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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.
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