SOLUTION: I need to solve for a system of three variables, and this is the first time I've ever came across this so I don't really know if I'm doing this right... So here is the system:

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Question 389508: I need to solve for a system of three variables, and this is the first time I've ever came across this so I don't really know if I'm doing this right...
So here is the system:
2x - y + 3z = 8
x - 6y - z = 0
-6x + 3y - 9z = 24
I decided to get x taken care of first so I took the middle equation and it made x = 6y + z
And then I took x and plugged it in to the top equation so it would be 2(6y+z) - y + 3z = 8 and then it went to 12y + 2z - y + 3z = 8 and then 11y + 5z = 8 and then I got y = 5/11z +8/11
Then to get z I went ahead and plugged in x and y into the third equation, (which is where I think I made the wrong move...) then I got -36y - 6z + 15/11z + 24/11 - 9z = 24
Added like terms, and then that's where I got stumped. -6z + 15/11z is 9/11z, and then when you add the -9z it makes 0/11z? and if I actually did solve for z I really wouldn't get the first solution...
Can someone please help me point out where I went wrong and help me through this problem??? Thanks!
Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website!
2x - y + 3z = 8......................1
x - 6y - z = 0.......................2
-6x + 3y - 9z = 24 ..................3
..
select (1) &(2) eliminate x
select (2)&(3) eliminate x
..
Then solve for y & z
multiply (2) by -2 and add to (1)
-2x+12y+2z=0
Add
+11y+5z=8......................(3)
...
consider (2)&(3) eliminate x
multiply (2)by 6 and add to (3)
6x-36y-6z=0
add
-33y-7z=24.....................(4)
..
multiply (3) by 3 & add to 4
33y+15z=24
add
+8z=48

plug z= 6 in (3)
11y+5*6=8
11y+30=8
11y=-22

...
plug y & z to get x
2x - y + 3z = 8......................1
2x+2+18=8
2x+20=8
2x=-12

...

m.ananth@hotmail.ca