system of equation with four variables.
x + y + z + w =  0
x + y - z - w =  1
x - y + z + w =  3
x - y - z + w = -1
We can eliminate y, z and w simply
by adding the second and third equations,
term by term
 x + y - z - w =  1
 x - y + z + w =  3
-------------------
2x             =  4
            2x =  4 
             x =  2
Substitute 2 for x in the
original equations.
2 + y + z + w =  0
2 + y - z - w =  1
2 - y + z + w =  3
2 - y - z + w = -1
 y + z + w = -2
 y - z - w = -1
-y + z + w =  1
-y - z + w = -3
The third equation is just the
second equation multiplied by
-1, so we can eliminate it:
 y + z + w = -2
 y - z - w = -1
-y - z + w = -3
We can eliminate both z and w
by adding the first and second
equation:
 y + z + w = -2
 y - z - w = -1
---------------
2y         = -3
        2y = -3
         y =  Substitute
  
Substitute  for y
in the three:
 for y
in the three: 
  + z + w = -2
 + z + w = -2
  - z - w = -1
 - z - w = -1
 - z + w = -3
 - z + w = -3
  + z + w = -2
 + z + w = -2
  - z - w = -1
 - z - w = -1
   - z + w = -3
Multiply each through by 2 to
eliminate the fractions:
  -3 + 2z + 2w = -4
  -3 - 2z - 2w = -2
   3 - 2z + 2w = -6
       2z + 2w = -1 
      -2z - 2w =  1
      -2z + 2w = -9
The second equation is just the
first equation multiplied by -1,
so we can eliminate it:
       2z + 2w = -1 
      -2z + 2w = -9
Now we can eliminate z by adding
those two equations:
       2z + 2w =  -1 
      -2z + 2w =  -9
     ---------------
            4w = -10
             w =
 - z + w = -3
Multiply each through by 2 to
eliminate the fractions:
  -3 + 2z + 2w = -4
  -3 - 2z - 2w = -2
   3 - 2z + 2w = -6
       2z + 2w = -1 
      -2z - 2w =  1
      -2z + 2w = -9
The second equation is just the
first equation multiplied by -1,
so we can eliminate it:
       2z + 2w = -1 
      -2z + 2w = -9
Now we can eliminate z by adding
those two equations:
       2z + 2w =  -1 
      -2z + 2w =  -9
     ---------------
            4w = -10
             w =  w =
             w =  So (x,y,z,w) = (2,
So (x,y,z,w) = (2, ,2,
,2, )
Edwin
)
Edwin