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Old 05-12-2008, 09:32 AM
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Default Reduction math problem
Hi, I'm trying to solve this very simple reduction problem, but I just can't get my head around it.

Here it goes:

xy + x^2 / xy

It's supposed to be reduced, and my best bet is that the solution is
x/y but I'm not sure at all..

Any help appreciated!
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Old 05-12-2008, 09:38 AM
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xy + x^2 / xy

I will assume that you mean xy + ((x^2) / xy)

If that is the case, xy + x/y -----> x(y + (1/y)) this last one not as nice..

Is this what you wanted? Explain please.

-Andy
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Old 05-12-2008, 09:55 AM
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Originally Posted by David T View Post
Hi, I'm trying to solve this very simple reduction problem, but I just can't get my head around it.

Here it goes:

xy + x^2 / xy

It's supposed to be reduced, and my best bet is that the solution is
x/y but I'm not sure at all..

Any help appreciated!
xy+\frac{x^2}{xy}\Rightarrow{xy+\frac{x}{y}}

Combining fractions we get \frac{xy^2+x}{y}=\frac{x(y^2+1)}{y}
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Old 05-12-2008, 10:14 AM
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Originally Posted by David T View Post
Hi, I'm trying to solve this very simple reduction problem, but I just can't get my head around it.

Here it goes:

xy + x^2 / xy

It's supposed to be reduced, and my best bet is that the solution is
x/y but I'm not sure at all..

Any help appreciated!
Ok, I won't do it the way the other ones did it

\frac{xy+x^2}{xy}=\frac{x(y+x)}{xy}=\frac{y+x}{y}=1+\frac xy

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\int \limits_{0}^{+\infty} {\cos\left(t^2\right) \ dt}=\sqrt{\frac{\pi}{8}} \quad \text{ and } \quad \int \limits_{-\infty}^{+\infty} {e^{-t^2} \cdot \cos( \alpha t) \ dt}=\sqrt{\pi} \cdot e^{-\alpha^2/4}
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Old 05-12-2008, 10:44 AM
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Originally Posted by Moo View Post
Ok, I won't do it the way the other ones did it

\frac{xy+x^2}{xy}=\frac{x(y+x)}{xy}=\frac{y+x}{y}=1+\frac xy
You know what? You are actually probably right in guessing that is what the poster meant. Sharp eyes
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Old 05-12-2008, 10:46 AM
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You know what? You are actually probably right in guessing that is what the poster meant. Sharp eyes
It's not me, it's you who look for complicated things

However, I agree on one point : the OP didn't write the text clearly...
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shinhidora production

\int \limits_{0}^{+\infty} {\cos\left(t^2\right) \ dt}=\sqrt{\frac{\pi}{8}} \quad \text{ and } \quad \int \limits_{-\infty}^{+\infty} {e^{-t^2} \cdot \cos( \alpha t) \ dt}=\sqrt{\pi} \cdot e^{-\alpha^2/4}
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Old 05-13-2008, 07:21 AM
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Thanks for all the input guys!

Moo your answer was what I was looking for, thanks a lot!
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