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4
Solving Systems of Linear Equations in
Three Variables
1. Determine if an ordered triple is a solution for a system
of equations.
2. Understand the types of solution sets for systems of
three equations.
3. Solve a system of three linear equations using the
elimination method.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Example
Determine whether (2, –1, 3) is a solution of the
x  y  z  4,
system
2 x  2 y  z  3,
4 x  y  2 z  3.
Solution In all three equations, replace x with 2, y
with –1, and z with 3.
2x – 2y – z = 3
x+y+z=4
2 + (–1) + 3 = 4
4=4
– 4x + y + 2z = –3
2(2) – 2(–1) – 3 = 3 – 4(2) + (–1) + 2(3) = –3
3=3
–3 = –3
Because (2, 1, 3) satisfies all three equations in
TRUE
TRUE
TRUE
the system, it is a solution for the system.
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Slide 9- 3
Example
Solve the system using elimination.
x  y  z  6,
x  2 y  z  2,
x  y  3z  8.
(1)
(2)
(3)
Solution
We select any two of the three equations and
work to get one equation in two variables. Let’s
add equations (1) and (2):
x yz6
(1)
(2)
x  2y  z  2
Adding to
(4)
2x + 3y
=8
eliminate z
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Slide 9- 4
Next, we select a different pair of equations and
eliminate the same variable. Let’s use (2) and (3) to
again eliminate z.
x  2y  z  2
x  y  3z  8
Multiplying
equation (2)
by 3
3x + 6y – 3z = 6
x – y + 3z = 8
4x + 5y
= 14.
(5)
Now we solve the resulting system of equations
(4) and (5). That will give us two of the numbers
in the solution of the original system,
(4)
2x + 3y = 8
(5)
4x + 5y = 14
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Slide 9- 5
We multiply both sides of equation (4) by –2 and
then add to equation (5):
–4x – 6y = –16,
4x + 5y = 14
–y = –2
y=2
Substituting into either equation (4) or (5) we find
that x = 1.
Now we have x = 1 and y = 2. To find the value
for z, we use any of the three original equations
and substitute to find the third number z.
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Slide 9- 6
Let’s use equation (1) and substitute our two
numbers in it:
x+y+z=6
1+2+z=6
z = 3.
We have obtained the ordered triple (1, 2, 3). It
should check in all three equations.
The solution is (1, 2, 3).
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Slide 9- 7
Solving Systems of Three Equations Using Elimination
1. Write each equation in the form Ax + By+ Cz = D.
2. Eliminate one variable from one pair of equations using the
elimination method.
3. If necessary, eliminate the same variable from another pair
of equations.
4. Steps 2 and 3 result in two equations with the same two
variables. Solve these equations using the elimination
method.
5. To find the third variable, substitute the values of the
variables found in step 4 into any of the three original
equations that contain the third variable.
6. Check the ordered triple in all three of the original
equations.
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Slide 9- 8
Example
Solve the system using elimination.
3x  9 y  6 z  3 (1)
2 x  y  z  2 (2)
x  y  z  2 (3)
Solution
The equations are in standard form.
Eliminate z from equations (2) and (3).
2 x  y  z  2 (2)
x  y  z  2 (3)
(4)
3x + 2y = 4
Adding
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Slide 9- 9
3x  9 y  6 z  3 (1)
2 x  y  z  2 (2)
Eliminate z from equations (1) and (2). x  y  z  2 (3)
continued
3x  9 y  6 z  3
2x  y  z  2
Multiplying
equation (2) by 6
3x  9 y  6 z  3
12 x  6 y  6 z  12
15x + 15y = 15
Adding
Eliminate x from equations (4) and (5).
3x + 2y = 4
15x + 15y = 15
Multiplying top
by 5
15x – 10y = 20
15x + 15y = 15
5y = 5
y = 1
Adding
Using y = 1, find x from equation 4 by substituting.
3x + 2y = 4
3x + 2(1) = 4
x=2
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Slide 9- 10
3x  9 y  6 z  3 (1)
2 x  y  z  2 (2)
x  y  z  2 (3)
continued
Substitute x = 2 and y = 1 to find z.
x+y+z=2
2–1+z=2
1+z=2
z=1
The solution is the ordered triple (2, 1, 1).
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Slide 9- 11
Example
At a movie theatre, Kara buys one popcorn, two drinks
and 2 candy bars, all for $12. Rebecca buys two
popcorns, three drinks, and one candy bar for $17.
Leah buys one popcorn, one drink and three candy
bars for $11. Find the individual cost of one popcorn,
one drink and one candy bar.
Understand We have three unknowns and three
relationships, and we are to find the cost of each.
Plan Select a variable for each unknown, translate the
relationship to a system of three equations, and then
solve the system.
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Slide 9- 12
continued
Execute: p = popcorn, d = drink, and c = candy
Relationship 1: one popcorn, two drinks and two candy
bars, cost $12
Translation: p + 2d + 2c = 12
Relationship 2: two popcorns, three drinks, and one
candy bar cost $17
Translation: 2p + 3d + c = 17
Relationship 3: one popcorn, one drink and three
candy bars cost $11
Translation: p + d + 3c = 11
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Slide 9- 13
continued
p  2d  2c  12
Our system: 2 p  3d  c  17
(Equation 1)
(Equation 2)
p  d  3c  11
(Equation 3)
Choose to eliminate p: Start with equations 1 and 3.
p + 2d + 2c = 12
p + d + 3c = 11 Multiply by
p + 2d + 2c = 12
-1
- p - d - 3c = -11
d c 1
(Equation 4)
Choose to eliminate p: Start with equations 1 and 2.
p + 2d + 2c = 12 Multiply by
2 p + 3d + c = 17
-2
-2 p - 4d - 4d = -24
2 p + 3d + c = 17
d  3c  7
(Equation 5)
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Slide 9- 14
continued
Use equations 4 and 5 to eliminate d.
d c 1
d  3c  7
4c  6
(Equation 4)
(Equation 5)
c  1.5
Substitute for c in d – c = 1
d c 1
d  1.5  1
d  2.5
Substitute
The for
costc of
and
one
d in
candy
p + dbar
+ is
3c$1.50.
= 11
The cost of one
$2.50.= 11
p +drink
2.5 +is3(1.5)
The cost of one popcorn is $4.00.
p + 7 = 11
p =4
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Slide 9- 15
Determine if (2, –5, 3) is a solution to
the given system.  x  5 y  2 z  7
a) Yes




