Systems of Linear Equations
(3  3)
A system of linear equations can consist of two equations with two variables, three equations
with three variables, four equations with four variables etc. To solve a system of three equations
with three unknowns, means to find one value for each variable that will satisfy each of the
equations. This process is usually done algebraically by using the process of elimination. The
steps involved in solving a 3  3 system of linear equations algebraically are:
 Set up two pair of two equations and eliminate the same variable from each pair.
 There is now a 2  2 system of linear equations to solve. The equations can be solved
using any of the known methods used to solve a 2  2 system.
 When the solution for the 2  2 system of linear equations has been determined, these
numbers must be substituted into one of the original equations of the given system of
linear equations. This new equation must be solved in order to find the value of the final
variable.
Example1: Solve the following system of linear equations algebraically:
2 x  y  z  1
3x  2 y  3z  4
x  3 y  z  4
2 x  y  z  1

3x  2 y  3z  4
4 x  2 y  2 z  2
2(2 x  y  z  1)

3x  2 y  3z  4
3x  2 y  3z  4
7 x  5z  2
2 x  y  z  1
 6 x  3 y  3z  3
 3(2 x  y  z  1)


x  3 y  z  4
x  3 y  z  4
x  3 y  z  4
 5x  4z  1
These are the two pair of two equations. In each pair, the variable y has been eliminated. There
is now a 2  2 system of linear equations which must be solved.
7 x  5z  2
 5x  4z  1
This 2  2 system is being solved by elimination.
4(7 x  5 z  2)
28x  20 z  8


5(5 x  4 z  1)
 25x  20 z  5
3x  3
3
3
x
3
3
x 1
7 x  5z  2
7(1)  5 z  2
7  5z  2
7  7  5z  2  7
5 z  5
5
5
z
5
5
z  1
2 x  y  z  1
2(1)  y  (1)  1
2  y  1  1
1  y  1
1  1  y  1  1
y  2
The solution is (x, y, z) = (1,-2,-1).
The answers for x and z were substituted into one of the equations of the system of linear
equations. It makes no difference which equation you substitute the values into since the
solution will satisfy all of the equations.
A 3  3 system of linear equations can also be solved by using the inverse matrix method.
Because this process is very long and cumbersome to do algebraically, it is normally done using
a graphing calculator. Solving a 3  3 system using technology will be presented in the
Appendix.
Example 2: Solve the following system of linear equations algebraically:
3a  b  c  10
2a  3b  2c  1
4a  2b  3c  15
3a  b  c  10
2(3a  b  c  10) 6a  2b  2c  20


2a  3b  2c  1 2a  3b  2c  1
2a  3b  2c  1
8a  5b  21
3a  b  c  10
 3(3a  b  c  10)
 9a  3b  3c  30


4a  2b  3c  15
4a  2b  3c  15
4a  2b  3c  15
 5a  5b  15
8a  5b  21
8a  5b  21
 5a  5b  15
3a  6
3
6
a
3
3
a  2
8(2)  5b  21
 16  5b  21
 16  16  5b  21  16
5b  5
5
5
b
5
5
b  1
3a  b  c  10
3(2)  (1)  c  10
 6  1  c  10
 7  c  10
 7  7  c  10  7
c  3
The solution is (a, b, c) = (-2, -1, -3).
Example 3: Solve the following system of linear equations algebraically:
x  2 y  3 z  14
2x  3y  z  2
3 x  2 y  4 z  11
x  2 y  3 z  14
 2 x  4 y  6 z  28
 2( x  2 y  3 z  14)


2x  3y  z  2
2x  3y  z  2
2x  3y  z  2
 y  7 z  26
x  2 y  3 z  14
 3x  6 y  9 z  42
 3( x  2 y  3z  14)


3 x  2 y  4 z  11 3x  2 y  4 z  11
3x  2 y  4 z  11
 8 y  5 z  31
 y  7 z  26
8 y  56 z  208
 8( y  7 z  26)


