a+b+c=717
b=3a⇒ a=b:3
b=c-17⇒ c=b+17
inlocuim a si c in prima ecuatie
b:3+b+b+17=717
b:3+2b=717-17
b:3+2b=700
se aduce la acelasi numitor (3)
b+3×2b=3×700
b+6b=2100
7b=2100
b=2100:7
b=300
a=b:3⇒a=300:3=100
⇒a=100
c=b+17=300+17=317
⇒c=317
deci a=100, b=300 si c=317
verificare: a+b+c=717⇒ 100+300+317=717⇒717=717
