<PRE><B>Question 2157</B>
X + 2Y + Z = 6      .............(1)
3X + Y + Z = 8      .............(2)
X + Y = 4           .............(3)
   Subtracting Eq.(2) from Eq.(1), we get
-2X + Y = -2        .............(4)
(3)-(4), we get
 3X = 6
So, X = 2 
and Y = 2    (By replacing the values of X an Y in Eq.3)
By replacing these values in Eq. No.2,
= 3*2 + 2 + Z = 8
= 8 + Z = 8
= Z = 0
So, X=2; Y=2; Z=0
 
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