Help for a medal???

- anonymous

Help for a medal???

- katieb

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- anonymous

Charlie needs to simplify the expression below before he substitutes values for x and y.
If x ≠ 0 and y ≠ 0, which of the following is a simplified version of the expression below?
x^9y^5
x^24y^16
x^6y^3 + x^3y^2
x^15y^8 + x^6y^4

##### 1 Attachment

- anonymous

@zepdrix Can you help?

- zepdrix

\[\Large\rm \frac{x^{18}y^{12}+x^9y^8}{x^3y^4}\]So um...
we have three x's multiplying in the bottom.
We'd like to find three x's in the top also, so we can cancel some stuff out.
But to do that we need to take the x's out of each term in the top.
Do you remember how to factor? :)
How do we take three x's out of each term in the top?\[\Large\rm x^{18}y^{12}+x^9y^8=x^3(\quad ?\quad+\quad?\quad)\]

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## More answers

- zepdrix

Hannah Bananaaaaa +_+ where you at girrrrrr

- anonymous

Sorry D: Pizza comes first math comes second! Do I add the Xs together?

- zepdrix

Ooo pizza noice :U
no.
You're dividing some x's out of each term.|dw:1434569099407:dw|

- anonymous

Ooooh, okay.

- anonymous

So would you divide x^3 by x^9 and x^18?

- zepdrix

mhm :o

- anonymous

Alright, give me a sec

- anonymous

x^6y^12 + x^3y^8

- zepdrix

See the first term, you're multiplying 18 x's together.
If you pull 3 of them out, then you're only multiplying 15 x's together, ya?
I don't think you wanted to divide the 18 by 3 silly :)

- anonymous

Oooooooooooooooooooohhhhhh. Alright, gotcha!

- zepdrix

So let's see what pulling out the x^3 does for us:\[\Large\rm \frac{x^{18}y^{12}+x^9y^8}{x^3y^4}=\frac{\color{orangered}{x^3}(x^{15}y^{12}+x^3y^{8})}{\color{orangered}{x^3}y^4}\]Ooo notice now that we have some matching x's on top and bottom!
So we can cancel out that junk.\[\Large\rm \frac{\cancel{\color{orangered}{x^3}}(x^{15}y^{12}+x^3y^{8})}{\cancel{\color{orangered}{x^3}}y^4}=\frac{x^{15}y^{12}+x^3y^8}{y^4}\]

- zepdrix

We'll have to do a similar step to deal with the y's.

- zepdrix

Any confusion on that step Bananz? :o
You have pizza in your hand don't you -_- lol

- anonymous

RIGHT

- zepdrix

prolly cheese everywhere -_-

- zepdrix

Since we have y^4 in the bottom, we'd like to try and pull out y^4 from each term in top.

- anonymous

Haha pizza is my life :D

- anonymous

Wait, what do you mean pull the y^4 out? Do you just get rid of it? Or do you gotta divide the y^4 by other Y's?

- anonymous

Or is it something else?

- zepdrix

yes, divide the other y's by y^4.

- anonymous

Alright

- anonymous

\[(x ^{15} y ^{3} + x^3 y^2) = x ^{15} y^{3} + x^3 y^2\]

- anonymous

That took a LONG TIME to type using the equation thing.

- zepdrix

woops! you did it again silly.
when you divide y's like that, you need to `subtract` the exponents.\[\Large\rm \frac{y^{12}}{y^4}=y^{12-4}=y^8\]

- anonymous

OOOOOOOOH

- anonymous

Alright, I'll do it again

- anonymous

\[(x^{15} y^{12} +x^3 y^{4}) = x^{15} y^{8} + x^3 y^4\]

- anonymous

OMG I FORGOT ABOUT THE y^12

- anonymous

This has proved difficult doing while eating pizza

- zepdrix

careful the way you're writing that +_+
don't forget about the y^4 out front.\[\Large\rm (x^{15} y^{12} +x^3 y^{4}) = \color{orangered}{y^4}(x^{15} y^{8} + x^3 y^4)\]Your work looks correct though :)

- zepdrix

\[\Large\rm \frac{x^{15}y^{12}+x^3y^8}{y^4}=\frac{\color{orangered}{y^4}(x^{15}y^{8} + x^3y^4)}{\color{orangered}{y^4}}\]

- anonymous

Haha, sorry about that XD

- zepdrix

Cool, so do you see which option you'll end up with? :)

- anonymous

D?

- anonymous

I dedicate this song to you https://www.youtube.com/watch?v=TjcOJmoJwpk skip to the 0:26 second mark please

- zepdrix

yay good job \c:/

- zepdrix

lolol -_-

- anonymous

I know cliche :D But still, thank you!

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