Question

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

Answer 1

I prefer a question at a time. I'd help you with the first.

Answer 2

Add all equations.

Answer 3

Lol it is a question, 9ut the last 8 sets are choices.

Answer 4

4 sets*

Answer 5

lol

Answer 6

Ha ha ha..

Answer 7

Start by adding the top two equations.

Answer 8

What do you get?? @Loujoelou

Answer 9

I know it's either C/D, cause 3x+3z=1 is what you get when you eliminate y from 1st/2nd equation.

Answer 10

nvm other way around. It's A/B

Answer 11

Cause you get 7x-3z=4 not what I said earlier.

Answer 12

Yep.

Answer 13

Now, you multiply the second equation by 2 and then add equation 1 and 2.

Answer 14

Now multiply second equation by 2 and then add second and third equation and let us see what you get.

Answer 15

K I get it now :) It's A cause you get 12x-9z=3. Was just a little confused about it.

Answer 16

Yep! :)

Answer 17

tyvm @waterineyes @ParthKohli I understand it perfectly now :) At least doing this by elimination method I mean.

Answer 18

You got the thing successfully. Congratulations, Lou!

Answer 19

You are doing this by elimination only @Loujoelou

Answer 20

Lol thx @ParthKohli and yes for this problem I had to @waterineyes

Answer 21

Okay..