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I have the formula $$f(x) = \frac{xy}{x+y}$$ How do I get the equation in terms of x or y?

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    That's not an equation. – 2017-02-20
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    @BobbieD Instead of being snarky, tell me what it actually is, then answer the question – 2017-02-20
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    I can't answer the question. To isolate $x$ or $y$ you need an equation (i.e. something equals $\frac{xy}{x+y}$). It seems you've left off the "something equals" part. – 2017-02-20
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    @BobbieD sorry, I've gotten used to being patronized and I overreacted. I will edit the question. – 2017-02-20
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    @script8man No problem. :-) – 2017-02-20
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    Um, it *is* in terms of x. What more do want. Do you mean if xa/(x+a) =c, solve for x? What are you asking? – 2017-02-20

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Note: your question doesn't exactly make sense as written. Here are two interpretations of it.

How to isolate $x$ in $z = \dfrac{xy}{x+y}$:

Let $z=\dfrac{xy}{x+y}$ where $x\ne -y$ (to avoid division by zero).

  1. Multiply both sides by $x+y$: $$z(x+y) = xy$$
  2. Distribute: $$xz + yz = xy$$
  3. Move all the terms with $x$ in them to the left and all the terms without $x$ to the right: $$xz - xy = -yz$$
  4. Factor out $x$: $$x(z-y) = -yz$$
  5. Assuming $z\ne y$ (to avoid division by zero), divide by $z-y$ on both sides: $$x = -\frac{yz}{z-y}\quad \Big(= \frac{yz}{y-z}\Big)$$

Rewrite the implicit functional equation $\frac{xy}{x+y} = c$, solving for $y$ as an explicit function of $x$:

Do the exact same steps as above, but solving for $y$ to get $$y = \frac{cx}{x-c}$$ for all $x\ne c$.

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    I like how you are much more explicit in each step than I was. Good idea for helping a beginner to algebraic manipulations - a +1 from me for sure! – 2017-02-20
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Let $z = f(x)$ $$z = \frac{xy}{x+y}$$ $$z(x+y) = xy$$ $$zx + zy= xy$$ $$zx = xy -zy$$ $$zx = y(x -z)$$ $$\frac{zx}{x-z} = y$$ By symmetry in the original equation, we can let $x \leftrightarrow y$ to get $$\frac{zy}{y-z} = x$$