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December 22, 2014

December 22, 2014

Posted by **Shreya** on Monday, January 9, 2012 at 10:13am.

5y/(2y-3)(3y+1)+(4y-3)(2y-3)(y-6)

i not get what it asking for.

- math-calculus -
**Reiny**, Monday, January 9, 2012 at 10:35am5y/( (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

- math-calculus -
**Shreya**, Monday, January 9, 2012 at 10:48amthanks very much reiny :)

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