Question 65845
Solve 
(1)   x + y + z = 5 
(2)   x - 2y + z = 8 
(3)   3x - y - z = 3

Lets add eq (1) & (3) and we get:

4x=8
x=2

Substitute x=2 in Eq (1) & (2) and we have:

(1)   2 + y + z = 5 
(2)   2 - 2y + z = 8 

subtract (2) from (1) getting:

3y=-3
y=-1

Substitute x=2 and y=-1 in (1) and we have:

(1)   2 + (-1) + z = 5 
1+z=5  subtract 1 from both sides:
z=4


Ck by substituting x=2, y=-1 and z=4 in the original eqs.


Hope this helps---ptaylor