Before you begin with this lesson, you should have some familiarity with linear systems in two variables, and how to solve them using elimination/combination and substitution techniques.
Linear systems in three variables are quite similar to those in two variables, except, obviously, we have an extra variable to work with. Just as in the two variable case, where we can only hope to find a unique solution when we have two equations to work with, for the three variable case we need usually need three equations.
A linear equation in three variables usually looks like this
ax +by +cz = d
where x, y, an z are our unknowns, and a, b c, and d are just numbers. And a three variable system usually looks like this:
a1x + b1y + c1z = d1
a2x + b2y + c2z = d2
a3x + b3y + c3z = d3
Just as a solution to a two variable system was an ordered pair, the solution to a three variable linear system is an ordered triple, (x0, y0, z0) such that all three equations are satisfied when x = x0, y = y0 and z = z0.
In the following example, we verify that an ordered triple works as a solution to a three variable linear system.
Great, we can check solutions. It would now be prudent to ask, how do we find them in the first place? This is very much like the 2 variable case, we just have an extra step or two because of the third equation. We continue with the previous example.
Here is another example that we work through from scratch. This one contains a little trick. It looks like we only have two equations with three variables, but in truth we have three. One of them contains a term whose coefficient is zero. This leads us to an interesting point - any equation is really an equation in as many variables as we like, but most of the coefficients happen to be zero. So we could say that the following is a system of equations in 53 billion variables, but that nearly all of the terms have zero coefficients.
We attempt to eliminate a variable by adding one equation to a constant multiple of another
Now we do the same thing with our two two variable equations in order to solve for one of our three unknowns, which immediately leads us to the solution for the other two.
