Solving Systems of Equations with Three Variables

Solving Systems of Equations with Three Variables

Recall the following about solving systems of equations:

• consistent system is a system that has at least one solution.
• An inconsistent system is a system that has no solution.

The equations of a system are dependent if ALL the solutions of one equation are also solutions of the other two equations. In other words, they end up being the same equation.

The equations of a system are independent if they do not share ALL solutions. They can have multiple solutions on a separate equation, but all three of them don't coincide.

When solving systems of equations in three variables, it is generally recommended to use the elimination method to eliminate a variable.  This will need to be repeated until all variables are determined.

Example

Solve the following system:

x + y + z = 4
x - 2y - z = 1
2x - y - 2z = -1

First, let's eliminate x.  This can be done by taking the equations two at a time and using the elimination method to cancel out x each time.

Start with the first and second equations.  Multiply the first equation by -1 and add it to the second equation to eliminate x.

-x -  y  - z = -4
x - 2y - z =  1
-3y - 2z = -3

Now let's use the first and third equations.  Multiply the first equation by -2 and add it to the third equation.

-2x - 2y - 2z = -8
2x -  y - 2z  = -1
-3y - 4z = -9

Next, use the two equations created by those steps to eliminate y.  Multiply the first equation by -1 and add it to the second.

3y + 2z = 3
-3y - 4z  = -9
-2z = -6

Solve the resulting equation for z.

-2z = -6
z = 3

Substitute 3 for z in one of the two y-z equations and solve for y.

-3y - 2z   = -3
-3y - 2(3) = -3
-3y -  6    = -3
-3y   = 3
y   = -1

Finally, substitute -1 for y and 3 for z in one of the original equations and solve for x.

x +   y +   z = 4
x + (-1) + 3 = 4
x + 2 = 4
x = 2

The answer is (2, -1, 3).