 8 y  5 z  31  8 y  5 z  31
 8 y  5 z  31
Expressed as a decimal z  3.47
51z  177
51
177
z
51
51
24
8
z  3 or 3
51
17
 y  7 z  26
 y  7(3.47)  26
 y  24.29  26
 y  24.29  24.29  26  24.29
 y  1.71
1
 1.71
y
1
1
y  1.71
x  2 y  3 z  14
x  2(1.71)  3(3.47)  14
x  3.42  10.41  14
x  13.83  14
x  13.83  13.83  14  13.83
x  0.17
The solution is (x, y, z)  (0.17, 1.71, 3.47).
The above example has a solution in which the values are in decimal form. These answers are
approximate values and this must be considered when the values are replaced for the variables in
the system of equations. The results will be approximate and not exact.
Example 4: Solve the following system of linear equations algebraically:
2r  3s  t  2
3r  4s  2t  5
2r  2s  t  12
2r  3s  t  2
2r  3s  t  2

2r  2s  t  12
2r  2s  t  12
4r  s  10
2r  3s  t  2
4r  6s  2t  4
2(2r  3s  t  2)


3r  4s  2t  5
3r  4s  2t  5
3r  4s  2t  5
7r  10s  1
4r  s  10
 40r  10s  100
 10(4r  s  10)


7r  10s  1
7r  10s  1
7r  10s  1
 33r  99
 33
 99
r
 33
 33
r 3
4r  s  10
4(3)  s  10
12  s  10
12  12  s  10  12
s  2
2r  2s  t  12
2(3)  2(2)  t  12
6  4  t  12
10  t  12
10  10  t  12  10
t2
The solution is (r, s, t) = (3,-2, 2).
Three by three systems of linear equations are also used to solve real-life problems. The given
problem is expressed as a system of linear equations and then solved to determine the value of
the variables. Sometimes, the system will consist of three equations but not every equation will
have three variables. Example three is one such problem.
Example 1: Solve the following problem using your knowledge of systems of linear
equations.
Jesse, Maria and Charles went to the local craft store to purchase supplies for making
decorations for the upcoming dance at the high school. Jesse purchased three sheets of craft
paper, four boxes of markers and five glue sticks. His bill, before taxes was $24.40. Maria
spent $30.40 when she bought six sheets of craft paper, five boxes of markers and two glue
sticks. Charles, purchases totaled $13.40 when he bought three sheets of craft paper, two
boxes of markers and one glue stick. Determine the unit cost of each item.
Let p represent the number of sheets of craft paper.
Let m represent the number of boxes of markers.
Let g represent the number of glue sticks.
Express the problem as a system of linear equations:
3 p  4m  5 g  $24.40
6 p  5m  2 g  $30.40
3 p  2m  g  $13.40
Solve the system of linear equations to determine the unit cost of each item.
3 p  4m  5 g  24.40
3 p  4m  5 g  24.40
3 p  4m  5 g  24.40


3 p  2m  g  13.40
 5(3 p  2m  g  13.40)
 15 p  10m  5 g  67.00
 12 p  6m  42.60
6 p  5m  2 g  30.40
6 p  5m  2 g  30.40
6 p  5m  2 g  30.40


