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What is 9y+3 divided by `y^2-9` times 3-y divided by 3y^2+y?

What is 9y+3 divided by `y^2-9` times 3-y divided by 3y^2+y?

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What is 9y+3 divided by `y^2-9` times 3-y divided by 3y^2+y?

Postby barnamah » Tue 05 April, 2011 20:46

so the question is to simplify `(9y+3)/(y^2_9)*(3-y)/(3y^2+y)`
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Re: What is 9y+3 divided by `y^2-9` times 3-y divided by 3y^

Postby Integr88 » Tue 05 April, 2011 22:21

so to simplify `(9y+3)/(y^2-9)*(3-y)/(3y^2+y)`
lets see `9y+3` can be factored and can be written as `3(3y+1)`
at the denominator we have `y^2-y` to factor it `(y-3)(y+3)`
also at the denominator we have `3y^2+y=y(3y+1)`

putting them together we have:
`(3(3y+1)(3-y))/((y-3)(y+3)y(3y+1))`

Now we have `(3y+1)`at the numerator and denominator so it cancels out and we have:

`(3(3-y))/((y-3)(y+3)y)`

at the numerator we have `3(3-y)=3(-y+3)` and if we factor out the `-` we have `3(-y+3)=-3(y-3)` so change the numerator to `-3(y-3)` and we have:
`(-3(y-3))/((y-3)(y+3)y)`

Now we can cancel the (y-3) from both numerator and denominator and left out with:
`(-3)/(y(y+3))`

so that is the final answer: `(-3)/(y(y+3))`
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