## Cancelling factors with multiplication problem

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melmel
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Joined: Wed Jun 01, 2016 3:53 pm
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### Cancelling factors with multiplication problem

Hello!
I'm having a little trouble understanding why I can't factor certain groups in this equation:

Original eq.: x^2+2xy+y^2/x^2-2xy+y^2 multiplied by (x-y)/5x+5y

I factored: (x+y)^2/(x-y)^2 multiplied by (x-y)/5(x+y)

Here is the trouble, I want to cancel out the exponents, and both pairs of the (x+y)s and (x-y) and only have the 5 remaining but that is not correct. The correct answer is: (x-y)/5(x+y) and I don't understand how. I hope this makes sense!

Mel

FWT
Posts: 153
Joined: Sat Feb 28, 2009 8:53 pm

### Re: Cancelling factors with multiplication problem

I'm having a little trouble understanding why I can't factor certain groups in this equation:

Original eq.: x^2+2xy+y^2/x^2-2xy+y^2 multiplied by (x-y)/5x+5y
What you've posted means this

$\left(x^2\, +\, 2xy\, +\, \dfrac{y^2}{x^2}\, -\, 2xy\, +\, y^2\right)\, \times\, \left(\dfrac{x\, -\, y}{5x}\, +\, 5y\right)$

Based on your factoring, I think you mean this

$\left(\, \dfrac{x^2\, +\, 2xy\, +\, y^2}{x^2\, -\, 2xy\, +\, y^2}\, \right)\, \times\, \left(\, \dfrac{x\, -\, y}{5x\, +\, 5y}\,\right)$
I factored: (x+y)^2/(x-y)^2 multiplied by (x-y)/5(x+y)

Here is the trouble, I want to cancel out the exponents....
What do you mean by "cancelling out the exponents"? I mean, you can do that if you're simplifying stuff with the same base, like:

$\dfrac{a^3}{a^5}\, =\, \dfrac{1}{a^{5-3}}\, =\, \dfrac{1}{a^2}$

But you can't do it with different bases, like

$\dfrac{a^3}{b^5}$

Are you saying that you think that the quotient of the squares

$\dfrac{(x\, +\, y)^2}{(x\, -\, y)^2}\, =\, \left(\dfrac{x\, +\, y}{x\, -\, y}\right)^2$

is the same as the quotient by itself?

$\dfrac{x\, +\, y}{x\, -\, y}$

This would be like saying that

$\dfrac{9}{16}$

equals

$\dfrac{3}{4}$

but it doesn't.