3 y  5 z  16
z  5
b) No
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Slide 9- 16
Determine if (2, –5, 3) is a solution to
the given system.  x  5 y  2 z  7
a) Yes




3 y  5 z  16
z  5
b) No
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Slide 9- 17
Solve the system.
2 x  4 y  z  1

5 x  2 y  z  9
3x  y  2 z  14

a) (–2, 2, –5)
b) (–5, 2, –2)
c) (–2, 5, 2)
d) no solution
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Slide 9- 18
Solve the system.
2 x  4 y  z  1

5 x  2 y  z  9
3x  y  2 z  14

a) (–2, 2, –5)
b) (–5, 2, –2)
c) (–2, 5, 2)
d) no solution
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Slide 9- 19
A company makes 3 types of cable. Cable A requires 3
black, 3 white, and 2 red wires. B requires 1 black, 2
white, and 1 red. C requires 2 black, 1 white, and 2 red.
They used 100 black, 110 white and 90 red wires. How
many of each cable were made?
a) 10 cable A, 30 cable B, 20 cable C
b) 20 cable A, 30 cable B, 10 cable C
c) 10 cable A, 103 cable B, 20 cable C
d) 10 cable A, 30 cable B, 93 cable C
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Slide 9- 20
A company makes 3 types of cable. Cable A requires 3
black, 3 white, and 2 red wires. B requires 1 black, 2
white, and 1 red. C requires 2 black, 1 white, and 2 red.
They used 100 black, 110 white and 90 red wires. How
many of each cable were made?
a) 10 cable A, 30 cable B, 20 cable C
b) 20 cable A, 30 cable B, 10 cable C
c) 10 cable A, 103 cable B, 20 cable C
d) 10 cable A, 30 cable B, 93 cable C
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Slide 9- 21
9.5
Solving Systems of Linear Equations
Using Matrices
1. Write a system of equations as an augmented matrix.
2. Solve a system of linear equations by transforming its
augmented matrix to echelon form.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Matrix: A rectangular arrow of numbers.
The following are examples of matrices:
1  1 6 9 1/ 2 0 
1 2    

