I have been trying to get the inverse of the following function for quite some time now: y = (x-2) / (x+2) Thought it would be seemingly easy but I am really having a hard time. Here is what I have thus far: y + (2y+2)/x = 1 Trying to solve for x, I think this is a good start, any help, I would like to get this implimented but I am horrible at simple algebra!

y       =  (x-2) / (x+2)xy+2y   =  x-2xy+2y-x = -2xy-x    = -2y-2(y-1)x  = -2y-2x       = -(2y+2) / (y-1) 

EDIT : Wrote it a little more by the book.

Have fun!
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Oooh, you found the horadric cube!

[edited by - Coincoin on September 4, 2002 8:24:24 PM]

y = (x-2) / (x+2), x ≠ -2
y(x+2) = x-2
xy + 2y - x = -2
xy - x = -2y - 2
x(y - 1) = -2y - 2
x = (-2y - 2) / (y - 1), y ≠ 1
x = (2y + 2) / (1 - y), y ≠ 1

HTH

quote:

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Original post by Beer Hunter
x = (-2y - 2) / (y - 1), y ‚ 1
x = (2y + 2) / (1 - y), y ‚ 1

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How about (2y+2)*(1-y)^-1

quote:

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Original post by darookie
How about (2y+2)*(1-y)^-1

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If you''re trying to avoid the y ≠ 1 restriction, you''re doomed to failure . If y = 1, then there is no real solution for x, no matter how you try to rephrase the equation.

let y = 1
((2y+2)*(1-y))^-1 = ((2y+2)*0)^-1 = 0^-1 = 1/0 = oops