# Solving Systems of Three Linear Equations in Three Variables - PowerPoint by tiw14488

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```									  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 = 11 3(–3) + 2(2) + 4(4) = 11 P
2x – y + 3z = 4   2(–3) – 2 + 3(4) = 4 P
5x – 3y + 5z = –1 5(–3) – 3(2) + 5(4) = –1P

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     x – 3y + 6z = 21
3x + 2y – 5z = –30   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)   (x – 3y + 6z = 21) (–2)
3x + 2y – 5z = –30        2x – 5y + 2z = –6

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

The solution is (–3, 2, 5).
It is very helpful to neatly organize your
work on your paper in the following manner.

(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)

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