y(x+3)+2x=7
y=(7-2x)/(x+3)∈Z x≠-3
(2x-7)/(x+3)∈Z
(2x+6-13)/(x+3)∈Z
2-13/(x+3)∈Z
cum 2∈Z, ramane ca
13/(x+3)∈Z
x+3∈D13Z= {1;-1;13'-13}
x+3=1 x=-2; y= (7-2x)/(x+3)=(7+4)/(-2+3)=11
x+3=-1 x=-4 y=(7-2x)/(x+3)=(7+8)/(-4+3)=-15
x+3=13 x=10 y=(7-2x)/(x+3)=(7-20)/(10+3)=-13/(13)=-1
x=3=-13 x=-16 y=(7-2x)/(x+3)= (7+32)/(-16+3)=39/(-13)=-3
(x;y)∈{(-2;11);(-4;-15);(10;-1);(-16;-3)}
verificare
-22-4+33=33-28=7
60-8-45=60-53=7
-10+20-3=10-3=7
48-32-9=48-41=7
toatecele 4 perechi verifica, problema e bine rezolvata
