mathcalculus
posted by Shreya .
(5y)/(6y^27y3)+(4y3)/(2y^215y+18)consider nonpermissible values for the addition of this equation, the product of the nonpermissible values is.
5y/(2y3)(3y+1)+(4y3)(2y3)(y6)
i not get what it asking for.

5y/( (2y3)(3y+1) )+(4y3)/((2y3)(y6) )
look at the denominators .
We know we cannot divide by zero, so the "nonpermissible" values of x would be all those that make either one of the denominators zero.
But you factored it, so that is good, all we have to find out which values make any of the factors equal to zero
so.....
2y3 ≠ 0
2y ≠ 3
y ≠ 3/2
3y+1 ≠ 0
3y ≠ 1
y ≠ 1/3
and
y ≠ 6
so the nonpermissible values of x are 1/3, 2/3 and 6 
thanks very much reiny :)