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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