1. Systems with Three Variables

2. Solve by elimination x – 3y +3z = -4 2x + 3y – z = 15 4x – 3y –z = 19
Note that the y values in this set are inverses. They will go away once you add the equations together.
x – 3y +3z = -4
2x + 3y – z = 15
4x – 3y –z = 19
Step 1: Pair the equation to eliminate y, because the y terms are already additive inverses
x – 3y +3z = -4
2x + 3y – z = 15
3x z = 11
2x + 3y – z = 15
4x – 3y –z = 19
6x z =34

3. 3x +2z = 11 6x – 2z = 34 9x = 45 x=5 3 (5) + 2z = 11 15 + 2z = 11
Step 2: Take the two new equations and use elimination to solve for the remaining two variables (in this case, x and z)
Plug in x into either equation to solve for z
3x +2z = 11
6x – 2z = 34
9x = 45
x=5
3 (5) + 2z = 11
15 + 2z = 11
2z = -4
z = -2

4. x=5, z =-2 x – 3y +3z = -4 5 – 3y + 3(-2)=-4 5 – 3y -6 = -4 -3y = -3
Step 3: Substitute the values for x and z back into one of the original equations (whichever seems easier) and solve for y
x=5, z =-2
x – 3y +3z = -4
5 – 3y + 3(-2)=-4
5 – 3y -6 = -4
-3y = -3
y=1
x – 3y +3z = -4
2x + 3y – z = 15
4x – 3y –z = 19
(5, 1, -2) is the solution

5. 3x3 Systems #1
2x + y – z = 5
3x - y + 2z = -1
x - y - z = 0

7. 3x3 Systems #2
x + 2y + z =10
2x – y + 3z = -5
2x – 3y – 5z = 27

9. 3x3 Systems #3
x – 3y + z =6
2x – 5y – z = -2
-x + y + 2z = 7

11. 3x3 Systems #4 6a + 12b – 8c = 24 9a + 18b – 12c = 30
Hint: can you factor?

13. 3x3 Systems #5
4x – 6y + 4z = 12
6x – 9y + 6z = 18
5x – 8y + 10z =20