Question

x^2 -y^2/xy /1/y - 1/x=

Answer 1

\[\Large \frac{\frac{x^2-y^2}{xy}}{\frac{1}{y}-\frac{1}{x}}\]This?

Answer 2

yes

Answer 3

okay, so I'd start with simplifying:
\[\frac{1}{y}-\frac{1}{x}\]

Answer 4

Do you know how to do this?

Answer 5

no

Answer 6

In order to subtract two fractions, they have to have a common denominator.

Answer 7

so is it 1

Answer 8

\[\frac{1}{y}-\frac{1}{x} = \frac{1}{y} \cdot \frac{x}{x} -\frac{1}{x} \cdot \frac{y}{y} = \frac{x}{xy} - \frac{y}{xy} = \frac{x-y}{xy}\]

Answer 9

Does what I did make sense to you?

Answer 10

We end up with:
\[\Large \frac{\frac{x^2-y^2}{xy}}{\frac{x-y}{xy}}\]

Answer 11

yes I was looking it over. So i have to pair them up inoder to get and answer... Sorry my son had just hit his head on the wall that is why a delay in responding back.

Answer 12

Do you know what to do from here?

Answer 13

I subtract and then divide it by xy then do the same for the bottom to get my answer

Answer 14

is that correct

Answer 15

yeah.

Answer 16

what did you get?

Answer 17

xy^2

Answer 18

That is not correct, how did you get that?

Answer 19

I thought it was simplifying..

Answer 20

to me i thought it was one the because both the top and the bottom equal out to me

Answer 21

Dividing by a fraction is the same as multiplying by its reciprocal.\[
\Large \frac{\frac{x^2-y^2}{xy}}{\frac{x-y}{xy}}
= \frac{x^2-y^2}{xy}\cdot \frac{xy}{x-y}
\]

Answer 22

In this case the xy cancel out leaving:
\[\frac{x^2-y^2}{x-y}\]

Answer 23

This can also be simplified more though, so it's not done.

Answer 24

We have to factor \(x^2-y^2\) to see if it has \(x-y\) as a factor.