Systems of Linear Equations (Algebraically)
(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).