3x+7y+5z=6
3x+3·3y-2y+5z=6
3(x+3y)-2y+5z=6
Notam x+3y=a∈Z
3a-2y+5z=6
3a-2y+2·2z+z=6
3a-2(y-2z)+z=6
Notam y-2z=b∈Z
3a-2b+z=6
z=6-3a+2b∈Z
y-2z=b⇒y=b+2z=b+12-6a+4b=-6a+5b+12
x+3y=a⇒x=a-3y=a+18a-15b-36=19a-15b-36
Concluzie: Ecuatia admite o infinitate de solutii. Solutiile sunt de forma:
x=19a-15b-36
y=-6a+5b+12
z=6-3a+2b, unde a, b sunt doua numere intregi oarecare.