1. 3.5 Solving Systems of Three
Linear Equations in Three
Variables
The Elimination Method
SPI 3103.3.8
Solve systems of three linear equations in three variables.

2. Solutions of a system with 3 equations
The solution to a system of three
linear equations in three
variables is an ordered triple.
(x, y, z)
The solution must be a solution
of all 3 equations.

3. Is (–3, 2, 4) a solution of this system?
3x + 2y + 4z = 3(–3) + 2(2) + 4(4) = 
11
11

2x – y + 3z = 4 2(–3) – 2 + 3(4) = 4 
5x – 3y + 5z = 5(–3) – 3(2) + 5(4) =
–1
–1
Yes, it is a solution to the system
because it is a solution to all 3
equations.

4. Methods Used to Solve Systems in 3 Variables
1. Substitution
2. Elimination
3. Cramer’s Rule
4. Gauss-Jordan Method
….. And others

5. Why not graphing?
While graphing may technically be
used as a means to solve a system
of three linear equations in three
variables, it is very tedious and
very difficult to find an accurate
solution.
The graph of a linear equation in
three variables is a plane.

6. This lesson will focus on the
Elimination Method.

7. Use elimination to solve the following
system of equations.
x – 3y + 6z = 21
3x + 2y – 5z = –30
2x – 5y + 2z = –6

8. Step 1
Rewrite the system as two smaller
systems, each containing two of the
three equations.

9. x – 3y + 6z = 21
3x + 2y – 5z = –30
2x – 5y + 2z = –6
x – 3y + 6z = 21
3x + 2y – 5z = –30
x – 3y + 6z = 21
2x – 5y + 2z = –6

10. Step 2
Eliminate THE SAME variable in each
of the two smaller systems.
Any variable will work, but sometimes
one may be a bit easier to eliminate.
I choose x for this system.

11. (x – 3y + 6z = 21) (–3)
3x + 2y – 5z = –30
(x – 3y + 6z = 21) (–2)
2x – 5y + 2z = –6
–3x + 9y – 18z = –63
3x + 2y – 5z = –30
–2x + 6y – 12z = –42
2x – 5y + 2z = –6
11y – 23z = –93
y – 10z = –48

12. Step 3
Write the resulting equations in two
variables together as a system of
equations.
Solve the system for the two
remaining variables.

13. 11y – 23z = –93
y – 10z = –48 (–11)
11y – 23z = –93
–11y + 110z = 528
87z = 435
z=5
y – 10(5) = –48
y – 50 = –48
y=2

14. Step 4
Substitute the value of the variables
from the system of two equations in
one of the ORIGINAL equations with
three variables.

15. x – 3y + 6z = 21
3x + 2y – 5z = –30
2x – 5y + 2z = –6
I choose the first equation.
x – 3(2) + 6(5) = 21
x – 6 + 30 = 21
x + 24 = 21
x = –3

16. Step 5
CHECK the solution in ALL 3 of the
original equations.
Write the solution as an ordered
triple.

17. x – 3y + 6z = 21
3x + 2y – 5z = –30
2x – 5y + 2z = –6

–3 – 3(2) + 6(5) = 21
3(–3) + 2(2) – 5(5) = –30
2(–3) – 5(2) + 2(5) = –6
The solution is (–3, 2, 5).



18. is very helpful to neatly organize you
k on your paper in the following mann
(x, y, z)

19. Try this one.
x – 6y – 2z = –8
–x + 5y + 3z = 2
3x – 2y – 4z = 18
(4, 3, –3)

20. Here’s another one to try.
–5x + 3y + z = –15
10x + 2y + 8z = 18
15x + 5y + 7z = 9
(1, –4, 2)