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Solving a System of Three Equations in Three Variables

Date: 06/23/2007 at 16:35:53
From: Christopher
Subject: Can you help me with the formula for this equation

My son is having problems with this equation and I have forgotten how
to do this it has been so long.  Can you help me to explain this to him?

x + y + z = 10
x - 2y + z = -2
x - y - z = -12




Date: 06/24/2007 at 23:27:51
From: Doctor Peterson
Subject: Re: Can you help me with the formula for this equation

Hi, Christopher.

I see that we don't seem to have an explanation of solving systems of
three equations in our archives.  I also see that this one has some
tricks to it that might confuse your son even if he's fairly good at
solving others.  I'm not sure, therefore, whether I should show you
the normal way or help your son with the specific issues in this
problem.  But since you wrote, not he, and you didn't show where he is
having trouble, I'll choose to help you with the general methods.  If
necessary, he can write back and show what goes wrong.

Let's take a different and simpler problem so we can focus on the method:

  x + 2y - z = -3   [1]
  x -  y + z = 4    [2]
  x - 2y + 2z = 7   [3]


This is known as a system of THREE equations in THREE variables.  You
may recall that you solve a system of TWO equations in TWO variables
by combining them to eliminate one variable, so you reduce the system
to ONE equation with ONE variable.

With a three equation and three variable system, our strategy will
again be to reduce the number of variables and equations--this time
first reducing to two equations in two variables and then solving that
by reducing to only one equation.

We want to eliminate one variable.  We can pick any of them; in this
case I'll eliminate x because it has the coefficient 1 everywhere.
Look at the first two equations:

  x + 2y - z = -3     [1]
  x -  y + z = 4      [2]

If we just subtract the second from the first (that is, add the
negative of the second), we eliminate x:

   x + 2y -  z = -3   [1]
  -x +  y -  z = -4  -[2]
  -----------------
       3y - 2z = -7   [4]

I gave the new equation the name [4] so we can keep track of it later.

We need another equation in the same two variables.  We can combine
the third equation with either of the first two and eliminate again--
making sure to eliminate x this time as well. I'll use the last two
equations:

   x -  y +  z =  4   [2]
  -x + 2y - 2z = -7  -[3]
  -----------------
        y -  z = -3   [5]

Now I've got two equations in two unknowns, which will both be true
when all three of the original equations are true.  Let's write them
as a system:

  3y - 2z = -7        [4]
   y -  z = -3        [5]

We can solve this by eliminating a variable. I'll eliminate y by
multiplying [5] by -3 and adding it to [4]:

   3y - 2z = -7       [4]
  -3y + 3z =  9     -3[5]
  -------------
         z =  2       [6]

I've found the value of z!  Now how do I find x and y?

Well, if you think about what we've done, we reduced the system to the
two equations [4] and [5]; we've solved THAT system for z, but to
finish solving it we have to find y.  We go back to either [4] or [5]
and plug in z = 2, then solve for y.  I'll use equation [5]:

  y - z = -3          [5]

  y - 2 = -3

      y = -1

Now we've finished solving this system, but we still need to find x,
the first variable we eliminated.  To find x, we have to go back to
any of the original three equations and plug in both y = -1 and z = 2.
I'll use [2]:

  x -  y + z = 4    [2]

  x - -1 + 2 = 4

       x + 3 = 4

           x = 1

So the solution is x = 1, y = -1, z = 2.  You should plug all three
into each of the original three equations in order to make sure we
didn't make a mistake anywhere.

I want to take a moment to look back at the whole process.

We started with three equations, [1], [2], and [3]; then we eliminated
x to get two new equations, [4] and [5], with only two variables. 
Then we started as if this were a new problem, solving it by
eliminating y and solving for z, then going back to one of the
equations to get y.  Finally we went back to the original system and
found the remaining variable.

I think of this process as like climbing Mt. Everest.  We start at the
bottom of the mountain (base camp) with a big ugly problem to solve.
We then climb part way up and set up a high camp with a simpler
problem to solve (only two variables).  Then we climb to the top,
solving for one variable.  On the way up we discarded some equipment
(the variables x and y).  On the way back down, we pick them back up:
y on our way back to the high camp, and x on the way to the bottom:

      [6] --> z
       ^      |
       |      |
      / \     v
    [4] [5]   --> y
     ^   ^        |
     |   |        |
    / \ / \       v
  [1] [2] [3]     --> x

In the process, we eliminated a variable three times, repeating the
same process you use to solve two equations.

Note that you could eliminate variables in any order--it doesn't have 
to be x then y then z, but whatever is easiest (or whatever you feel 
like).  The important thing is that once you eliminate the first 
variable, you don't think about it again until you've finished solving
the system of two equations completely; then you find the remaining
variable by substituting in one of the original equations.

If you have any further questions, feel free to write back.  If you
have trouble with the actual problem, show me your work (showing at
least the equations [1] through [6] that you get, or whatever comes
out oddly) and I'll help work out the kinks.


- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Basic Algebra
High School Linear Equations
Middle School Algebra
Middle School Equations

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