3 p  2m  g  13.40
 2(3 p  2m  g  13.40)
 6 p  4m  2 g  26.80
m  3.60
 12 p  6m  42.60
 12 p  6(3.60)  42.60
 12 p  21.60  42.60
 12 p  21.60  21.60  42.60  21.60
 12 p  21
 12
 21
p
 12
 12
p  1.75
3 p  2m  g  13.40
3(1.75)  2(3.60)  g  13.40
5.25  7.20  g  13.40
12.45  g  13.40
12.45  12.45  g  13.40  12.45
g  .95
The unit cost of each item is: 1 sheet of craft paper = $1.75
1 box of markers = $3.60
1 glue stick = $0.95
Example 2: Solve the following problem using your knowledge of systems of linear
equations.
Rafael, an exchange student from Brazil, made phone calls within Canada, to the United
States, and to Brazil. The rates per minute for these calls vary for the different countries.
Use the information in the following table to determine the rates.
Month
September
October
November
Time within
Canada (min)
90
70
50
Time to the
U.S. (min)
120
100
110
Time to Brazil
(min)
180
120
150
Charges
($)
$252.00
$184.00
$206.00
Let c represent the rate for calls within Canada.
Let u represent the rate for calls to the United States.
Let b represent the rate for calls to Brazil.
Express the problem as a system of linear equations:
90c  120u  180b  $252.00
70c  100u  120b  $184.00
50c  110u  150b  $206.00
2(90c  120u  180b  252.00)
90c  120u  180b  252.00

 3(70c  100u  120b  184.00)
70c  100u  120b  184.00

180c  240u  360b  504.00
 210c  300u  360b  552.00
 30c  60u  48.00
 5(70c  100u  120b  184.00)
70c  100u  120b  184.00

4(50c  110u  150b  206.00)
50c  110u  150b  206.00

 350c  500u  600b  920.00
200c  440u  600b  824.00
 150c  60u  96.00
 30c  60u  48.00
30c  60u  48.00
 1(30c  60u  48.00)


 150c  60u  96.00
 150c  60u  96.00
 150c  60u  96.00
 120c  48.00
 120
 48.00
c
 120
 120
c  .40
 30c  60u  48.00
 30(.40)  60u  48.00
 12.00  60u  48.00
 12.00  12.00  60u  48.00  12.00
 60u  36.00
 60
 36.00
u
 60
 60
u  .60
70c  100u  120b  184.00
70(.40)  100(.60)  120b  184.00
28.00  60.00  120b  184.00
88.00  120b  184.00
88.00  88.00  120b  184.00  88.00
120b  96.00
120
96.00
b
120
120
b  .80
The cost of minutes within Canada is $0.40/min. The cost of minutes to the United States
is $0.60/min. The cost of minutes to Brazil is $0.80/min.
Example 3: Solve the following problem using your knowledge of systems of linear
equations.
Calculate the number of minutes that Carlos called within Canada, to the United States, and
to Mexico during the month of December. The charges are 28¢/min within Canada, 30¢/min
to the U.S., and 84¢/min to Mexico if the following conditions applied:
 His total bill for the month was $90.96
 He talked twice as long to Mexico as he did to the U.S.
 The total number of minutes spent talking within Canada and to Mexico was 122.
Let c represent the number of minutes within Canada
Let u represent the number of minutes to the United States
Let m represent the number of minutes to Mexico
Express the problem as a system of linear equations:
.28c  .30u  .84m  $90.96
2u  m
c  m  122
Solve c  m  122 for c  c  122  m
Replace m with 2u in c  122  m  c  122  2u
Now replace c and m in the first equation to get an equation in terms of the variable u.
34.16  34.16  1.42u  90.96  34.16
.28(122  2u )  .30u  .84(2u )  90.96
1.42u  56.80
34.16  .56u  .30u  1.68u  90.96
 1.42
56.80
34.16  1.42u  90.96
u
1.42
1.42
u  40
c  m  122
2u  m
c  80  122
2(40)  m
80  m
c  80  80  122  80
c  42
Carlos talked for 42 minutes within Canada, to the U.S. for 40 minutes and to Mexico for
80 minutes.
Example 4: Solve the following problem using your knowledge of systems of linear
equations.
Tracy, Danielle and Sherri bought snacks for a girls’ sleepover. They each bought the items
shown in the following table at the local convenience store:
Number of bags of
potato chips
4
3
2
Number of litres of
pop
4
2
3
Number of
chocolate bars
6
10
4
Cost
($)
21.00
20.88
13.17
Calculate the unit price of each snack purchased by the girls.
Let c represent the unit cost of the potato chips.
Let p represent the unit cost of the pop.
Let b represent the unit cost a chocolate bar.
Express the problem as a system of linear equations:
4c  4 p  6b  21.00
3c  2 p  10b  20.88
2c  3 p  4b  13.17
4c  4 p  6b  21.00

