College Algebra Lecture Notes Section 5.2 Page 1 of 8
Section 5.2: Linear Systems in Three Variables with Applications
Big Idea: A system of three linear equations in three unknowns is the next level of difficulty in solving systems of linear equations. We will use elimination to get the system into triangular form, which is then very simple to solve.
Big Skill: You should be able to convert a system of equations to triangular form using the method of elimination, and then solve it using back substitution. This is called Gaussian elimination .
A. Visualizing Systems in Three Dimensions
Example of a system of three linear equations in three unknowns:


 2 x x x



2
2 y y y



3 z z z
 

12
 
2
9
A linear equation with three variables describes a two-dimensional plane embedded in three dimensions:
You can sketch a plane in three dimensions by plotting the intercepts and connecting them to represent a triangular portion of the plane.
Practice:
1.
Sketch the plane described by the equation 2 x y z 6 .

College Algebra Lecture Notes Section 5.2 Page 2 of 8
B. Solutions to a System of Three equations in Three Unknowns
When two planes intersect, the intersection is usually a line and when three planes intersect, the intersection is usually a point: y x
Practice: x
1.
Verify (-1, 2, -3) is the solution to the system 
 2 x x



2
2 y y y



3 z z z
 

2
12
 
9
.
The number of solutions to a system of linear equations in two variables falls under one of three cases:

Zero solutions : The planes never intersect (i.e., they are parallel to one another), or they intersect along different parallel lines and thus the system has no solutions. This type of system is called inconsistent .

One solution : The planes intersect at one point, and thus the system has exactly one solution. This type of system is called consistent and the equations are called independent . o Picture: See tope of page…

College Algebra Lecture Notes Section 5.2 Page 3 of 8

Infinitely many solutions : The planes are coincident (i.e., the equations are equivalent), or the planes intersect along a common line, and thus the system has infinitely many solutions. This type of system is called consistent and the equations are called dependent .
C. Solving Systems of Three Equations using Elimination
Example of a system of three linear equations in three unknowns that is in triangular form:



 x

2 y y


2 z z z



1
5
3
Notice: the name triangular form comes from the “blank” triangular space in the lower left corner due to no x or y variables. Also, this system is really easy to solve using backsubstitution: y 2 3 5 x y 6 5 y
 
1 x

2 y
 
2
 
3 1
1 x x

2
The goal of this section is to learn how to convert any given system of linear equations into triangular form. We will do this using the method of elimination, which involves adding constant multiples of equations together to eliminate variables.

College Algebra Lecture Notes Section 5.2 Page 4 of 8
Key fact behind the technique of elimination:

Multiplying an equation in a system by a constant, or adding two equations in a system together results in a new system of equations called a transformed system , and the solution to the transformed system is the same as the solution to the original system.
Steps for Solving a System of Three Linear Equations in Three Unknowns Using
Elimination

Eliminate the x variable from the second and third equations using elimination. If one of the equations has an x coefficient of 1, things are easier if you make it the first equation.

Eliminate the y variable from the third equation using elimination.


Solve for
Substitute z z
in the third equation.
into the second equation to find the solution for y , then substitute y and z into the first equation to find the solution for x .
Rules for showing your work:

Draw an arrow from one transformed system to the next, and write on the arrow what you did to transform the system.

Any equation that is unchanged gets copied from one system to the next.
Example: x


 2 x x

 y
2 y
2 y



3 z z z
 

12
 
2
9



2 x


2
4 y y y



2
4 z z z


4
14
 
5

(4)(eqn. #2) is placed in row#2









 x
 x

2 x



2 x

2
2 y y y




2
12
 
9
 x

 2 x


2 y y y



4 z z z


2
14
 
9

(2)(eqn. #1) is placed in row#1
2 y
3 z z z
2 z

2 x

2 x


2
2 y y y



4
2 z z z


4
14
 
9

4 y y



4 z z

4

14
 
5









2 x



2 x


2 x

2
4
4
2
4 y y y y y
2 y
4 y








2
16
2
16
17 z z z z z z
2 z
16 z z



Substitute into equation #2:
4 y

16
 
56
 y

2


4
56
 
5


4
56
 
3
Substitute into equation #1:
     
4
4
56
51 x 1
#3

College Algebra Lecture Notes Section 5.2
Practice:
2.
Solve 



3 x x x


 y
2 y
3 y


 z z
2 z


6
 
5 using elimination.
14
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College Algebra Lecture Notes Section 5.2 x
3.
Solve 
 3 x

3 y
2 y


3 z
4 z
 
3
 
5 using elimination.

6
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College Algebra Lecture Notes Section 5.2
2 x
4.
Solve 

4 x
2 x



2 y
3 y
3 y



3 z
2 z
7 z



6
0 using elimination.
1
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College Algebra Lecture Notes Section 5.2

5.
Solve 
 3 x x x

 y
2 y
2 y


 z
3 z
7 z

1
 
4 using elimination.

0
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