Question

Help ASAP would be awesome c:

Answer 1

I need to solve
x - 3y + 2z = 1
-x + 4y - 4z = 1
5x - y + 6z = 9

Answer 2

Favorite method?

Answer 3

So far what I did is

x - 3y + 2z = 1
+(-x + 4y - 4z = 1)

and I got y - 2z = 1

So then I multiplied -x + 4y - 4z = 1 by 5 to get -5x + 20y - 20z = 5
Then I did this

-5x + 20y - 20z = 5
+ 5x - y + 6z = 9
to get 19y - 14z = 14

Am I correct so far?

I don't have a favorite method

Answer 4

1+1 = 2
isn't it ?

Answer 5

ahh I missed that, yes it does ><

Answer 6

that fails miserably I know >.>

Answer 7

2nd equation is correct
19y-14z=14

Answer 8

Looks good to me. Now you just need to take the two variable equations and solve by substitution or elimination.

Answer 9

y-2z = 2
so, y = 2+2z
plug this in 2nd correct equation

Answer 10

So what I'm doing next is

multiplying y - 2z = 2 by -7 so that it's -7y + 14z = 14 so that I can eliminate "z" in this:
19y - 14z = 14
+ -7y + 14z = -14
which gives me
12y = 0

Is that right?

Answer 11

yes, y comes out to be 0
good :)

Answer 12

now you can find z easily

Answer 13

put y= 0 in y-2z=2

Answer 14

I need to find x as well, can I find z first or do I have to find x and then y?

Answer 15

you will be able to find z first, not x
because you have one equation in y and z

Answer 16

**two equations in y and z
use any

Answer 17

I'll put y into this equation "y-2z = 2"
so
y-2z = 2
0 - 2z = 2
-2z = 2
divide by -2 on both sides

z = -1?

Answer 18

yes, z=-1 is correct :)
now find x using any equation in x

Answer 19

I have to use an original equation this time correct?

Answer 20

yes, any 1 of the 3 original equations...

Answer 21

Okay I'll use "5x - y + 6z = 9"
So
5x - y + 6z = 9
5x - 0 + 6(-1) = 9
5x - 6 = 9
+6 +6
5x = 15
divide by 5 on both sides
x = 3

Answer 22

correct! :)

Answer 23

Awesome! Thank you! :3