Question

Daca:

x+y=162
y+z=354
x+z=408

Gasiti valorile pentru fiecare necunoscuta, si va rog explicati cum ati rezolvat. :)

Answer 1

x+y=162
y+z=354⇒y=354-z
x+z=408⇒x=408-z
Inlocuim in prima ecuatie
x+y=162⇒408-z+354-z=162⇒ -2z=162-762⇒ -2z= -600⇒z=300
y=354-z=354-300=54
x=408-z=408-300=108

Answer 2

x+y=162
y+z=354
x+z=408

2x+2y+2z=162+354+408
2(x+y+z)=924
x+y+z=462

z=462-162
z=300

x=462-354
x=108

y=462-408
y=54