The Alienware Server costs $5768.
This is how I got that answer:
IBM = a
DELL = b
Alienware = c
Custom = x
We know that:
a + b + c + x = 16480
---------------------------------
(I've summerized that paragraph into these true values: )
a = b + .5x
a = 3x
b = c - x
c = b + x
After that we can find that since a = 3x that b must equal 2.5x
So since c = b + x we can sub the b with 2.5x giving us c = 3.5x
So now we know:
a = 3x
b = 2.5x
c = 3.5x
So we sub in those values into the main equation (a+b+c+x = 16480)
This leads us to 10x = 16480
16480 divided by 10 equals 1648, so we now know that:
x = 1648
And since c = 3.5x we multiply 3.5 by 1648 which gives us 5768
So my final answer is Alienware (c) = $5768
Now my 8th grade math teacher always got on me about checking my work, so here it goes (these are the costs for all the servers):
a = 4944
b = 4120
c = 5768
x = 1648
-------------- (add those all up and you get....)
16480
God, it's 4 AM over here and I'm really tired but I can't stand not being able to do a math equation so I probably wouldn't get much sleep anyway if I didn't do this.