SOLUTION: x+y+z=30000 .02x+.03y+.06z=1110 .02x+150=.03y : solve for x,y, and z. The problem im having with this problem is the "z" variable. It is only on 2 of the 3 equations. Iv

Question 106811: x+y+z=30000
.02x+.03y+.06z=1110
.02x+150=.03y : solve for x,y, and z.
The problem im having with this problem is the "z" variable. It is only on 2 of the 3 equations. Ive been trying to solve by using the method of elimination, but I cant seem to isolate the variables. thank you.

Answer by HyperBrain(694) (Show Source): You can put this solution on YOUR website!
Elimination method is perfect for this system!

First let's eliminate x! To do that, let's use and .
Let's multiply by 2!

Now, let's multiply by 100!

It's elimination time!

Line up the equations
_____________________________2x+3y+6z=111000
_____________________________2x+2y+2z=60000

If you subtract
from ,
you'll get

Solving .

Now, let's eliminate y! we'll do that by using and

Solving ,

Multiplying this by 100,

Multiplying by 100,

it's elimination time!

Line up the equations!

______________________________2x+3y+6z=111000
______________________________2x-3y =-15000
If you add these, you'll get:

Solving this,

Substituting and to ,

Yahoo! we found the value of z at last!
But wait, there's more!
Remember that . So,

Also,
. So,

therefore, x=13000, y=11000, and z=10000.

Power up,
HyperBrain!
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