tristatefabricatorsinc
Junior Member
- Joined
- Jan 31, 2006
- Messages
- 60
I am trying to solve the following equation using elimination
. . .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!
. . .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!