1. 5.2 Systems of Linear Equations in Three Variables

2. Example: Solve x + 2y + z = 4 4y – 3z = 1 5z = 5
This system is in Triangular Form (its equations follow a stair-step pattern).
Example: Solve x + 2y + z = y – 3z = z = 5
Solve for one of the variables, somewhere
5z = z = 1
Now substitute into the other equations
x + 2y = y – 3(1) = 1

3. Solve the 2nd equation for y
4y = 4 y = 1
Now, solve the 1st equation for x
x + 2(1) + 1 = 4
x = x = 1
(1, 1, 1)

4. Example: Solve: x + y + z = 0 2x + 2y – 3z = 5 x – 4y – 3z = –11
This system is not in triangular
form. Use elimination to get it
in triangular form, or linear
combinations to produce a system
of 2 equations in 2 variables...
Example: Solve: x + y + z = x + 2y – 3z = x – 4y – 3z = –11
We will use linear combinations in the 1st and 2nd equations to eliminate the z terms :
3 (x + y + z = 0) x + 3y + 3z = 0
2x + 2y – 3z = x + 2y – 3z = x + 5y = 5

5. We will do the same for the 2nd and 3rd equations:
2x + 2y – 3z = x + 2y – 3z = 5
–1(x – 4y – 3z = –11) –x + 4y + 3z = 11
x + 6y = 16
We now have now produced a system of 2 equations in 2 variables…
5x + 5y = 5
x + 6y = 16

6. x + 6y = x + 5y = 5
–5x – 30y = – x y = 5
–25y = –75
y = 3

7. Now, find x: x + 6(3) = x = 16
x = –2
Finally, go back to one of the original equations to find z: x + y + z = 0
– z = 0
1 + z = 0
z = –1 (–2, 3, –1)