-fie x,y,z cele 3 numere necunoscute
[tex] \frac{x+y+z}{3}=2,2 \\ \frac{x+y}{2}=1,15 [/tex] atunci z=?
[tex] \frac{x+y+z}{3}=2,2=\ \textgreater \ x+y+z=2,2\cdot3=\ \textgreater \ x+y+z=6,6 \\ \frac{x+y}{2}=1,15=\ \textgreater \ x+y=1,15\cdot2=\ \textgreater \ x+y=2,3 =\ \textgreater \ 2,3+z=6,6 \\ z=6,6-2,3 \\ z=4,3[/tex]
