(2y+1)/(y^2-y-6)

Question

ruling out value for y.
which numbers can not be used in place of (y).

Answer 1

we cannot divide by zero
so the the problem arises when
y^2 - y - 6 = 0
(y-3)(y+2) = 0
y = 3 or y = -2

so y CANNOT be 3 or -2