Question

Determina numerele x,y si z daca x+y+z=132 x-z = y+z iar z=y :4

Answer 1

x+y+z=132
z=y:4⇒y=4z
x-z=y+z⇔ x-z=4z+z
x-z=5z
x=z+5z=6z
x=6z
se inlocuieste in prima relatie xsi y cu z: x+y+z=132
6z+4z+z=132
1z=132
z=132:11
z=12
y=4z=4x12=48
y=48
x=6z=6x12=72
x=72
12+48+72=132