a) 2x -  y + z =  8
    x + 2y + z = -3
    x - 2y - z =  7

Add the 1st and 3rd equations and the z's
will cancel out

   2x -  y + z =  8
    x - 2y - z =  7
 ------------------
   3x - 3y     = 15

And what's more, that can be divided through
by 3

        x - y  =  5


Next add the 2nd and 3rd equations and the z's
will also cancel out

     x + 2y + z = -3
     x - 2y - z =  7
   ------------------
    2x          = 4
              x = 2

Substitute in 

        x - y = 5
        2 - y = 5
           -y = 3
            y = -3  
  
Substitute in this original equation:

    2x -  y + z = 8
2(2) - (-3) + z = 8
      4 + 3 + z = 8
          7 + z = 8 
              z = 1

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

-----------------------------------------

b) 3x -  y - 2z = 11
    x - 2y + 3z = 12
    x +  y - 2z =  5

Add the 1st and 3rd equations and the y's
will cancel out



   3x -  y - 2z = 11
    x +  y - 2z =  5
   -----------------
   4x      - 4z = 16

And what's more, that can be divided through
by 4  (that rhymes)

    x      -  z = 4


To make the y's cancel in the 2nd and 3rd
original equations which are:

    x - 2y + 3z = 12
    x +  y - 2z =  5

we will have to multiply the bottom one through by 2 
and add them:

    x - 2y + 3z = 12
   2x + 2y - 4z = 10
  -------------------
   3x      -  z = 22

Now we have a system of two equations in two unknowns:

    x - z =  4
   3x - z = 22

Multiply the 1st one through by -1 and the z's will cancel
when you add them:

   -x + z = -4
   3x - z = 22
:  -----------
   2x     = 18
        x = 9

Substitute in

    x - z = 4
    9 - z = 4
       -z = 4-9
       -z = -5
        z = 5

 Substitute in the original equation:

    x +  y - 2z = 5
 (9) + y - 2(5) = 5
     9 + y - 10 = 5
          y - 1 = 5
              y = 6

(x,y,z) = (9,6,5)

Edwin