3c  2 p  10b  20.88
4c  4 p  6b  21.00
4c  4 p  6b  21.00

 2(3c  2 p  10b  20.88)
 6c  4 p  20b  41.76
 2c  14b  20.76
3c  2 p  10b  20.88
 3(3c  2 p  10b  20.88)
 9c  6 p  30b  62.64


2c  3 p  4b  13.17
2(2c  3 p  4b  13.17)
4c  6 p  8b  26.34
 5c  22b  36.30
 2c  14b  20.76

 5c  22b  36.30
 5(2c  14b  20.76)

2(5c  22b  36.30)
 2c  14b  20.76
 2c  14(1.20)  20.76
 2c  16.80  20.76
 2c  16.80  16.80  20.76  16.80
 2c  3.96
2
 3.96
c
2
2
c  1.98
10c  70b  103.80
 10c  44b  72.60
26b  31.20
26
31.20
b
26
26
b  1.20
2c  3 p  4b  13.17
2(1.98)  3 p  4(1.20)  13.17
3.96  3 p  4.80  13.17
3 p  8.76  13.17
3 p  8.76  8.76  13.17  8.76
3 p  4.41
3
4.41
p
3
3
p  1.47
The price of one bag of potato chips is $1.98.
The price of one litre of pop is $1.47
The price of one chocolate bar is $1.20
Exercises:
1. Solve the following systems of linear equations algebraically:
3a  2b  3c  1
a) 4a  3b  c  7
2a  b  2c  16
3 p  3q  2r  7
b) p  2q  r  3
2 p  2q  r  8
2x  y  z  3
c) x  2 y  2 z  4
3x  y  z  9
c  2d  3e  13
d) 3c  2d  e  1
2c  d  2e  4
2. A local computer company sells three types of laptop computers to three nearby
stores. The number of laptops ordered by each store and the amount owing to the
company for the order is shown in the following table:
Store
Laptop A
Laptop B
Laptop C
Wal-Mart
Sears
Target
10
7
8
8
9
4
6
5
3
Amount
Owing($)
21 200
18 700
13 000
Write a system of equations to represent the above information and determine the unit
price of each type of laptop computer.
3. Cory, Josh and Dan went shopping for Halloween treats. Cory bought 3 chocolate
pumpkins, 4 masks and 8 candy witches. He spent $36.65. Josh bought 5 chocolate pumpkins,
3 masks and 10 candy witches. He spent $37.50. Dan bought 4 chocolate pumpkins, 5 masks
and 6 candy witches. He spent $43.45. Write a system of equations to represent this problem
and algebraically calculate the unit price of each item purchased.
4. Janet, Larry and Sam bought decorations to decorate the clubhouse for a
Christmas party. The number of items bought by each person is given in the table
along with the total cost of each purchase. Write a system of equations to represent
this problem and algebraically calculate the unit price of each item. Each item was
bought at the same store.
Name of the
shopper
Janet
Larry
Sam
Number of
rolls of
garland
2
3
3
Number of
wreaths
Number of
poinsettias
Cost
4
2
3
2
4
1
$49.50
$57.75
$38.50
Solutions:
3a  2b  3c  1
4a  3b  c  7
2a  b  2c  16
1.a)
3a  2b  3c  1
4a  3b  c  7

3a  2b  3c  1
 3(4a  3b  c  7)

4a  3b  c  7
2a  b  2c  16

2(4a  3b  c  7)
2a  b  2c  16

 9a  7b  22
10a  5b  30

5(9a  7b  22)
7(10a  5b  30)

