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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.

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