• April 13th 2013, 09:29 PM
CHUBBS
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.
• April 13th 2013, 09:33 PM
CHUBBS
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)
• April 13th 2013, 09:42 PM
ibdutt
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)
• April 14th 2013, 01:22 AM
CHUBBS
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.
• April 14th 2013, 04:14 AM
topsquark
Quote:

Originally Posted by CHUBBS
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)

Don't worry. The question belongs here because it has nothing to do with Pre-Calculus/Calculus.

-Dan
• April 14th 2013, 07:03 AM
Soroban
Hello, CHUBBS!

Quote:

According to Wolfram Alpha: $\frac{x+y}{x-y}$ can also be written $\frac{-2x}{y-x}-1$

Their answer is correct . . . but really stupid!

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

We have: . $\frac{x+y}{x-y}$

Multiply by $\tfrac{-1}{-1}\!:\;\;\frac{-1}{-1}\cdot\frac{x+y}{x-y} \;=\;\frac{-x-y}{-x+y} \;=\;\frac{-y-x}{y-x}$

Long division:

. . $\begin{array}{ccccc}&&&-1 \\ && -- & -- \\ y-x & | & -y & -x \\ && -y & +x \\ && -- & -- \\ &&& -2x\end{array}$

Therefore: . $\frac{-y-x}{y-x} \;=\;-1 - \frac{2x}{y-x}$

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

Why not divide "normally"? . $\frac{x+y}{x-y}$

. . $\begin{array}{cccccc} &&&&1 \\ && --&--&-- \\ x-y & | & x & + & y \\ && x & -& y \\ && --&--&-- \\ &&&& 2y \end{array}$

Therefore: . $\frac{x+y}{x-y} \;=\;1 + \frac{2y}{x-y}$