Question

Please Help! I need an explanation for how to do this... Step-by-step directions would be great!

Simplify. Assume the divisor is not zero.

\[(8x^2y^2 – 4xy^2) \over 4xy\]

Answer 1

Answer choices include:
A. xy – 1
B. 2xy
C. 2xy – 1
D. 2xy – y

Answer 2

First look at the numerator of this fraction - can you see a common constant number that will divide into both terms in the numerator?

Answer 3

I believe four or two....

Answer 4

4 is indeed the correct value, so now you can write your fraction as:\[\frac{8x^2y^2 – 4xy^2}{4xy}=\frac{4(2x^2y^2 – xy^2)}{4xy}\]Now look at the \(x\) terms in the numerator - what power of \(x\) is common to both terms in the numerator?

Answer 5

2 or 1... I am not quite sure. :(

Answer 6

I believe the final answer is D. 2xy - y. Is it correct? @asnaseer

Answer 7

yes :)

Answer 8

Really? Wow! Thank you! I have two more problems, may you help with those as well?

Answer 9

so we can write this as:\[\frac{8x^2y^2 – 4xy^2}{4xy}=\frac{4(2x^2y^2 – xy^2)}{4xy}=\frac{4x(2xy^2 – y^2)}{4xy}=\frac{4xy(2xy – y)}{4xy}=\frac{\cancel{4xy}(2xy – y)}{\cancel{4xy}}\]\[=2xy-y\]

Answer 10

yw :)

Please post each question separately and I am sure there will be plenty of people that will try and help you out (if I am busy with other things).

Answer 11

Yes, I have a similar equation written in my notebook, but your equation definitely looks nicer. :)

Answer 12

And yes, I have posted the next question. May I tag you in the question?

Answer 13

@asnaseer ^

Answer 14

sure - but, as I said, I am not always free but will try my best :)

Answer 15

Well, if you cannot help, then do not worry. :)