How to solve for $x$ in $\frac{200}{x+10} = \frac{200}{x} -1$?

I am really confused how to calculate the following. The answer is $40$, but what are the steps to get $40$? $$\frac{200}{x+10} = \frac{200}{x} - 1.$$ Thank you.

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I had no intentions to dig up this thread; I came to this page via related thread while reading another question; The algebra tag was obsolete and 200/x still seems odd in my eyes so without much though I edited this, I didn't noticed the asking date before (which I should have noticed). I apologize for any if anybody become frustrated due to this action. Thanks. –  Quixotic Dec 8 '11 at 13:02

Hint: Just multiply everything by $x(x+10)$ to clear the denominators. Then solve the quadratic equation you get. There are two possible answers...

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\begin{align}{200 \over x + 10} &= {200 - x \over x} \\ \\ \\ 200x &=(x + 10)(200 - x) \\ \\ \\ 200x&=-x^2 + 190x + 2000 \\ \\ \\ x^2 + 10x - 2000&=0\end{align}Solve the quadratic equation and you get your solutions.

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We have our equation: $$\frac{200}{x+10}=\frac{200}{x}-1$$ First, multiply both sides by $x(x+10)$. $$x(x-10)\left(\frac{200}{x+10}\right)=x(x+10)\left(\frac{200}{x}-1\right)$$ $$200x=200(x+10)-x(x+10)$$ $$200x=200x+2000-x^2+-10x$$ $$x^2+200x-200x-2000+10x=0$$ $$x^2+10x-2000=0$$ $$(x-40)(x+50)=0$$ $$x=40, \ -50$$ Therefore $x=40, \ -50$

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What exactly was the purpose of necrobumping this post? By the way, it's $x+10$ in the denominator, not $x-10$. You had a convenient algebra mistake where you turned $x^2 - 10x + 2000$ (which has no real solutions) into $x^2 + 10x - 2000$. –  Soke Mar 18 at 3:25
@user92774 First, it showed up on the "Related" list next to a similar question I answered, and I have to say I did not look at the time this was posted –  JChau Mar 18 at 5:15