7
6
7 5  ,  4  ,  4 3 1


   
 0   6 5 9 1 2 
The individual numbers are called elements.
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Slide 9- 23
The rows of a matrix are horizontal, and the
columns are vertical.
1
4

 6
9
1
9
0

6

2 
column 1
column 2
column 3
row 1
row 2
row 3
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Slide 9- 24
Augmented Matrix: A matrix made up of the
coefficients and the constant terms of a system. The
constant terms are separated from the coefficients by
a dashed vertical line.
Let’s write this system as an augmented matrix:
3x  y  7

 x  3 y  1
3 1 7 
 1 3 1 


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Slide 9- 25
Row Operations
The solution of a system is not affected by the
following row operations in its augmented matrix:
1. Any two rows may be interchanged.
2. The elements of any row may be multiplied (or
divided) by any nonzero real number.
3. Any row may be replaced by a row resulting from
adding the elements of that row (or multiples of
that row) to a multiple of the elements of any other
row.
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Slide 9- 26
Echelon form: An augmented matrix whose
coefficient portion has 1s on the diagonal from
upper left to lower right and 0s below the 1s.
1 1 1 1 
0 1
1 2 


0 0 1
1
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Slide 9- 27
Example
Solve the following linear system by transforming
its augmented matrix into echelon form. 3x  y  7

 x  3 y  1
Solution
We write the augmented matrix. 3
1 7
 1 3 1 


We perform row operations to transform the
matrix into echelon form.
3R2 + R1
3 1 7 
0 10 10


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Slide 9- 28
3 1 7 
0 10 10


continued
3 1 7 
0 1 1 


3x  y  7

The resulting matrix represents the system: 
R2  10

y 1
Since y = 1, we can solve for x using substitution.
3x + y = 7
3x + 1 = 7
3x = 6
x=2
The solution is (2, 1).
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Slide 9- 29
Example
2 x  y  z  8
Use the echelon method to solve 
x  y  z  1
 x  2 y  z  2

Solution
Write the augmented matrix.
1 1 8 
2
1

1 1 1


 1 2 1 2 
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Slide 9- 30
continued
Our goal is to transform
1 8 
2 1
1
1 1 1  into the form


 1 2 1
2 
a b c
0 e f

0 0 h
d
g .

i 
1 1 1 
1
Interchange Row 1 and Row 2
2 1

1 8


 1 2 1
2 
1 1 1 
1
0 3 3 6 


 1 2 1
2 
–2R1 + R2
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Slide 9- 31
continued
1 1 1 1 
0 1
1 2 


0 1
2 1
(1/3)R2
R1 + R3
1 1 1 1 
0 1
1 2 


0 0 3 3
–R2 + R3
1 1 1 1 
0 1
1 2 


0 0 1
1
(1/3)R3
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Slide 9- 32
continued
z = –1.
Substitute z into y – z = 2
y –(1) = 2
y+1=2
y=1
Substitute y and z into x – y + z = 1
x–1–1=1
x–2=1
x=3
The solution is (3, 1, –1).
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Slide 9- 33
Replace R2 in
 8 6 4 
 2 4 0 


a)
 10 10 4 
 10 10 4 


c)
 8 14 2
 2 6 0 


b)
d)
with R1 + R2.
 8 6 4 
 10 10 4 


 10 10 4 
 2 4 0 


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Slide 9- 34
Replace R2 in
 8 6 4 
 2 4 0 


a)
 10 10 4 
 10 10 4 


c)
 8 14 2
 2 6 0 


b)
d)
with R1 + R2.
 8 6 4 
 10 10 4 


 10 10 4 
 2 4 0 


Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 9- 35
Solve by transforming the
augmented matrix into
echelon form.
x  y  z  0

 x  y  4 z  11
4 x  y  z  6

a) (2, 1, 3)
b) (3, 1, 2)
c) (3, 2, 1)
d) No solution
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Slide 9- 36
Solve by transforming the
augmented matrix into
echelon form.
x  y  z  0

 x  y  4 z  11
4 x  y  z  6

a) (2, 1, 3)
b) (3, 1, 2)
c) (3, 2, 1)
d) No solution
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Slide 9- 37