Question 1088768
Solve the system:
  x + y + z = 4
 3x – y + z = 2
rearrange the 3rd equation
  y – z = -2 
  y = z - 2
In the 2nd equation, replace y with (z-2)
x - (z-2) + z = 2
x + 2 = 2
x = 0
replacing x with 0 in the first two equations we have
y + z = 4
-y +z = 2
------------Addition eliminates y, find z
2z = 6
z = 3
then
y = 3 - 2
y = 1
We have x=0; y=1; z = 3
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you can check these solutions in the original equations