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Hi, According to wolfram alpha x+y / x-y can also be represented as <-(2x)/(y-x)> -1When 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.

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)

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)

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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.

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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)

According to Wolfram Alpha: (displaystyle frac{x+y}{x-y}) can also be written (displaystyle frac{-2x}{y-x}-1)

Their answer is correct . . . but really stupid!I tried to figure out their reasoning . . . weird!We have: .(displaystyle frac{x+y}{x-y})Multiply by (displaystyle frac{-1}{-1}!:;;frac{-1}{-1}cdotfrac{x+y}{x-y} ;=;frac{-x-y}{-x+y} ;=;frac{-y-x}{y-x})Long division:. . (displaystyle egin{array}{ccccc}&&&-1 \ && — & — \ y-x & | & -y & -x \ && -y & +x \ && — & — \ &&& -2xend{array})Therefore: .(displaystyle frac{-y-x}{y-x} ;=;-1 – frac{2x}{y-x})~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~Why not divide “normally”? .(displaystyle frac{x+y}{x-y}). . (displaystyle egin{array}{cccccc} &&&&1 \ && –&–&– \ x-y & | & x & + & y \ && x & -& y \ && –&–&– \ &&&& 2y end{array})Therefore: .(displaystyle frac{x+y}{x-y} ;=;1 + frac{2y}{x-y})

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