Solving Systems Of Linear Equations in Three Variables - College Level

Solve the following system of equations:

x + y + z = 6 ………………….. (1)
2x – y + z = 3 ………………….. (2)
x + 2y – 3z = -4 ……………….. (3)


Answer : (1,2,3)
X = 1
Y = 2
Z = 3

Solution:
Since we have three variables, we need to eliminate one of the variables form each equation in order to reduce them to two variables, which will be easier to handle. Once that is done, we solve for those two variables first. Once we have the first ordered pair, we can substitute those in our original equations and solve for the third and final variable.
Here are the steps:

Step 1.
Add equations (1) and (2) to eliminate the y variable as follows:
x + y + z = 6 ………………….. (1)
2x – y + z = 3 ………………….. (2)
-------------------
3x + 2z = 9 ………………….. (4)

Step 2.
In order to eliminate the y variable from equation (2) and (3) we must first multiply equation (2) by 2 to make the y coefficients equal and then adding the two equations as follows:
4x – 2y + 2z = 6 (the result of multiplying equation (2) above by the number 2)
x + 2y – 3z = -4 ………………….. (3)
-----------------------
5x - z = 2 ………………….. (5)

Step 3.
Equations (4) and (5) above have only two variables. If we add the two we
can eliminate one of the variables and solve for the other one. In order to do that,
we must first multiply equation (5) by the number 2 to eliminate z as follows:
3x + 2z = 9 ………………….. (4)
10x - 2z = 4 (the result of multiplying equation (5) above by the number 2
-------------------
13x = 13
x = 1 (we solve for x by dividing both sides of the equation by 13)

Step 4.
We can now substitute the value for x in either equation (4) or (5) to find z as as follows:
(3) (1) + 2z = 9
2z = 9-3
2z = 6
z = 3

Step 5.
Now that we have the values for both x and z we can substitute in any of the original equations to find y as follows:
1+ y + 3 = 6
y = 6 – 3 –1
y = 2

Therefore the triple points that will satisfy the equation is (1,2,3). We can check the accuracy of our answer by substituting the above points in any of the original equations and finding out that the right side of the equation will equal the left side.

The above problem appears in Intermediate Algebra Text/Workbook by Charles McKeague. 3rd. Edition

©Copyright 2007.Najwa S. Hirn. All rights reserved.