Solving a System of Three Linear Equations by Elimination
Solving a System of Three Linear Equations by Elimination
The techniques for solving a
system of linear equations in three variables are similar to those used on systems of
linear equations in two variables. We eliminate variables by either substitution or
addition.
Example 1
A linear system with a single solution
Solve the system:
(1) x + y - z = -1
(2) 2x - 2y + 3z = 8
(3) 2x - y + 2z = 9
Solution
We can eliminate z from Eqs. (1) and (2) by multiplying Eq. (1) by 3 and adding it
to Eq. (2):
3x
+ 3y
- 3z
= -3
Eq. (1) multiplied by 1
2x
- 2y
+ 3z
= 8
Eq.(2)
(4)
5x
+ y
= 5
Now we must eliminate the same variable, z, from another pair of equations.
Eliminate z from (1) and (3):
2x
+ 2y
- 2z
= -2
Eq. (1) multiplied by 2
2x
- y
+ 2z
= 9
Eq. (3)
(5)
4x
+ y
= 7
Equations (4) and (5) give us a system with two variables. We now solve this
system. Eliminate y by multiplying Eq. (5) by -1 and adding the equations:
5x + y
= 5
Eq. (4)
-4x - y
= -7
Eq. (5) multiplied by -1
x
= -2
Now that we have x, we can replace x by -2 in Eq. (5) to find y:
4x + y
4(-2) + y
-8 + y
y
= 7
= 7
= 7
= 15
Now replace x by -2 and y by 15 in Eq. (1) to find z:
x + y - z
-2 + 15 - z
13 - z
-z
z
= -1
= -1
= -1
= -14
= 14
Check that (-2, 15, 14) satisfies all three of the original equations. The solution set
is {(-2, 15, 14)}.
Helpful Hint
Note that we could have chosen
to eliminate x, y, or z first
in Example 1.You should solve
this same system by eliminating
x first and then by eliminating
y first. To eliminate x
first, multiply the first equation
by -2 and add it with the second
and third equations.
In the next example we use a combination of addition and substitution.
