# How do I solve this rational function in terms of $x$?

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?

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