6x - 4y + 5z = 31
5x + 2y + 2z = 13
 x +  y +  z =  2

1. Pick a letter to eliminate.
2. Pick a pair of equations and 
   eliminate that letter
3. Pick a different pair of equations
   and eliminate the SAME letter you
   eliminated in step 1.
4. Now you have a system of two 
   equations in two unknowns. Solve
   for the two unknowns.
5. Substitute these values in one of
   the original equations to find
   the value of the first unknown
   eliminated in step 1.


6x - 4y + 5z = 31
5x + 2y + 2z = 13
 x +  y +  z =  2

1. Pick a letter to eliminate.

   I arbitrarily pick y.

2. Pick a pair of equations and 
   eliminate that letter

   I arbitrarily pick the first
   and third equations to eliminate 
   y.

To eliminate y I multiply the 3rd
equation through by 4 to make the
y-term become +4y so it will cancel
out with the -4y term in the 1st
equation:

  6x - 4y + 5z = 31
4[ x +  y +  z =  2]

  6x - 4y + 3z = 31
  4x + 4y + 4z =  8
 -------------------
 10x      + 9z = 39

3. Pick a different pair of equations
   and eliminate the SAME letter you
   eliminated in step 1.

   I arbitrarily pick the second
   and third equations to eliminate 
   the same letter y.

To eliminate y I multiply the 3rd
equation through by -2 to make the
y-term become -2y so it will cancel
out with the +2y term in the 2nd
equation:

   5x + 2y + 2z = 13
-2[ x +  y +  z =  2]

   5x + 2y + 2z = 13
  -2x - 2y - 2z = -4
 --------------------
   3x           =  9
              x = 3 


4. Now you have a system of two 
   equations in two unknowns. Solve
   for the two unknowns.

The system we have is 

      10x  + 9z = 39
              x = 3

So we substitute x = 3 in

       10x + 9z = 39
     10(3) + 9z = 39
        30 + 9z = 39
             9z = 9
              z = 1  
    
5. Substitute these values in one of
   the original equations to find
   the value of the first unknown
   eliminated in step 1. 

I arbitrarily pick the 3rd original
equation and substitute x = 3 and
z = 1 into:

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

So the solution is:

(x, y, z) = (3, -2, 1)

Edwin