The tutor shows an example of solving a three-variable linear system algebraically.

Example: Solve the following system of equations:

1. -3x + 2y + z = 16
2. x – y – 3z =-19
3. 2x + y + 7z = 30

The first idea is to eliminate the same variable in two ways, leaving a system of two equations in two variables.
We might add the first with twice the second:
-3x + 2y + z = 16
2x – 2y – 6z =-38

4. -x -5z = -22

Similarly, we can add the second to the third:
x – y – 3z =-19
2x + y + 7z = 30

5. 3x + 4z = 11

Next, we can eliminate x by adding 3(IV) to V:
-3x – 15z = -66
3x + 4z = 11
– 11z = -55

Therefore, z=5.

We can substitute z=5 back into V, for instance, to get x:

3x + 4(5) = 11 ⇒ 3x + 20 = 11 ⇒ 3x = -9 ⇒x = -3

Now, subbing both x=-3 and z=5 into III we get

2(-3) + y +7(5) = 30 ⇒ y = 1

Apparently the solution is

x=-3, y=1, z=5

HTH:)

Source:

Travers, Kenneth et al. Using Advanced Algebra. Toronto: Doubleday Canada, 1977.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.