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How would I solve for $x$ where $y=(1-x)/(1+x)$?. I have tried multiplying both sides by $(1+x)$, but I couldn't get far with that, as I had a pesky $xy$ on one side, which I couldn't figure out how to take out. How would I solve this?

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4 Answers 4

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This is a useful trick to learn; it comes up frequently. $$y=\frac{1-x}{1+x} \implies y(1+x)=1-x \implies y+xy=1-x$$ Now remember we want to solve for $x$, so collect all terms involving $x$: $$xy+x=1-y \implies x(1+y)=1-y \implies x=\frac{1-y}{1+y}$$ The reason these expressions for $x$ in terms of $y$ and $y$ in terms of $x$ are the same is that the function $f(x)=\frac{1-x}{1+x}$ is its own inverse - if you put in one number and get out a second, putting that back in will return you to your first number. You may be interested to know that your expression is a special case of the Möbius transformation, and these in general have very interesting properties.

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  • $\begingroup$ Thanks, this is exactly what I wanted. I didn't know where to go from y+xy+x=1 $\endgroup$
    – scrblnrd3
    Aug 27, 2013 at 21:28
  • $\begingroup$ You're welcome, I remember when I first learnt the very same manipulation. I've added some extra information you may be interested in. $\endgroup$ Aug 27, 2013 at 21:31
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Your approach is just fine. Multiplying by $1+x$ on both sides gets us $$(1+x)y=1-x,$$ and distributing on the left gets us $$y+xy=1-x.$$ Now, gather the terms with $x$ factors on the left side and terms without $x$ factors on the other, getting us $$x+xy=1-y,$$ whence factoring out the $x$ on the left gets us $$(1+y)x=1-y.$$ Finally, divide by $1+y$ to get $$x=\frac{1-y}{1+y}.$$

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A mechanical way of doing it is to first multiply both sides by $1+x$. We get $$y(1+x)=1-x.$$ Rewrite this as $$y+xy=1-x.$$ Now bring all the $x$ stuff to one side, and the rest to the other side. We get $$x+xy=1-y.$$ Thus $$x(1+y)=1-y,$$ and therefore $$x=\frac{1-y}{1+y}.$$

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Since $$y=\frac{1-x}{1+x}=\frac{2-(1+x)}{1+x}=\frac{2}{1+x}-1$$ we have $$x=\frac{2}{1+y}-1.$$

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