# Solving linear systems with three variables?

My Algebra 2 teacher has been out all week and we have had a sub. The sub doesn't really explain it well and I can't figure out solving linear systems with three variables using substitution or elimination. Can someone do these problems for me and show me the steps. You can at least do one and show me the steps. This hw is do tomorrow and it's getting late.

4x+y+3z=0

2x-2y-z=10

3x-2y+2x=11

Actually I'll only give you all one to help me with and show me the steps. Thanks

### 2 Answers

- Julius NLv 79 years agoFavorite Answer
4x+y+3z = 0; 2x-2y-z = 10; 5x-2y = 11;

4 1 3 0

2 -2 -1 -10

5 -2 0 -11

1 1/4 3/4 0

0 -5/2 -5/2 -10

0 -13/4 -15/4 -11

1 0 1/2 -1

0 1 1 4

0 0 -1/2 2

1 0 0 1

0 1 0 8

0 0 1 -4

--> x=-1; y=-8; z= 4;

- weiLv 43 years ago
i'm going to relabel the equations as follows a. 8x + 37 - 6z = 4, b. x-2y-z=2, c. 4x+y-2z=-4. 2b + c = 6x - 4z = 0 a -3c = -4x = sixteen --> x = -4 placed that into 6(-4) - 4z = 0 --> z = -6 and use this records to unravel for y in the two a b or c (-4) - 2y - (-6) = 2 --> y = 0 i'm going to do 2 a lots extra uncomplicated way utilising matricies. each and each extensive form is a separate get entry to and the '/' shows a clean row. A = [ 2 -a million 3 / 5 -4 -2 / 3 3 2 ] B = [ 7 / 3 / -8 ] x = [ x / y / z ] So the equipment could nicely be written as Ax = B to unravel multiply form the left with A^-a million A^-a million A x = A^-a million b x = A^-a million b x = [ -a million / -3 / 2 ] --> x = -a million, y = -3, z = 2.