This will likely have infinitely many solutions,
since there are more letters, variables, than equations.

2w - 3x + 4y +  z =  7
 w -  x + 3y - 5z = 10
3w +  x - 2y - 2z =  6



Swap rows 1 and 2, to get a 1 in the
upper left corner:

abbreviated R1<->R2



To get a 0 where the 2 is, multiply row 1 
by -2 and add it to row 2:

abbreviated -2R1+R2->R2



To get a 0 where the bottom left 3 is, 
multiply row 1 by -3 and add it to row 3:

abbreviated -2R1+R2->R2



To get a 1 where the -1 is on Row 2, 
multiply row 2 by -1

abbreviated -R2->R2


 
To get a 0 where the 4 is, 
multiply row 2 by -4 and add it to row 3:

abbreviated -4R2+R3->R3



To get a 1 where the -19 is on Row 3, 
multiply row 3 by -1/19



Then we change the matrix back to equations:

 w -  x + 3y -  5z = 10
      x + 2y - 11z = 13
           y -  3z =  4

Solve each equation for the first letter

                 w = 10 +  x -  3y +  5z
                 x = 13 - 2y + 11z  
                 y =  4 + 3z

Now we do back substitution:

Sunstitute the expression for y in the middle equation:

                 x = 13 - 2(4 + 3z) + 11z
                 x = 13 - 8 - 6z + 11z
                 x =  5 + 5z

Substitute the expressions for x and y in the top equation:

                 w = 10 +  x -  3y +  5z
                 w = 10 + (5 + 5z) - 3(4 + 3z) + 5z
                 w = 10 + 5 + 5z - 12 - 9z + 5z
                 w = 3 + z               

So the solution is

  (w,x,y,z) = (3+z, 5+5z, 4+3z, z)

Some teachers will tell you to use a different letter than z for z, 
such as "a", or "k".  If they use "a" the solution would be:

  (w,x,y,z) = (3+a, 5+5a, 4+3a, a)

Edwin