SOLUTION: I think that my problem falls under this topic. I am in Algebra 2 and I am stuck on this one problem. Solving Systems of Equations in Three Variables. Here's the problem: 1.)2x-y

Question 235195: I think that my problem falls under this topic. I am in Algebra 2 and I am stuck on this one problem. Solving Systems of Equations in Three Variables.
Here's the problem:
1.)2x-y+z=1
2.)x+2y-4z=3
3.)4x+3y-7z=-8
So here's what I tried to do:
First, I put equations 1 and 2 together and used the elimination method and I got -5y+9z=-5
so that would be the 4th equation
Then, I put equations 2 and 3 together and used elimination again and got -5y+9z=-20
which would be the 5th equation
I tried to put equations 4 and 5 together since they had like terms but it just didn't seem to work out well at all. Asking for your help and thanking you in advance, bthnydyson
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
1.)2x - y + z = 1
2.)x + 2y -4z = 3
3.)4x +3y -7z =-8
:
Multiply the 1st equation by 2 and add to the 2nd equation
4x - 2y + 2z = 2
x + 2y - 4z = 3
-------------------addition eliminates y
5x - 2z = 5
;
Multiply the 1st equation by 3 and add to the 3rd equaiont
6x - 3y + 3z = 3
4x + 3y - 7z = -8
--------------------addition eliminates y again
10x - 4z = -5
:
You can see right now that this system has no solution
If you multiply eq 4 by 2 and add to the above, you can see
10x - 4z = 10
10x - 4z = -5
:
No wonder you were having trouble
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