Solving using Back Substitution

Sample Equation:

10x + 9y – z = 0
      15y – 5z = 20
                   z = 2

It is fairly easy to see how to proceed in this case. I'll just back-substitute the z-value from the third equation into the second equation, solve the result for y, and then plug z and y into the first equation and solve the result for x.

15y – 5(2) = 20                  
15y – 10 = 20
15y = 30

Therefore, y = 2

10x + 9(2) – (3) = 0
10x + 18 – 3 = 0
5x + 15 = 0
5x = –15

Therefore, x = -3

Then the solution is:
x =-3
y = 2
z = 2

The reason this system was easy to solve is that the system was "triangular"; this refers to the equations having the form of a triangle, because of the lower equations containing only the later variables.