Part 1: Explain, in complete sentences, which method you would use to solve the following system of equations. (1 point)
Part 2: Explain why you chose that method (1 point)
Part 3: Provide the solution to the system. (2 points)
x – 2y + z = 0
2x – 3y – 4z = –9
x + 2y – 5z = 0
1. I used elimination method
2. I used it cause it's easier than the rest.
3. I got ((97/8,10/3,23/18)
Can anyone double-check my solution?

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- anonymous

- anonymous

Sure why not..

- anonymous

Yeah I will also prefer elimination here..

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- anonymous

Reason is:
The first and third equation consists same x and 2y so that they can be cancelled out by just addition or subtraction..

- anonymous

Let me solve this please wait.

- anonymous

K no problem :)

- anonymous

Add first and third equation you will get:
2x - 4z = 0
2x = 4z ---------------------1
Now multiply first equation with 3 and second equation by 2:
3x – 6y + 3z = 0
4x – 6y – 8z = –18
Now subtract them:
-x + 11z = 18
-2z + 11z = 18
9z = 18
z = 2

- anonymous

I got z = 2 check your solution once again @Loujoelou

- anonymous

K. I thought my solution was a bit weird.

- anonymous

K I see how you got z, and I realize x=4 correct?

- anonymous

@waterineyes Now I got (4,3,2)

- anonymous

x = 2z
if z = 2 then surely x will be 4..

- anonymous

Yes you are right now..

- anonymous

May I suggest you something?

- anonymous

Of course.

- anonymous

You can verify your answer..
See you got 4, 3 and 2,
Just plug in these values in the equation given and check whether they are coming equal..
Like for first:
x – 2y + z = 0
4 - 2(3) + 2 = 0
6 - 6 = 0
0 = 0
This means your answer is correct..
Now similarly put in second and third you will get 0 in each case..
This is how we verify our answers..

- anonymous

K :D Tyvm @waterineyes

- anonymous

Welcome Dear.. \(\huge \checkmark\)

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