Topic 7.2
Solving Systems of
Linear Equations in
Three Variables
Using the Elimination
Method
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Topic 7.2,
1.1, Slide 1
OBJECTIVES
1. Verifying a Solution of a System of Linear Equations
in Three Variables
2. Solving a System of Linear Equations Using the
Elimination Method
3. Solving Consistent, Dependent Systems of Linear
Equations in Three Variables
4. Solving Inconsistent Systems of Linear Equations in
Three Variables
5. Solving Applied Problems Using a System of Linear
Equations Involving Three Variables
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Topic 7.2,
1.1, Slide 2
Verifying a Solution of a System of
Linear Equations in Three Variables
Definition: Linear Equation in Three Variables
A linear equation in three variables is an equation that
can be written in the form Ax + By + Cz = D, where A, B, C,
and D are real numbers, and A, B, and C are not all equal
to 0.
Definition: System of Linear Equations in Three Variables
A system of linear equations in three variables is a
collection of linear equations in three variables considered
simultaneously. A solution to a system of linear equations
in three variables is an ordered triple that satisfies all
equations in the system.
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Topic 7.2,
1.1, Slide 3
Verifying a Solution of a System of
Linear Equations in Three Variables
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Topic 7.2,
1.1, Slide 4
Verifying the Solution of a System
of Linear Equations in Three
Variables
EXAMPLE
2x + 3 y + 4x =
12
x − 2 y + 3z =
0
− x + y − 2 z =−1
Verify that the ordered triple (1, 2, 1) is
a solution to the following system of
linear equations:
?
?
2 (1) + 3 ( 2 ) + 4 (1) = 12 → 2 + 6 + 4 = 12 → 12 = 12
(1) − 2 ( 2 ) + 3 (1=)
?
?
0
?
→ 1− 4 + 3= 0
→ 0= 0
?
− (1) + ( 2 ) − 2 (1) =− 1 → −1+ 2 − 2 =− 2 → −1 =−1
All three statements are true. This implies that the ordered triple
(1, 2, 1) is a solution to this system.
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Topic 7.2,
1.1, Slide 5
Verifying the Solution of a System
of Linear Equations in Three
Variables
Systems of equations that have at least one solution are called
consistent systems. A system that has exactly one solution is called
a consistent, independent system.
It is possible that a system of three linear equations in three variables
has infinitely many solutions. These systems are called consistent,
dependent systems.
It is also possible for a system of three linear equations in three
variables to have no solution at all. These systems are called
inconsistent systems.
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Topic 7.2,
1.1, Slide 6
Verifying the Solution of a System
of Linear Equations in Three
Variables
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Topic 7.2,
1.1, Slide 7
Guidelines for Solving a System of Linear
Equations in Three Variables by Elimination
Step 1. Write each equation In standard form. Write each equation in the
form Ax + By + Cz = D lining up the variable terms. Number the
equations to keep track of them.
Step 2. Eliminate a variable from one pair of equations. Use the
elimination method to eliminate a variable from any two of the
original three equations, leaving one equation in two variables,
Step 3. Eliminate the same variable again. Use a different pair of the
original equations and eliminate the same variable again, leaving one
equation in two variables.
Step 4. Solve the system of linear equations in two variables. Use the
resulting equations from steps 2 and 3 to create and solve the
corresponding system of linear equations in two variables by
substitution or elimination.
Step 5. Use back substitution to find the value of the third variable.
Substitute the results from step 4 into any of the original equations to
find the value of the remaining variable.
Step 6. Check the solution. Check the proposed solution in each equation of
the system and write the solution set.
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Topic 7.2,
1.1, Slide 8
Solving a System of Linear Equations
Using the Elimination Method
2x + 3 y + 4z =
12
0
Solving the following system: x − 2 y + 3 z =
− x + y − 2 z =−1
EXAMPLE
Step 1. The equations are already in
standard form and all the variables are
lined up. We rewrite the system and
number each equation.
Step 2. We can eliminate any of the
variables. For convenience, we will
eliminate the variable
x from equations (1) and (2). We can do
this by multiplying equation (2) by -2 and
adding the equations together.
2x + 3 y + 4z =
12 (1)
x − 2 y + 3z =
0 (2)
− x + y − 2 z =−1 (3)
2x + 3 y + 4z =
12 (1)
−2 x + 4 y − 6 z =
0 (2) x -2
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7 y − 2z =
12 (4)
Topic 7.2,
1.1, Slide 9
Solving a System of Linear Equations
Using the Elimination Method
2x + 3 y + 4z =
12
0
Solving the following system: x − 2 y + 3 z =
− x + y − 2 z =−1
EXAMPLE continued
Step 3. We need to eliminate the same
variable, x, from a different pair of
equations. We will use equations (2) and
(3) for the second pair.
0 (2)
x − 2 y + 3z =
− x + y − 2 z =−1 (3)
Step 4. Combining equations (4) and (5), we
form a system of linear equations in two
variables.
