Question 639177
<pre>
(1)   2x - 3y +  z =  5
(2)    x + 3y + 8z = 22
(3)   3x -  y + 2z = 12

The smart thing to notice is that if you add (1) and (2)
the -3y and the +3y will eliminate the y's

(1)   2x - 3y +  z =  5
(2)    x + 3y + 8z = 22
-----------------------
      3x      + 9z = 27

and that can be divided through by 3

(4)         x + 3z = 9

Now we need another equation that does not contain y.

We multiply (3) by 3 and add it to (2)

(3)   3x -  y + 2z = 12

Multiply it by 3:

      9x - 3y + 6z = 36
(2)    x + 3y + 8z = 22
-----------------------
     10x      +14z = 58

We can divide that through by 2 and get

(5)        5x + 7z = 29

Now we put (4) and (5) together and we now
have a system of only 2 equations in only
2 unknowns.

(4)         x + 3z =  9
(5)        5x + 7z = 29

Multiply (4) through by -5 and add to (5)

         -5x - 15z = -45
(5)       5x +  7z =  29
-------------------------
               -8z = -16
(6)              z = 2

Substitute z = 2 into (4)

(4)         x + 3z = 9
          x + 3(2) = 9
             x + 6 = 9
(7)              x = 3

Substitute x=3 and z=2 into (3)

(3)   3x -  y + 2z = 12
    3(3)- y + 2(2) = 12
         9 - y + 4 = 12
            13 - y = 12
                -y = -1
(8)              y = 1

From (7), (8), and (6),

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

Edwin</pre>