Consider the system of equation,
x+y+z=a …… (1)
x−y+bz=2 …… (2)
2x+3y−z=1 …… (3)
As these equations have infinitely many solutions, therefore,
[]=0 1(1−3b)−1(−1−2b)+1(3+2)=0 −b=−7 b=7 From equation (1) and (2),
2x+8z=a+2 x+4z= …… (4)
From equation (1) and (3), we get
x+4x=3a−1 …… (5)
From equation (4) and (5), we get
=3a−1 a+2=6a−2 5a=4 So, the required value is,
b−5a=7−4 =3