Question 377901
x+y+z=18 <--First equation
x+y=15 <--Second equation
x=3z <--Third equation


Subtract second equation from the first one.

x+y+z=18
x+y=15
z=3


Substitute z=3 into the third equation.
x=3z
x=3(3)
x=9


Substitute x=9 into the second equation.
x+y=15
(9)+y=15
y=6


Numbers are 9, 6, and 3