3.5 solving systems of equations in three variables

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.

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.

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.

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

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.

This lesson will focus on the
Elimination Method.

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

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

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

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.

(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

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

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

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

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

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

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).




is very helpful to neatly organize you
k on your paper in the following mann

(x, y, z)

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

(4, 3, –3)

Here’s another one to try.
–5x + 3y + z = –15
10x + 2y + 8z = 18
15x + 5y + 7z = 9

(1, –4, 2)

3.5 solving systems of equations in three variables

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3.5 solving systems of equations in three variables