If the system of equations x+y=1,(c+2)x+(c+4)y=6,(c+2)2x+(c+4)2y=36
are consistent, then the value of c can be
For system of equations to be consistent:
Δ=0
⇒Δ=⎡⎢⎣11−1(c+2)(c+4)−6(c+2)2(c+4)2−36⎤⎥⎦=0
By C1→C1−C2 and C2→C2+C3,
⇒⎡⎢⎣00−1(−2)(c−2)−6−4(c+3)(c−2)(c+10)−36⎤⎥⎦=0
Taking (−2) common from C1,(c−2) common from C2 and (−1) common from C3, we get
⇒(−2)⋅(c−2)⋅(−1)⎡⎢⎣0011162(c+3)(c+10)36⎤⎥⎦=0
∴c=2,4