Posted by Shreya on Monday, January 9, 2012 at 10:13am.
5y/( (2y-3)(3y+1) )+(4y-3)/((2y-3)(y-6) )
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.....
2y-3 ≠ 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 :)
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