7 y − 2z =
12 (4)
− y + z =−1 (5)
To solve this system, we use the
elimination method by multiplying
equation (5) by 7 and adding the result to
equation (4) to eliminate the variable y.
− y + z =−1
(5)
7 y − 2z =
12 (4)
−7 y + 7 z =
−7 (5)x7
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5z = 5
z =1
Topic 7.2,
1.1, Slide 10
Solving a System of Linear Equations
Using the Elimination Method
2x + 3 y + 4z =
12
0
Solving the following system: x − 2 y + 3 z =
− x + y − 2 z =−1
EXAMPLE continued
Since z = 1, we back-substitute this
into (5) to solve for y.
− y + z =−1
− y + (1) =−1
− y =−2
y=2
Step 5. Substitute 2 for y and 1 for z in any
of the original equations, (1), (2), or (3),
and solve for x. We will back-substitute
using equation (2).
The solution to the system is the ordered
triple (1, 2, 1).
x − 2 y + 3z =
0
x − 2(2) + 3(1) =
0
x−4+3 =
0
x −1=
0
x =1
Step 6. Be sure to check your answer.
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Topic 7.2,
1.1, Slide 11
Solving Consistent, Dependent
Systems Linear Equations
If a system of equations has at least one solution, then the
system is consistent. If the system has infinitely many solutions, then
the system is consistent and dependent. For the two-variable case, a
consistent, dependent system can be geometrically
represented by a pair of coinciding lines. For the three-variable case,
consistent, dependent systems can be geometrically represented by
three planes that intersect at a line or by three equations that all
describe the same plane.
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Topic 7.2,
1.1, Slide 12
Solving Consistent, Dependent
Systems Linear Equations
EXAMPLE
Solve the following system.
10 (1)
x + 2 y + 3z =
7 (2)
x+ y+z =
3x + 2 y + z =
18 (3)
Eliminating the variable x using equations (1) and (2)
results in the equation -y - 2z = -3. Eliminating the
variable x again using equations (1) and (3) results in
the equation -4y - 8z = -12. Combining these two
equations gives the two-variable system:
We can multiply equation (4) by -4 to get
the new equation (6) and add (6) and (5):
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− y − 2 z =−3 (4)
−4 y − 8 z =
−12 (5)
4 y + 8z =
12 (6)
−4 y − 8 z =
−12 (5)
0=0
Topic 7.2,
1.1, Slide 13
Solving Consistent, Dependent
Systems Linear Equations
EXAMPLE continued
Solve the following system.
10 (1)
x + 2 y + 3z =
7 (2)
x+ y+z =
3x + 2 y + z =
18 (3)
We see that adding equations (6) and (5) results in
the identity 0 = 0. Therefore, this two-variable
system has infinitely many solutions and the system
is said to be dependent. Solve either equation
(6) or (5) for either variable. Solving for the variable y
we get y = 3 - 2z.
Since y = 3 - 2z, we back-substitute this into
(1) to solve for x.
4 y + 8z =
12 (6)
−4 y − 8 z =
−12 (5)
0=0
x + 2(3 − 2 y ) + 3 z =
10
x + 6 − 4 y + 3z =
10
x= 4 + z
The solution to this dependent system can be written in ordered
triple notation as (4 + z, 3 - 2z, z) where z is free to be any real
number.
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Topic 7.2,
1.1, Slide 14
Solving Inconsistent Systems of
Linear Equations in Three Variables
A geometric representation of inconsistent systems in three variables.
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Topic 7.2,
1.1, Slide 15
Solving Inconsistent Systems Linear
Equations in Three Variables
EXAMPLE
Solve the following system.
x − y + 2z =
5
3x − 3 y + 6z =
15
−2 x + 2 y − 4 z =
7
Step 1. The equations are already
in standard form with all the variables
lined up. We rewrite the system and
number each equation.
x − y + 2z =
5 (1)
3x − 3 y + 6z =
15 (2)
−2 x + 2 y − 4 z =
7 (3)
Step 2. For convenience, we will eliminate
the variable x from equations (1) and (2).
To do this, we multiply equation (1) by -3
and add the equations together.
−3 x + 3 y − 6 z =
−18 (1) x -3
3x − 3 y + 6z =
15 (2)
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0=0
Topic 7.2,
1.1, Slide 16
Solving Inconsistent Systems Linear
Equations in Three Variables
EXAMPLE continued
Solve the following system.
x − y + 2z =
5
3x − 3 y + 6z =
15
−2 x + 2 y − 4 z =
7
The last line, 0 = 0, is an identity. In the two-variable case, we would stop and say
that the system had an infinite number of solutions. However, in systems of three
variables, this is not necessarily the case.
Step 3. Let’s continue our process and
eliminate the variable x from the pairing of
equations (1) and (3). To do this, we
multiply equation (1) by 2 and add the
equations.
