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Coolmath Algebra
Systems of Equations 3x3's Lesson 2 - Solving by Elimination  
(page 1 of 8)

---- This algebra lesson explains how to solve a system of 3 equations and 3 unknowns by the elimination (addition) method.

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Solving 3 x 3 systems of equations works a lot like the 2 x 2 method of elimination -- with a little substitution mixed in.  It really isn't hard, but it takes good organization skills.  The problems are a bit long, but I'll show you how to work through them and how to keep things organized.

Instead of starting out by working through an entire problem (which would look pretty creepy), let's start with some that are almost finished already.

Remember that our goal is to find the values for x, y and z.  (or whatever letters you have.)

Let's finish solving this guy...  As you can see, we already know what z is!

2x - y + 6z = 20

5y - 3z = -2

z = 4

What we're going to do is called "back substitution" because we're going to work our way back up through the equations, starting at the bottom with
that
z.

Take the value for z and stick it into the equation with two variables... This will pop the y!

2x - y + 6z = 20 ... 5y - 3z = -2 ... z = 4 ...stick the 4 back into the equation 5y - 3z = -2 5y - 3 ( 4 ) = -2 ... 5y - 12 = -2 ... 5y = 10 ... y = 2

Continued on the next page

 The printing and distribution and/or downloading of these lessons is strictly prohibited.

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