2
$\begingroup$

$ 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?

1 Answers 1

3

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 $y=\frac{\left(20+\frac{x}{100}\right)\pm\sqrt{\left(20+\frac{x}{100}\right)^2-\frac{2x}{5}}}{2}=\frac{1}{200}\left(2000+x\pm\sqrt{(x+2000)^2-4000x}\right)=$ $\frac{1}{200}\left(2000+x\pm\sqrt{x^2+4000000}\right)$

  • 0
    Tremendously helpful. Thanks a lot to both of you for your time.2011-10-10