Question

Can you help? @zepdrix

r^2-4r+3/8-7r-r^2 divided by 3r+15/r+8

Answer 1

\[\Large\rm \frac{r^2-4r+3}{8-7r-r^2}\div\frac{3r+15}{r+8}\]The restrictions are going to be a bit more difficult since we have division :)
It will result in more restrictions on our variable.

Let's start this problem a little different though, let's fully factor all of the pieces first.
Top left what two factors do you get?
You know how to factor all of these?

Answer 2

Yes. The top left I got (x-3)(x-1)/-1(x+8)(x-1)

Answer 3

\[\Large\rm \frac{(r-3)(r-1)}{-(r+8)(r-1)}\div\frac{3r+15}{r+8}\]Ok looks good so far!

Answer 4

Okayy. I factored the other side and got 3(r+5)/R+8

Answer 5

\[\Large\rm \frac{(r-3)(r-1)}{-(r+8)(r-1)}\div\frac{3(r+5)}{(r+8)}\]K great, fully factored.

Answer 6

Imagine that I hadn't written this with the ugly division symbol.
But that I had written it as a fraction over a fraction.\[\Large\rm \frac{\left(\frac{a}{b}\right)}{\left(\frac{c}{d}\right)}\]Remember we can't have a zero in the BOTTOM of a fraction.
Whether it's an inner fraction, or the outermost one, it just can't be allowed.
So, in my little example here, which letters would cause a problem for us?

Answer 7

D?

Answer 8

D is a denominator, yes D is a problem area.
But also B, yes? B is the bottom of the top fraction, ya?

Answer 9

There is also another that might not seem as obvious.
What happens when C=0?

C/D = 0/D = 0. But then we have (a/b)/(0). Division by 0 again! bad bad bad.
So in our last problem, we had to deal with two bad places, two denominators.
Here, when we divide like this, it gives us three bad places.
Lemme relate it to our problem by putting the fractions back into the form we're using,\[\Large\rm \frac{a}{b}\div\frac{c}{d}\]We run into a problem when we get any of these situations:\[\Large\rm \frac{a}{0}\div\frac{c}{d},\quad\qquad \frac{a}{b}\div\frac{c}{0},\qquad\quad \frac{a}{b}\div\frac{0}{d}\]

Answer 10

In our last problem,\[\Large\rm \frac{r^2-4r+3}{\color{orangered}{8-7r-r^2}}\div\frac{\color{royalblue}{3r+15}}{\color{orangered}{r+8}}\]The orange were giving us our restrictions.
For this problem, the blue will also give us another set of restrictions.

Answer 11

Woops I should have colored the factored one :\ my bad

Answer 12

Okayy, so i set those to zero and solve just like in multiplication?

Answer 13

Yes :)

Answer 14

I got x cannot equal 1 or -8

Answer 15

Ok, but you forgot some! :o

Answer 16

Not just the orange in this problem, you gotta deal with the blue also.
Set that one equal to zero

Answer 17

Well I thought once you have everything factored you have to multiply be the reciprocal of the 2nd fraction. that puts the 3(x+5) on the bottom and it cannot cancel with anything right??

Answer 18

Don't worry about cancelling yet! :)
Yes, you'll have to apply your `keep change flip` so you can cancel and all that.
But we're not to that point yet.

You didn't find all of your restrictions.

Answer 19

-5 as well then

Answer 20

Mmmm ok good!
\(\Large\rm x\ne-8,\quad x\ne-5,\quad x\ne1\)

And our equation:\[\Large\rm \frac{(r-3)(r-1)}{-(r+8)(r-1)}\div\frac{3(r+5)}{(r+8)}\]becomes,\[\Large\rm \frac{(r-3)(r-1)}{-(r+8)(r-1)}\cdot\frac{(r+8)}{3(r+5)}\]

Answer 21

And you're right, it looks like the one with the -3 and the 5 don't cancel with anything :)

Answer 22

Final Answer:\[\Large\rm -\frac{(r-3)}{3(r+5)},\qquad r\ne-8,-5,~1\]Yay good job team \c:/