3a  2b  3c  1
 12a  9b  3c  21
 9a  7b  22
8a  6b  2c  14
2a  b  2c  16
10a  5b  30
 45a  35b  110
70a  35b  210
25a  100
25
100
a
25
25
a4
10a  5b  30
10(4)  5b  30
2a  b  2c  16
2(4)  2  2c  16
40  5b  30
40  40  5b  30  40
 5b  10
5
 10
b
5
5
b2
10  2c  16
10  10  2c  16  10
 2c  6
2
6
c
2
2
c  3
The solution is (a, b, c) = (4, 2, -3)
3 p  3q  2r  7
p  2q  r  3
2 p  2q  r  8
b)
3 p  3q  2r  7
p  2q  r  3

3 p  3q  2r  7
2( p  2q  r  3)

3 p  3q  2r  7
2 p  4q  2r  6
5 p  7q  1
p  2 q  r  3
2 p  2q  r  8

 1( p  2q  r  3)
2 p  2q  r  8

 p  2q  r  3
2 p  2q  r  8
p  4q  11
5 p  7q  1
p  4q  11

5 p  7q  1
 5( p  4q  11)

5 p  7q  1
 5 p  20q  55
27q  54
27
 54
q
27
27
q  2
p  2 q  r  3
3  2( 2 )  r  3
3  4  r  3
 1  r  3
 1  1  r  3  1
 r  2
1
2
r
1
1
r2
p  4q  11
p  4(2)  11
p  8  11
p  8  8  11  8
p3
The solution is (p, q, r) = (3, -2, 2)
2x  y  z  3
x  2 y  2 z  4
3x  y  z  9
c)
2x  y  z  3
x  2 y  2 z  4

 2(2 x  y  z  3)
x  2 y  2 z  4

 4 x  2 y  2 z  6
x  2 y  2 z  4
 3x  4z  10
x  2 y  2 z  4
3x  y  z  9

x  2 y  2 z  4
2(3 x  y  z  9)

x  2 y  2 z  4
6 x  2 y  2 z  18
7 x  14
7
14
x
7
7
x2
 3 x  4 z  10
 3(2)  4 z  10
2x  y  z  3
2( 2)  y  1  3
 6  4 z  10
 6  6  4 z  10  6
 4 z  4
4
4
z
4
4
z 1
4  y 1  3
5 y  3
55 y  35
y  2
The solution is (x, y, z) = (2, -2, 1)
c  2d  3e  13
3c  2d  e  1
2c  d  2e  4
d)
c  2d  3e  13
3c  2d  e  1

 3(c  2d  3e  13)
3c  2d  e  1

c  2d  3e  13
2c  d  2e  4

 2(c  2d  3e  13)
2c  d  2e  4

 4d  10e  38
 5d  4e  22

 5(4d  10e  38)
4(5d  4e  22)
 4d  10e  38
 4d  10(3)  38
 4d  30  38
 4d  30  30  38  30
 4d  8
4
8
d
4
4
d  2

 3c  6d  9e  39
3c  2d  e  1
 4d  10e  38
 2c  4d  6e  26
2c  d  2e  4
 5d  4e  22
20d  50e  190
 20d  16e  88
34e  102
34
 102
e
34
34
e  3
c  2d  3e  13
c  2(2)  3(3)  13
c  4  9  13
c  13  13
c  13  13  13  13
c0
The solution is (c, d, e) = (0, -2, -3)
2. Let a represent the price of Laptop A.
Let b represent the price of Laptop B.
Let c represent the price of Laptop C.
10a  8b  6c  21200
7a  9b  5c  18700
8a  4b  3c  13000
10a  8b  6c  21200

7a  9b  5c  18700
 5(10a  8b  6c  21200)
 50a  40b  30c  106000

6(7 a  9b  5c  18700)
42a  54b  30c  112200
 8a  14b  6200
10a  8b  6c  21200
10a  8b  6c  21200
10a  8b  6c  21200


