Here's one exactly like it.  Just put your numbers instead:

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



The object is to get the matrix like this:



where there will be 1's on the diagonal, 0's underneath the
diagonal, and numbers where the @'s are.

Start with:



Swap rows 1 and 2, because it's easier when there's a 1
in the upper left corner.

R1 <-> R2


 
Multiply row 1 by -2
-2R1->R1



Add Row 1 and Row 2 and put that in place of Row 2

R1+R2->R2 



Restore Row 1 both 

R1/(-2) -> R1



Multiply Row 1 by -1



Add row 1 to Row 3



Restore row 1

-R1 <-> R1



Multiply row 2 by -1

-R2 <-> R2




Multiply Row by -2



Add row 2 and Row 3 and put that sum in row 3

R2+R3 -> R3



Restore Row 2

R2/(-2) -> R2



Divide row 3 by -5



R3/(-5)( -> R3

Rewrite as a system of equations:

1x + 1y + 1z = 2
0x + 1y + 3z = 2
0x + 0y + 1z = 1

Remove the understood 1's and 0's

x + y +  z = 2
    y + 3z = 2
         z = 1
         
The bottom equation of the system is already solved for z

         z = 1

Substitute in the middle equation of the system for y:

    y + 3z =  2
  y + 3(1) =  2
     y + 3 =  2
         y = -1
 
 Substitute in the top equation of the system for z

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

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

Edwin