This document discusses solving systems of three linear equations with three variables. It introduces Gaussian elimination as a method for solving such systems. Gaussian elimination involves eliminating variables one at a time to solve for the remaining variables in reverse order. The document provides an example of using Gaussian elimination to solve a 3x3 system. It also discusses the different outcomes possible for 3x3 systems, such as a single solution, infinitely many solutions, or no solution if the system is inconsistent or dependent.

1. 7.2 Systems of Linear
Equations: Three Variables
Chapter 7 Systems of Equations and Inequalities

2. Concepts & Objectives
⚫ The objectives for this section are
⚫ Solve systems of three equations in three variables.
⚫ Identify inconsistent systems of equations containing
three variables.
⚫ Express the solution of a system of dependent
equations containing three variables.

3. Three Equations, Three Variables
⚫ To solve a system of three equations and three variables
(usually called a “three-by-three system”), we can
certainly use the methods of substitution and
elimination that we used on two-variable systems.
⚫ Unfortunately, unless you know how to use a 3-D
graphing utility, graphing a three-by-three system isn’t
really practical.
⚫ The nSpire calculators will solve it for you, but
unfortunately, I’m not always going to let you get away
with just entering the answer. 😄

4. Gaussian Elimination
⚫ Sometimes, it can be helpful to have a structure to use
when trying to solve a system. One of the most
commonly-used systems is Gaussian elimination.
⚫ Gaussian elimination may not always be the most
efficient method, but as you become more proficient, you
can find ways to make it work better.
⚫ The main idea of Gaussian elimination is to eliminate the
x variable, then the y, which lets you solve for z. Then,
you substitute in the reverse direction (y, then x) to
solve.

5. Gaussian Elimination (cont.)
Example: Solve ( )
2 3 9 1
x y z
− + =
( )
( )
3 6 2
2 5 5 17 3
x y z
x y z
− + − = −
− + =

6. Gaussian Elimination (cont.)
Example: Solve
To eliminate x, we can combine equations 1 and 2.
( )
2 3 9 1
x y z
− + =
( )
( )
3 6 2
2 5 5 17 3
x y z
x y z
− + − = −
− + =
( )
2 3 9
3 6
2 3 4
x y z
x y z
y z
− + =
− + − = −
+ =

7. Gaussian Elimination (cont.)
Example: Solve
Since equation 2 already has a negative x, we can multiply
it by 2 to combine with equation 3:
( )
2 3 9 1
x y z
− + =
( )
( )
3 6 2
2 5 5 17 3
x y z
x y z
− + − = −
− + =
( )
( )
2 6 2 12 2 2
2 5 5 17
3 5 5
x y z
x y z
y z
− + − = − 
− + =
+ =

8. Gaussian Elimination (cont.)
Example (cont.)
Now, we use the two new equations to eliminate y.
Since the coefficient of y in both equations is 1, we can
multiply the top equation by –1 to solve for z:
2 3
3 5
y z
y z
+ =
+ =
2 3
3 5
2
y z
y z
z
− − = −
+ =
=

9. Gaussian Elimination (cont.)
Example (cont.):
Since we’ve solve for z, we can pick either equation 4 or 5
to solve for y:
And finally, we can solve for x:
( )
2 2 3
4 3
1
y
y
y
+ =
+ =
= −
( ) ( )
2 1 3 2 9
8 9
1
x
x
x
− − + =
+ =
= ( )
Solution: 1, 1, 2
−

10. Solutions
⚫ The solution set to a three-by-three system is an ordered
triple (x, y, z). This triple defines the point that is the
intersection of three planes in space. (Think of the
corner of a room where the walls and ceiling meet.)
⚫ Just as with two-by-two systems, however, sometimes a
three-by-three system will not have a single solution.
Compare the two pictures:

11. Solutions (cont.)
Just as with two-by-two systems, however, sometimes a
three-by-three system will not have a single solution.
Compare the two pictures:
Single point  single solution
Line  infinitely many solutions
(0 = 0)

12. Solutions (cont.)
These three pictures represent different scenarios for the
system to have no solution:
a) The three planes intersect with each other, but not at a
common point or line.
b) Two planes are parallel.
c) All three planes are parallel.

13. Inconsistent Systems
⚫ Consider the following system:
If we use the procedures we used earlier, we end up
with
3 4
2 5 3
5 13 3 8
x y z
x y z
x y z
− + =
− + − =
− + =
4 7
2 8 12
y z
y z
− − =
+ = −

14. Inconsistent Systems (cont.)
When we combine them to eliminate y, we get:
The final solution is a contradiction, so the system of
equations is inconsistent and has no solution.
( )
2 8 14 mult. by 2
2 8 12
0 2
y z
y z
− − =
+ = −
=

15. Dependent Systems
⚫ Now, consider the system
As before, we first work to eliminate x by multiplying the
first equation by ‒2 and adding it to the second
equation.
2 3 0
4 2 6 0
0
x y z
x y z
x y z
+ − =
+ − =
− + =
4 2 6 0
4 2 6 0
0 0
x y z
x y z
− − + =
+ − =
=
The true statement, 0 = 0,
tells us we have an
infinite number of
solutions.

16. Dependent Systems (cont.)
When a system is dependent, we can find general
expressions to the solutions. If we add the first and third
equation from the system, we get
We then solve the resulting equation for z:
2 3 0
0
3 2 0
x y z
x y z
x z
+ − =
− + =
− =
2 3
3
2
z x
z x
=
=

17. Dependent Systems (cont.)
We then back-substitute this expression for z into one of
the equations and solve for x:
And we get a general solution for the system of
3
2 3 0
2
9
2 0
2
9 5
2
2 2
x y x
x y x
y x x x
 
+ − =
 
 
+ − =
= − =
5 3
, ,
2 2
x x x
 
 
 

18. Classwork
⚫ College Algebra 2e
⚫ 7.2: 8-24 (4); 7.1: 42-50 (even); 6.7: 38-52 (even)
(omit 46,48)
⚫ 7.2 Classwork Check
⚫ Quiz 7.1