SOLUTION: Is there an easy way to solve systems with 3 variables? I'm having a hard time understanding exactly what to do & the step-by step solutions.... PLEASE HELP !! Example: solve the

Question 31591: Is there an easy way to solve systems with 3 variables? I'm having a hard time understanding exactly what to do & the step-by step solutions.... PLEASE HELP !!
Example:
solve the system:
3x-3y+z=1
x+6y-z=5
2x-9y+2z=-8
AND
x+y=0
x+z=1
2x+y+z=2

Thank you very much for your help

Dianne
Answer by longjonsilver(2297) (Show Source): You can put this solution on YOUR website!
if you have not done matrices then you will have to a more long-winded algebraic approach:

I shall do the second example...it looks easier:-). You can then try the first one, in the same way.

First off, we will choose one variable in one of the 3 equations and then sub this into the other 2 equations, leaving us with 2 equations with just unknowns in:

eqn1 - x+y=0
eqn2 - x+z=1
eqn3 - 2x+y+z=2

Looking at this, eqn1 already is just 2 variables...x and y. So my approach will be to get z=? in either of eqn2 or eqn3 and then sub that back into eqn3 or eqn2, depending upon which i chose.

So... i will chose to do the following:
eqn2 - x+z=1
z = 1-x

Sub this into eqn3: 2x+y+z=2
--> 2x+y+(1-x)=2
2x+y+1-x=2
x+y+1=2
x+y=1

So we have x+y=0 --eqn1
and also x+y=1

Looking at these, there can be no solution, since a number added to another number cannot simultaneously be equal to 0 and 1...so no solution.

Hopefully, the other example will lead to a solution.

jon.
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