Question

What is the solution of the system? Use elimination.

x+2y+z=-2
3x-2y+z=10
x-4y+3z=3

Answer 1

Need help starting this, I don't know how to do this since there is 3 equations instead of 2.

Answer 2

Get a system in terms of just z and x by looking at two systems at a time.
Look at x+2y+z=-2 x-4y+3z=3
3x-2y+z=10 x+2y+z=-2

Answer 3

Take any of the two equations, I picked eq 1 and 2, and eliminate a variable, I chose to eliminate z.
x + 2y + z = -2
3x - 2y + z = 10 -->(-1)
-----------------
x + 2y + z = -2
-3x + 2y - z = -10 (result of multiplying by -1)
------------------add
-2x + 4y = -12

now take the equation you did not do, eq 3, and pick another equation, I chose eq 2, and do what you have to do to eliminate z. Just remember, what you eliminate in the first two equations has to be the same thing that you eliminate in the last two equations.
3x - 2y + z = 10 --> (-3)
x - 4y + 3z = 3
-------------------
-9x + 6y - 3z = -30 (result of multiplying by -3)
x - 4y + 3z = 3
-------------------add
-8x + 2y = - 27

now take the " answers " you got and eliminate a variable
-2x + 4y = -12
-8x + 2y = -27 -->(-2)
---------------
-2x + 4y = -12
16x - 4y = 54 (result of multiplying by -2)
--------------add
14x = 42
x = 3

now sub 3 in for x in one of the two equations we just worked on
-8x + 2y = -27
-8(3) + 2y = -27
-24 + 2y = -27
2y = -27 + 24
2y = -3
y = -3/2

now sub in 3 for x and -3/2 in for y in one of the original equations
3x - 2y + z = 10
3(3) - 2(-3/2) + z = 10
9 + 3 + z = 10
12 + z = 10
z = 10 - 12
z = -2

check...
x - 4y + 3z = 3
3 - 4(-3/2) + 3(-2) = 3
3 + 6 - 6 = 3
9 - 6 = 3
3 = 3 (correct)

x = 3 ,, y = -3/2 and z = -2

I tried to do it step by step so you could see how I did it. If you have any questions, please ask.

Answer 4

Wow, thank you so much! You have helped in a lot and I appreciate it. I understand a lot better now.