Răspuns:
Explicație pas cu pas:
x(3y+1)=6y-3
x= (6y-3)/(3y+1) = (6y+2-5)/(3y+1)=2-5/(3y+1)
2∈Z⇒5/(3y+1)∈Z⇒3y+1∈{-5;-1;1;5}⇒3y∈{-6;-2;0;4}⇒
⇒y∈{-2; -2/3;0;4/3}∩Z= {-2;0}
pt y=-2...x=3
py y=0...x=-3
b)alegem y=1/3 obtinem x=-1/2
c)fie y1=π. obtinem x= (6π-3)/(3π+1)∈R\Q
42 fiind o suma de module minimul este 0+0=0
3x+4y=-1
x-2y=2
3x+4y=-1
2x-4y=4
5x=3
x=3/5
y=(x-2)/2= (3/5-2)/2=-7/10
care verifica