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$$ 100\left(y + \frac{100}{10-y}\right) = x + 1000 $$

I'm looking to have the equation in the form $y(x) = ...$ Seems pretty simple, but I'm just not seeing it.

The answer is:

$$ y(x) = \frac{1}{200}\sqrt{x^2 + 4000x} − x $$

Wolfram|Alpha returns a slightly different answer: $$ y(x) = \frac{1}{200} (\sqrt{x^2+4000000}+x+2000) $$

There must be something I'm missing. Thoughts?

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Wolfram Alpha is correct. Starting with $$100\left(y + \frac{100}{10-y}\right) = x + 1000$$ and dividing both sides by $100$: $$y+\frac{100}{10-y}=\frac{x}{100}+10$$ multiplying both sides by $(10-y)$: $$-y^2+10y+100=\left(\frac{x}{100}+10\right)(10-y)$$ collecting everything on one side: $$y^2-\left(20+\frac{x}{100}\right)y+\frac{x}{10}=0$$ and then applying the quadratic formula gives $$y=\frac{\left(20+\frac{x}{100}\right)\pm\sqrt{\left(20+\frac{x}{100}\right)^2-\frac{2x}{5}}}{2}=\frac{1}{200}\left(2000+x\pm\sqrt{(x+2000)^2-4000x}\right)=$$ $$\frac{1}{200}\left(2000+x\pm\sqrt{x^2+4000000}\right)$$

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    Thanks! Clever, I didn't realize one could us the quadratic formula for this purpose. Makes sense.2011-10-10
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    My textbook lists the first solution as the correct solution. Any chance you see how to go from the WolframAlpha solution to my textbook's solution? They don't seem equivalent to me.2011-10-10
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    @Gortaur: I'm afraid I don't see how that demonstrates they are equivalent. If we write $$\frac{1}{200}\left(2000+x\pm\sqrt{x^2+4000000}\right) = \frac{1}{200}\sqrt{x^2 + 4000x} − x$$ then we get $$2000+201x\pm\sqrt{x^2+4000000}=\sqrt{x^2+4000x}$$ and squaring both sides, $$x^2+4000000+2(2000+201x)\sqrt{x^2+4000000}+(2000+201x)^2=x^2+4000x$$ which is impossible (there is a non-rational function on one side, and a [rational function](http://en.wikipedia.org/wiki/Rational_function) on the other). So, David, this demonstrates that they are indeed *not* equivalent.2011-10-10
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    @ZevChonoles: you're right, I made a thinko in my calculations.2011-10-10
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    Tremendously helpful. Thanks a lot to both of you for your time.2011-10-10