#### tristatefabricatorsinc

##### Junior Member

- Joined
- Jan 31, 2006

- Messages
- 60

. . .2x + 4y + 6z = 2

. . .x . . .. . .+ 2z = 0

. . .2x + 3y – z = -5

I want to start off by getting rid of the x variable - starting with equation 2, I did the following...

. . .-2(x + 2z = 0) => -2x - 2z = 0

Take this and add it to equation 1:

. .. .2x + 4y +6z = 2

. . .-2x. . . . . -2z = 0

. . .-------------------

. . .. . . .4y + 4z = 2

Now I will do the same thing to equation 2 and add it to equation 3:

. . .-2(x + 2z = 0) => -2x -2z = 0

. .. .2x + 3y - z = -5

. . .-2x . .. . - 2z = 0

. . .-------------------

. . .. .. . .3y - 3z = -5

Now I take my two answers and combine them:

. . .4y + 4z = 2: multiply by 3

. . .3y - 3z = -5: multiply by -4

to get...

. . . .12y + 12z = 6

. . .-12y + 12z = 20

. . .-------------------

. . .. . .. . .24z = 26

I am confused. It is not coming out even. Am I doing it wrong? Any help would be greatly appreciated. Thank you!