Question

WILL GIVE MEDAL
Question below (in comments)

Answer 1

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

Answer 2

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

Answer 3

What is Keep Change Flip?

Answer 4

I wasn't taught that :(

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

Am I doing it right so far?

Answer 9

\[\large\rm \frac{\left(\frac{ 18x-6 }{ 9x^5 }\right)}{\left(\frac{15x+5}{21x^2}\right)}\quad=\frac{18x-6}{9x^5}\cdot\frac{21x^2}{15x+5}\]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.

Answer 10

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

Answer 11

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.

Answer 12

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

Answer 13

can i have a medal

Answer 14

Okay. I think that everything is simplified.

Answer 15

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.

Answer 16

We can factor something out of each of them.

Answer 17

They are both divisible by three!

Answer 18

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

Answer 19

How about 6?

Answer 20

\[\large\rm \frac{\color{orangered}{6(3x-1)}}{9x^5}\cdot\frac{21x^2}{15x+5}\]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.

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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.

Answer 26

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

Answer 27

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

Answer 28

Okay, can I cancel 3x and 3x?

Answer 29

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 :)

Answer 30

Okay, what should I cancel. I'm kind of confused…

Answer 31

\[\large\rm \frac{6(3x-1)}{9x^5}\cdot\frac{21x^2}{5(3x+1)}\]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,\[\large\rm \frac{6(3x-1)}{9\cancel{x} \cdot \cancel{x} \cdot x \cdot x \cdot x}\cdot\frac{21\cancel{x}\cdot \cancel{x}}{5(3x+1)}\]

Answer 32

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

Answer 33

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

I think 6 and 9 have something in common.

Answer 34

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 \[\frac{2x(3x-1) * 7}{x^3*5x(3x+1)}.\]

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

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.

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

Correct?

Answer 44

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.\[\large\rm \frac{14(3x-1)}{5x^3(3x+1)}\]

Answer 45

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

Answer 46

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

Answer 47

XD

Answer 48

yay we did it!
np c: