Question

Simplify:
x+y / x-y + x / y-x - y / y² - x²

Please help! Thanks

Answer 1

Are there parentheses around the x+y terms? For example, in the first part, do you mean (x+y)/(x-y)?

Answer 2

I think yes.
\[\frac{x+y}{ x-y} + \frac{x}{ y-x} -\frac{ y}{ y^2 - x^2}\]

Answer 3

You should rewrite this way,
\[-\frac{x+y}{ y-x} + \frac{x}{ y-x} +\frac{ y}{ (y - x)(y+x)}\]
And then sum the fractions.

Answer 4

I would do it this way,
\[ \frac{-y}{ y-x} +\frac{ y}{ (y - x)(y+x)}=\frac{-y(y+x)+y}{ (y - x)(y+x)}\]
And that's the end.

Answer 5

I don't get why you put a minus out the front when you rewrote it ...

Answer 6

And I also don't get why you put a plus out the front of the last fraction.

Answer 7

@John_ES I don't think you quite got it right:

\[\frac{x+y}{x-y}+\frac{x}{y-x}-\frac{y}{y^2-x^2} = \frac{-1(x+y)}{y-x}+\frac{x}{y-x}+\frac{y}{y^2-x^2}\]
(multiplying the numerator of the first fraction by -1 allows us to reverse the terms in the denominator)

Now combine the first two terms which have a common denominator already:

\[ = \frac{x-(x+y))}{y-x}-\frac{y}{y^2-x^2} = -\frac{y}{y-x}-\frac{y}{y^2-x^2}\]
Let's get rid of the pesky - signs:
\[ -\frac{y}{y-x}-\frac{y}{y^2-x^2} = \frac{y}{x-y} + \frac{y}{x^2-y^2}\]
\[\frac{y(x+y)}{(x-y)(x+y)} + \frac{y}{x^2-y^2} = \frac{y(x+y)}{x^2-y^2} + \frac{y}{x^2-y^2}\]
\[\frac{y(x+y)}{x^2-y^2} + \frac{y}{x^2-y^2} = \frac{y+y(x+y)}{x^2-y^2} = \frac{y(1+x+y)}{x^2-y^2}=\frac{y(1+x+y)}{(x+y)(x-y)}\]

Answer 8

Sorry, this was my error.

Answer 9

It should be a minus, as you said.

Answer 10

@whpalmer is right ;)

Answer 11

There's a good reason why ancient mathematicians refused to write equations that had minus signs in them! :-)

Answer 12

I changed the sign of the fraction in the first line because I wanted a common denominator. That's the only convenient way to go about it. -1(x-y) = -1x -(-1y) = -1x+1y = y-x.

Answer 13

Let me format that for legibility:
\[-1(x-y) = -1x -(-1y) = -1x+1y = y-x\]

Answer 14

ok, but what about how you changed the negative to a plus outside the last fraction?

Answer 15

you mean right under where I said "let's get rid of the pesky - signs"?

Answer 16

No, at the start

Answer 17

Oh, that was just a typo in that line, I put it back correctly in the next line. Sorry, I can see how that would be confusing!

Answer 18

Should have been\[\frac{x+y}{x-y}+\frac{x}{y-x}-\frac{y}{y^2-x^2} = \frac{-1(x+y)}{y-x}+\frac{x}{y-x}-\frac{y}{y^2-x^2}\]

Answer 19

Good eye for spotting that, medal awarded :-)

Answer 20

ohhh, that makes sense, thankyou
and one last question, how do you get y(1+x+y) from y+y(x+y) near the end?

Answer 21

\[y+y(x+y) = y*1 + y(x+y) = y*1 + y*x + y*y = y(1+x+y)\]

Answer 22

it's coming up on 3 AM here, so I think I'm going to call it a night :-)

Answer 23

Ok thank you soooo much for your help ! :)