On a test recently I ended up having to solve this for y:

$$ x = \frac{2y}{y + 1} $$

But I kept getting stuck in cirlces...

$$ \begin{aligned} x(y + 1 ) = 2y \\ xy + x = 2y \\ \frac{xy + x}{2} = y \\ \end{aligned} $$

That didn't get me anywhere, so then I started over and tried multiplying both sides by the reciprocal:

$$ \begin{aligned} (\frac{y+1}{2y})(x) = 1 \\ \frac{xy+x}{2y} = 1 \\ \end{aligned} $$

And still I can't see a way to isolate y.

The worst part of this is that I remember being specifically taught a trick for this particular conundrum, but I can't remember the trick!

Wolfram gives the answer as

$$ y = - \frac{x}{x-2} $$

but it doesn't show the steps.

  • 1
    $\begingroup$ You haven't isolated y. Start from xy + x = 2y and subtract xy from both sides. $\endgroup$ Dec 12, 2010 at 0:46
  • $\begingroup$ That just gets me $ x = 2y - xy $ . From there I can divide by 2 or by x but I still can't isolate y. :( $\endgroup$
    – friedo
    Dec 12, 2010 at 0:53
  • 1
    $\begingroup$ You can factor $y$ from the RHS. $\endgroup$ Dec 12, 2010 at 0:55
  • 3
    $\begingroup$ But now you can factor the RHS, then divide by what multiplies y (assuming it isn't zero-as Arturo Magidin says, you don't want the universe to explode) $\endgroup$ Dec 12, 2010 at 0:56

2 Answers 2


Thanks to Timothy and Yuan for pointing out how to factor it. Despite my best efforts, I have never, ever, ever, ever been able to magically see factors appear the way everybody else seems to do so immediately. Sigh.

So that gives

$$ \begin{aligned} xy + x = 2y \\ x = 2y - xy \\ x = y( 2 - x ) \\ y = \frac{x}{2-x} \\ \end{aligned} $$

which is the same as the answer given by Wolfram. All is right with the world again.


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    $\begingroup$ It might help to plug in an actual number for x. Experienced manipulators learn to "see x" or "see y" as the variable and everything else as constants to move around, but often nobody explicitly teaches you how to do this, and actually setting all the other variables to concrete numbers is a good first step. $\endgroup$ Dec 12, 2010 at 1:38

Another way to arrive at the conclusion, if you don't "see" factors jumping out at you is this: When you see a fraction like $\frac{2y}{y+1}$, I am sure you feel the urge to split into summands. Now, of course, you cannot do it with the sum in the denominator, so that suggests inverting the whole thing (assuming that everything that you want to be non-zero is non-zero): $$ \frac{1}{x} = \frac{y+1}{2y} = \frac{y}{2y} + \frac{1}{2y} = \frac{1}{2} + \frac{1}{2y}. $$ Now subtract 1/2 and invert back again to get the result.


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