Question

I need help with solving these systems for the x,y,and z. I know the basic steps but I need help working it out!
x - 3y + 4z = -9
2x + y + 3z = 6
x - 4y + 3z = -9
Please do not just give me the answer I need to learn this.

Answer 1

@ChmE pleaseee

Answer 2

Go through the same steps as http://openstudy.com/study#/updates/50b7ff75e4b0c789d50ff722

Answer 3

I'll check your answer

Answer 4

okay thanks I need a second though

Answer 5

which method would i use substitution? elimination?

Answer 6

elimination til you get down to 2 equations. Then I prefer to substitute. But you could eliminate again to solve for one variable.

Answer 7

I started out by multiplying the first and third equation by 2 is that right

Answer 8

keep going

Answer 9

the equations I got to start with are
1.2x-6y+8z=-18
2.2x+y+3z=6
3.2x+8y+6z=18
And then I added ( because my instructor told me to add only)
1. & 2. and got:
4x-5y+11z=-12, but I don't understand I think I have to subtract to get x by its self

Answer 10

when I subtract I get -7y+5z=-24

Answer 11

multiply the 1st equation by -1. Then add equations one and two

Answer 12

exactly

Answer 13

x - 3y + 4z = -9
2x + y + 3z = 6
x - 4y + 3z = -9

2x - 6y + 8z = -18
2x + y + 3z = 6
2x - 8y + 6z = -18

2x - 6y + 8z = -18
-2x + -y + -3z = -6
-7y + 5z = -24

2x - 6y + 8z = -18
-2x + 8y + -6z = 18
2y + 2z = 0

-7y + 5z = -24
2y + 2z = 0

-7y + 5z = -24
y = -z

-7(-z) + 5z = -24
7z + 5z = -24
12z = -24
z = -2

y = -z
y = -(-2)
y = 2

x - 3y + 4z = -9
x - 3(2) + 4(-2) = -9
x - 6 - 8 = -9
x - 14 = -9

x = 5
y = 2
z = -2

Answer 14

In the 4th step how did you get a positive 18

Answer 15

oh u multiplied by 1

Answer 16

-1

Answer 17

that is from the third equation which was multiplied by -1, correct

Answer 18

Okay I just worked it out and it matched thanks so much!Again lol,

Answer 19

yup. You should be able to do all the rest now. I, like your teacher, prefer adding. It is harder to make a stupid mistake with all the negative signs floating around