Hi,

According to wolfram alpha x+y / x-y can also be represented as [-(2x)/(y-x)] -1

When it is in this form it works for me I just can't seem to get it in that form,

any help (or suggestions or basic rules to apply) would be greatly appreciated.

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- Apr 13th 2013, 10:29 PMCHUBBSPlease help simplify (x+y) / (x-y)
Hi,

According to wolfram alpha x+y / x-y can also be represented as [-(2x)/(y-x)] -1

When it is in this form it works for me I just can't seem to get it in that form,

any help (or suggestions or basic rules to apply) would be greatly appreciated.

- Apr 13th 2013, 10:33 PMCHUBBSRe: Please help simplify (x+y) / (x-y)
Sorry I just realised this is in the wrong section, it is for a limit question but the question itself is not. (I don't know how to move this into the right section)

- Apr 13th 2013, 10:42 PMibduttRe: Please help simplify (x+y) / (x-y)
We want to express the express (x+y)/(x-y) as [ -2x / (y-x) ] - 1.

In order to form it into that form we will first take out -1 common from the denominator to get

(x+y)/(x-y) = - ( x+y)/ (y-x) . Now divide the numerator by the denominator, quotient being 1 and remainder 2x.

Thus we get: (x+y)/(x-y) = - ( x+y)/ (y-x) = - [ 1 + 2x / ( y-x) ] = -1 - 2x / ( y-x) - Apr 14th 2013, 02:22 AMCHUBBSRe: Please help simplify (x+y) / (x-y)
Thank you for the quick reply, but I have no idea how you get from:

"Now divide the numerator by the denominator, quotient being 1 and remainder 2x.

Thus we get: - ( x+y)/ (y-x) = - [ 1 + 2x / ( y-x) ] = -1 - 2x / ( y-x)"

I haven't done algebra in 4-5 years, could you please explain this step by step? (as I have forgotten "basic math rules")

It's - ( x+y)/ (y-x) = - [ 1 + 2x / ( y-x) ] that I don't understand. - Apr 14th 2013, 05:14 AMtopsquarkRe: Please help simplify (x+y) / (x-y)
- Apr 14th 2013, 08:03 AMSorobanRe: Please help simplify (x+y) / (x-y)
Hello, CHUBBS!

Quote:

According to Wolfram Alpha: can also be written

Their answer is correct . . . but*really stupid!*

I tried to figure out their reasoning . . . weird!

We have: .

Multiply by

Long division:

. .

Therefore: .

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Why not divide "normally"? .

. .

Therefore: .