SOLUTION: (15ab/4)(8/9a^2b^2)= ? (6a/8b)/(10c/12d)= ? (2x^2+x-3/9)[(x+1)^2/2x^2+5x+3)= ? (1/25-y^2)/(6-3y/y^2-7y+10)= ? Thanks!

Question 203063: (15ab/4)(8/9a^2b^2)= ?
(6a/8b)/(10c/12d)= ?
(2x^2+x-3/9)[(x+1)^2/2x^2+5x+3)= ?
(1/25-y^2)/(6-3y/y^2-7y+10)= ?
Thanks!
Answer by PRMath(133) (Show Source): You can put this solution on YOUR website!
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I won't know how to put this in nice fractions for you, so I'll explain, as best as I can, what I'm doing............

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The 15 and the 9 can each be reduced when divided by 3. They become 5 and 3.
The 4 and the 8 can each be reduced when divided by 4. They become 1 and 2
The ab in the numerator can be canceled, but that changes the and the in the denominator to:

SO when we reduce and cancel, we get:

/ .........OR.........

NEXT QUESTION:

The above is how I am reading your question. If I am correct, then we would do that problem the way we would do any division problem when we work with fractions. In other words, when you divide with a fraction, you MULTIPLY by the reciprocal. So your problem is now........

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The 6 and the 10 can be reduced when divided by 2. They become 3 and 5.
The 12 and the 8 can be reduced when divided by 4. They become 3 and 2.

Your problem is now............

* and this is:

NEXT QUESTION:

/*/

If I have written your problem correctly, then it can be factored like this:

/

So the (2x + 3) in the numerator can be canceled with the (2x+3) in the denominator.
The (x + 1) in the numerator can be canceled with the (x + 1) in the denominator.

That leaves you with:

I am sorry to say that your last problem I don't think I'm seeing correctly. Can you repost it, please? It's this part of the problem I'm not sure about:
(1/25-y^2)

I am not sure you have written it correctly. The second part is easily factored. The (6 - 3y) can be factored to 3(2 - y) and the can be easily factored to (y-5)(y-2). I just am not certain your first part of the problem is written correctly.

I hope this helps .......
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