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.