Avem,t+m+f1+f2+f3=440
t=f1+f2+f3
m=3t/4
f1=t/2
f3=(f1+f2+f3)/8
Din,prima,relatie,avem
2(f1+f2+f3)+3(f1+f2+f3)/4=440
Aducem,la,numitor,comun,si,avem
8(f1+f2+f3)+3(f1+f2+f3)=4x440
11(f1+f2+f3)=4x440
f1+f2+f3=160=t
m=3x160/4=120
f1=t/2=160/2=80
f3=160/8=20
f2=160-80-20=60