1. 3.4 Solving Systems with 3 variables

2. Systems of equations with 3 variables need 3 equations
Systems of equations with 3 variables need 3 equations. We will pick 2 equations and try to eliminate a variable. Pick 2 other equations and try to eliminate the same variable creating 2 equations with 2 variables. We will then solve for those 2 variables and substitute into an original equation to find the third variable.

3. Ex.
2x + y – z = 6
2x – 4y + z = 4
6x +2y –z = -4
Take the first and second equation and add them
+ 2x – 4y + z = 4
________________
4x -3y = 10

4. Now take the 2nd and last equation and eliminate the same z
2x – 4y + z = 4
+ 6x +2y –z = -4
_____________
8x -2y = 0
Now take those equations and make another system:
4x -3y = times x +6y = -20
8x -2y = x -2y = 0
____________
4y = -20
y = -5

5. Plug y back in to find x
8x – 2(-5) = 0
8x + 10 = 0
8x = -10
x =
Now plug y and x back into original equation to get z
2 ( ) z = 6
z = 6
z = 12 2
-z = z = ( , -5, )