Need help!! 5x-xy/6x3y2 times 2/25-y2 ???

[Gretchen Vivanco]

Hi! I need help with a math problem

5x-xy/6x3y2 x 2/25-y2

and the 6x3y2 is 6, x cubed, y squared

and 25- y squared

I hope this makes sense. The part I don't get is the 25-y2. I can't remember how you do that with the negative. I think you switch the -y2 and 25 around so somehow it comes out to be (y+5)(y-5), but the negative is throwing me off!!

Please help!! This is for an exam!!!

Thank you!!

 

[stapel]

5x-xy/6x3y2 x 2/25-y2

and the 6x3y2 is 6, x cubed, y squared and 25- y squared

I'm sorry but, lacking grouping symbols, there is little way to guess what is meant by the above. Please review the "Kurts Notes" article (in the "Forum Help" pull-down menu at the very top of this page), and repost using standard web-safe formatting. Thank you.

This is for an exam!

You're hoping we'll finish your exam for you...? I'm assuming you don't mean that, but I don't know what you do mean. Why would an exam being involved be important...?

In any case, lacking instructions, it is difficult to know how to proceed. Please read the indicated article, reply with a clarified version of the exercise, and include the instructions and a clear listing of everything you have tried so far.

Thank you.

Eliz.

 

[Gretchen Vivanco]

sorry! I think I fixed it

No, I meant studying for an exam!! I wouldn't cheat!!!

ok.....I think I understand how to write it....

(5x-xy)/(6x^3y^2) * 2/(25-y^2)

Does that look right?

Thank you for your help!

 

[skeeter]

\(\displaystyle \L \frac{5x-xy}{6x^3y^2} \cdot \frac{2}{25-y^2} =\)

\(\displaystyle \L \frac{x(5-y)}{6x^3y^2} \cdot \frac{2}{(5-y)(5+y)} =\)

\(\displaystyle \L \frac{2x(5-y)}{6x^3y^2(5-y)(5+y)} =\)

\(\displaystyle \L \frac{1}{3x^2y^2(5+y)}\)

 

[Gretchen Vivanco]

Thank you!

Thank you so much!! I was trying to complicate it, and the 25-y^2 was so simple!