SOLUTION: solve the system of three linear equations with three variables 4x+y-z=8 x-y+2z=3 3x-y+z=6

Always write systems of equations so that like terms
line up vertically, like this:

4x + y -  z = 8
 x - y + 2z = 3
3x - y +  z = 6

We can pick any letter to eliminate and any two equations
to eliminate it from, but we look for the easiest possible.

We can eliminate the y's from the first two equations just
by adding them since the y-terms are opposites:

4x + y -  z = 8
 x - y + 2z = 3
---------------
5x     +  z = 11

Now we must elimiante the same letters y from another
pair of equations, and again we look for the easioest
possible way.

We can eliminate the y's from the first and third equations 
just by adding them since the y-terms are opposites:

4x + y -  z =  8
3x - y +  z =  6
----------------
7x          = 14

We didn't plan on the z's canceling but they did and that
was a bonus.  Now we have the system of two equations in
two variables:

5x + z = 11
7x     = 14

We solve the second equation for x

7x = 14
 x = 2

Then we substitute in 

  5x + z = 11
5(2) + z = 11
  10 + z = 11
       z = 1

Now we need y, so we substitute for x and z in 
any one of the original equations.  The easiest
one is

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

So the solution is (x,y,z) = (2,1,1)

Edwin