WILL GIVE MEDAL
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- 0487308

WILL GIVE MEDAL
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- schrodinger

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- 0487308

\[\frac{\frac{ 18x-6 }{ 9x^5 }}{\frac{15x+5}{21x^2}}\]

- zepdrix

Remember how to deal with fraction division?
Maybe you were taught `Keep Change Flip` or something similar? :)

- 0487308

What is Keep Change Flip?

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## More answers

- 0487308

I wasn't taught that :(

- zepdrix

\[\large\rm \frac{\left(\frac{a}{b}\right)}{\left(\frac{c}{d}\right)}\quad=\frac{a}{b}\cdot\frac{d}{c}\]We `Keep` the numerator as is,
We ` Change` the operation from division to multiplication,
We `Flip` the bottom fraction.

- zepdrix

That's our first step that we'll want to apply :)

- 0487308

Wow, that's really helpful! Does that mean that \[\frac{ 18x-6 }{ 9x^5 } * \frac{ 15x + 5 }{ 21x^2 }\]

- 0487308

Am I doing it right so far?

- zepdrix

\[\large\rm \frac{\left(\frac{ 18x-6 }{ 9x^5 }\right)}{\left(\frac{15x+5}{21x^2}\right)}\quad=\frac{18x-6}{9x^5}\cdot\frac{21x^2}{15x+5}\]You applied the first two steps correctly,
`Keeping` the numerator as is,
`Changing` the operation.
It looks like you forgot to `flip` the other fraction though.

- 0487308

Oh, sorry. Thanks for the correction. Is the next step to cross-multiply?

- zepdrix

The next step is to look for cross-cancellations.
We don't want to multiply and mix everything together.
That's going to cause more trouble for us.

- zepdrix

Before looking for cancellations though,
let's make sure everything is factored down as far as it can go.

- jacob11

can i have a medal

- 0487308

Okay. I think that everything is simplified.

- zepdrix

Let's look at this first numerator a sec,\[\large\rm \frac{\color{orangered}{18x-6}}{9x^5}\cdot\frac{21x^2}{15x+5}\]Both terms share something in common.

- zepdrix

We can factor something out of each of them.

- 0487308

They are both divisible by three!

- zepdrix

True :)
I think they're divisible by a larger number though.

- 0487308

How about 6?

- zepdrix

\[\large\rm \frac{\color{orangered}{6(3x-1)}}{9x^5}\cdot\frac{21x^2}{15x+5}\]6 seems good.
Dividing 6 out of 18x leaves us with 3x for the first term,
and dividing 6 out of 6 leaves us with a 1 in the second.

- 0487308

Okay. 5 is a factor of 15x + 5.

- 0487308

So now I think I have \[\frac{ 6(3x-1) }{ 9x^5 } â€˘ \frac{ 21x^2 }{ 5(3x + 1) }\]

- zepdrix

Cool, everything is factored all the way,
now we can proceed to look for cancellations.

- 0487308

Okay. Just to verify, is a cross-cancellation factoring something out that is common in both of the 'corners' of the equation?

- zepdrix

No, I should have been more clear about that.
Cross or bottom/top cancel is acceptable.
Keep in mind that we could multiply across and have this:\[\large\rm \frac{6(3x-1)}{9x^5}\cdot\frac{21x^2}{5(3x+1)}\quad= \frac{6(3x-1)\cdot21x^2}{9x^5\cdot 5(3x+1)}\]And now it's just one big numerator on top of one big denominator,
so maybe it's a little easier to see why we're allowed to do normal cancellation, but also cross-cancellation.

- 0487308

Okay, should I multiply before canceling, or should I immediately cancel?

- zepdrix

Immediately cancel :)
Otherwise you're just giving yourself more work.

- 0487308

Okay, can I cancel 3x and 3x?

- zepdrix

No no no.
3x is grouped with -1 in the numerator,
3x is grouped with +1 in the denominator,
(3x-1) and (3x+1) are not the same, they can not be cancelled out unfortunately.
And to answer your question directly, no we can't magically take the 3x out of the brackets :)

- 0487308

Okay, what should I cancel. I'm kind of confusedâ€¦

- zepdrix

\[\large\rm \frac{6(3x-1)}{9x^5}\cdot\frac{21x^2}{5(3x+1)}\]Well I notice that the top has a couple x's being multiplied together,
while the denominator has 5 of them, ya?\[\large\rm \frac{6(3x-1)}{9x \cdot x \cdot x \cdot x \cdot x}\cdot\frac{21x\cdot x}{5(3x+1)}\]So we can cancel out ... two of them from the top and bottom,\[\large\rm \frac{6(3x-1)}{9\cancel{x} \cdot \cancel{x} \cdot x \cdot x \cdot x}\cdot\frac{21\cancel{x}\cdot \cancel{x}}{5(3x+1)}\]

- 0487308

Okay, thanks. Now I believe that I have \[\frac{ 6x(3x-1) * 21 }{ 9x^3 * 5x(3x+1) }\]

- zepdrix

\[\large\rm \frac{6x(3x-1)\cdot 21}{9x^3\cdot5x(3x+1)}\]K looks good.
What else?
I think 6 and 9 have something in common.

- 0487308

6 and 9 have 3 in common, so I have \[\frac{2x(3x-1) * 21}{3x^3*5x(3x+1)}.\]Next, 21 and 3 have 7 in common, so I factor 7 out and have \[\frac{2x(3x-1) * 7}{x^3*5x(3x+1)}.\]

- zepdrix

Woops, 21 and 3 have 3 in common.
You misspoke, but you applied the step correctly, taking the 3 out.

- zepdrix

Looks like there is oooone more thing we can cancel out.
See it?

- zepdrix

Oh wait wait, I think we made a boo boo somewhere.

- 0487308

Yes, 2x and x^3 have x in common, right?

- zepdrix

We had this:\[\large\rm \frac{6(3x-1)}{9x^5}\cdot\frac{21x^2}{5(3x+1)}\]There shouldn't be these red x's,\[\large\rm \frac{6\color{red}{x}(3x-1)}{9x^5}\cdot\frac{21x^2}{5\color{red}{x}(3x+1)}\]I'm not sure where those came from.

- zepdrix

So your problem should look like this right now:\[\large\rm \frac{2(3x-1)\cdot7}{x^3\cdot5(3x+1)}.\]

- 0487308

Oh yeah, I think that I added those out of no where by accident, sorry :P

- 0487308

Now we have nothing to cancel out on, so we multiply, I think, so the problem should look like this: \[\frac{ 7(6x-2) }{x^3(15x+5)}\]

- 0487308

Correct?

- zepdrix

Yes good :)
But the final answer might look a little better if we leave it fully factored.
So don't distribute the 2 back into the brackets, and the same with the 5.\[\large\rm \frac{14(3x-1)}{5x^3(3x+1)}\]

- zepdrix

I multiplied the 2 and the 7.
Multiplication is `commutative`, so you're allowed to rearrange the multiplication in that way.

- 0487308

Okay, thanks for your patience with me! Please never become a qualified helper, or I'll never be able to ask you questions :)

- zepdrix

XD

- zepdrix

yay we did it!
np c:

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