2x − 2 y + 4z =
10 (1) x 2
−2 x + 2 y − 4 z =
7 (3)
0 = 17
The last line, 0 = 17, is a contradiction, so
the system is inconsistent and has no
solution.
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Topic 7.2,
1.1, Slide 17
Solving Applied Problems Using a System
Linear Equations in Three Variables
EXAMPLE
While playing a real-time strategy game, Joel created military units to defend
his town: warriors, skirmishers, and archers. Warriors require 20 units of food
and 50 units of gold. Skirmishers require 25 units of food and 35 units of wood.
Archers require 32 units of wood and 32 units of gold. If Joel used 506 units of
gold, 606 units of wood, and 350 units of food to create the units, how many of
each type of military unit did he create?
We want to find the number of each type of military unit created. There are three
types of units: warriors, skirmishers, and archers. Each unit requires a certain
amount of gold, wood, and food. We know how much of each resource is needed
for each unit, and we know the total amount of each resource that is used.
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Topic 7.2,
1.1, Slide 18
Solving Applied Problems Using a System
Linear Equations in Three Variables
EXAMPLE continued
While playing a real-time strategy game, Joel created military units to defend
his town: warriors, skirmishers, and archers. Warriors require 20 units of food
and 50 units of gold. Skirmishers require 25 units of food and 35 units of wood.
Archers require 32 units of wood and 32 units of gold. If Joel used 506 units of
gold, 606 units of wood, and 350 units of food to create the units, how many of
each type of military unit did he create?
Let W, S, and A represent the number of warriors, skirmishers, and archers
respectively. We can obtain the three equations based on the total units of gold,
wood, and food use
GOLD:
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Topic 7.2,
1.1, Slide 19
Solving Applied Problems Using a System
Linear Equations in Three Variables
EXAMPLE continued
While playing a real-time strategy game, Joel created military units to defend
his town: warriors, skirmishers, and archers. Warriors require 20 units of food
and 50 units of gold. Skirmishers require 25 units of food and 35 units of wood.
Archers require 32 units of wood and 32 units of gold. If Joel used 506 units of
gold, 606 units of wood, and 350 units of food to create the units, how many of
each type of military unit did he create?
Let W, S, and A represent the number of warriors, skirmishers, and archers
respectively. We can obtain the three equations based on the total units of gold,
wood, and food use
WOOD:
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Topic 7.2,
1.1, Slide 20
Solving Applied Problems Using a System
Linear Equations in Three Variables
EXAMPLE continued
While playing a real-time strategy game, Joel created military units to defend
his town: warriors, skirmishers, and archers. Warriors require 20 units of food
and 50 units of gold. Skirmishers require 25 units of food and 35 units of wood.
Archers require 32 units of wood and 32 units of gold. If Joel used 506 units of
gold, 606 units of wood, and 350 units of food to create the units, how many of
each type of military unit did he create?
Let W, S, and A represent the number of warriors, skirmishers, and archers
respectively. We can obtain the three equations based on the total units of gold,
wood, and food use
FOOD:
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Topic 7.2,
1.1, Slide 21
Solving Applied Problems Using a System
Linear Equations in Three Variables
EXAMPLE continued
While playing a real-time strategy game, Joel created military units to defend
his town: warriors, skirmishers, and archers. Warriors require 20 units of food
and 50 units of gold. Skirmishers require 25 units of food and 35 units of wood.
Archers require 32 units of wood and 32 units of gold. If Joel used 506 units of
gold, 606 units of wood, and 350 units of food to create the units, how many of
each type of military unit did he create?
This gives us the following system:
50W + 0 S + 32 A =
506 (1)
0W − 35 S − 32 A =
−606 (2) x -1
50W − 35 S
=
−100 (4)
50W + 0 S + 32 A =
506 (1)
0W + 35 S + 32 A =
606 (2)
20W + 25 S + 0 A =
350 (3)
100W + 125 S + 0 A =
1750 (3) x 5
200 (4) x − 2
−100W + 70 S + 0 A =
195S
= 1950
S = 10 skirmishers
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Topic 7.2,
1.1, Slide 22
Solving Applied Problems Using a System
Linear Equations in Three Variables
EXAMPLE continued
While playing a real-time strategy game, Joel created military units to defend
his town: warriors, skirmishers, and archers. Warriors require 20 units of food
and 50 units of gold. Skirmishers require 25 units of food and 35 units of wood.
Archers require 32 units of wood and 32 units of gold. If Joel used 506 units of
gold, 606 units of wood, and 350 units of food to create the units, how many of
each type of military unit did he create?
50W − 35(10) =
−100 (4)
50W
= 250
W = 5 warriors
50W + 0 S + 32 A =
506 (1)
50(5) + 32 A =
506
250 + 32 A =
506
32 A = 256
A = 8 archers
5 warriors, 10 skirmishers and 8 archers
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Topic 7.2,
1.1, Slide 23