\Bbb{Z}$. O solutie este, deci, $(x_1, y_1)=(3,-1)$.

Cazul 3. $y=0$. Ecuatia devine: $x-1=0\implies x=1\in\Bbb{Z}$. O solutie este, deci, $(x_1, y_1)=(1,0)$.

Solutii: {(2,-2),(3,-1),(1,0)}.