Răspuns :

x/3=y/7=z/11=k⇒x=3k,  y=7k,  z=11k  si dupa ce inlocuim in relatia din enunt, avem
3k+14k+33k=200
50k=200
k=4⇒x=12;  y=28;  z=44
Antoniadeboraid
[tex]x= \frac{3y}{7} [/tex]
[tex]y= \frac{7z}{11} [/tex]
rezulta :[tex]x= \frac{3 ( \frac{7z}{11})}{7} => x= \frac{21z}{11} ( \frac{1}{7} ) = x = \frac{21z}{77} = \frac{3z}{11} [/tex]

Inlocuim:
3z/11 + 2X(7z/11) + 3z=200
3z/11 + 14z/11+ 3z =200

Aducem la acelasi numitor si vine:
3z/11+14z/11+ 33z/11 = 2200/11
si ramane 3z+14z+33z=2200
50z = 2200
z = 2200/50
z = 44

Apoi inlocuim:
y = 7z/11
y = 7 X 44/11
y = 7 X 4 
y=28

x= 3y/7
x= 3 X 28 / 7
x = 3 X 4
x= 12