 2(8a  4b  3c  13000)
8a  4b  3c  13000
 16a  8b  6c  26000
 6a  4800
6
 4800
a
6
6
a  800
 8a  14b  6200
 8(800)  14b  6200
 6400  14b  6200
 6400  6400  14b  6200  6400
14b  12600
14
12600
b
14
14
b  900
Laptop A costs $800.00.
10a  8b  6c  21200
10(800)  8(900)  6c  21200
8000  7200  6c  21200
15200  6c  21200
15200  15200  6c  21200  15200
6c  6000
6
6000
c
6
6
c  1000
Laptop B costs $900.00.
3. Let p represent the unit price of the chocolate pumpkin.
Let m represent the unit price of the mask.
Let w represent the unit price of the candy witch.
3 p  4m  8w  36.65
5 p  3m  10 w  37.50
4 p  5m  6 w  43.45
Laptop C costs $1000.00
3 p  4m  8w  36.65
 9 p  12m  24 w  109.95
 3(3 p  4m  8w  36.65)


5 p  3m  10 w  37.50
20 p  12m  40 w  150.00
4(5 p  3m  10 w  37.50)
11 p  16w  40.05
5 p  3m  10 w  37.50
 25 p  15m  50 w  187.50
 5(5 p  3m  10w  37.50)


4 p  5m  6 w  43.45
12 p  15m  18w  130.35
3(4 p  5m  6w  43.45)
 13 p  32w  57.15
11 p  16 w  40.05

 13 p  32 w  57.15
22 p  32 w  80.10
2(11 p  16 w  40.05)

 13 p  32 w  57.15
 13 p  32 w  57.15
9 p  22.95
9
22.95
p
9
9
p  2.55
11 p  16 w  40.05
11(2.55)  16 w  40.05
28.05  16 w  40.05
28.05  28.05  16 w  40.05  28.05
16 w  12.00
16
12.00
w
16
16
w  0.75
4 p  5m  6 w  43.45
4(2.55)  5m  6(0.75)  43.45
10.20  5m  4.50  43.45
14.70  5m  43.45
14.70  14.70  5m  43.45  14.70
5m  28.75
5
28.75
m
5
5
m  5.75
Each chocolate pumpkin costs $2.55.
Each mask costs $5.75.
Each candy witch costs $.75.
4. Let g represent the cost of the garland.
Let w represent the cost of a wreath.
Let p represent the cost of a poinsettia.
2 g  4 w  2 p  49.50
3 g  2 w  4 p  57.75
3 g  3w  p  38.50
2 g  4 w  2 p  49.50
2 g  4w  2 p  49.50
2 g  4 w  2 p  49.50


3 g  3w  p  38.50
 2(3 g  3w  p  38.50)
 6 g  6 w  2 p  77.00
 4 g  2 w  27.50
3 g  2 w  4 p  57.75
3g  2w  4 p  57.75
3g  2w  4 p  57.75


 4(3 g  3w  p  38.50)
 12 g  12 w  4 p  154.00
3 g  3w  p  38.50
 9 g  10w  96.25
 4 g  2 w  27.50
20 g  10 w  137.50
 5(4 g  2 w  27.50)


 9 g  10 w  96.25
 9 g  10 w  96.25
 9 g  10 w  96.25
11g  41.25
11
41.25
g
11
11
g  3.75
 4 g  2 w  27.50
 4(3.75)  2 w  27.50
 15.00  2 w  27.50
 15.00  15.00  2 w  27.50  15.00
 2w  12.50
2
 12.50
w
2
2
w  6.25
2 g  4 w  2 p  49.50
2(3.75)  4(6.25)  2 p  49.50
7.50  25.00  2 p  49.50
32.50  2 p  49.50
32.50  32.50  2 p  49.50  32.50
2 p  17.00
2
17.00
p
2
2
p  8.50
The cost of each roll of garland is $3.75.
The cost of each wreath is $6.25.
The cost of each poinsettia is $8.50.