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If you were to use the elimination method to solve the following system, choose the new system of equations that would result after the variable y is eliminated in the first and second equations, then the second and third equations. 3x – y + 3z = 1 4x + y – 6z = 3 4x – 2y + 3z = –3 7x – 3z = 4 12x – 9z = 3 7x – 3z = 4 4x – 6z = 3 3x + 3z = 1 12x – 9z = 3 3x + 3z = 1 4x – 6z = 3

Mathematics
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I prefer a question at a time. I'd help you with the first.
Add all equations.
Lol it is a question, 9ut the last 8 sets are choices.

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Other answers:

4 sets*
lol
Ha ha ha..
Start by adding the top two equations.
What do you get?? @Loujoelou
I know it's either C/D, cause 3x+3z=1 is what you get when you eliminate y from 1st/2nd equation.
nvm other way around. It's A/B
Cause you get 7x-3z=4 not what I said earlier.
Yep.
Now, you multiply the second equation by 2 and then add equation 1 and 2.
Now multiply second equation by 2 and then add second and third equation and let us see what you get.
K I get it now :) It's A cause you get 12x-9z=3. Was just a little confused about it.
Yep! :)
tyvm @waterineyes @ParthKohli I understand it perfectly now :) At least doing this by elimination method I mean.
You got the thing successfully. Congratulations, Lou!
You are doing this by elimination only @Loujoelou
Lol thx @ParthKohli and yes for this problem I had to @waterineyes
Okay..

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