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Old May 12th, 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 May 12th, 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 May 12th, 2008, 09:55 AM
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Quote:
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 May 12th, 2008, 10:14 AM
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Quote:
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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Old May 12th, 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 May 12th, 2008, 10:46 AM
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Originally Posted by Mathstud28 View Post
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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Old May 13th, 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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