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Mathematics 24 Online
OpenStudy (anonymous):

Algebra equation! simplify completely (x^2 + x - 12) / (x^2 - x - 20) divided by (3x^2 -24x + 45) / (12x^2 - 48x - 60)

OpenStudy (anonymous):

@adrynicoleb @Ahsome @amydh @bebe365 @badboy18 @ChrystalC @CoconutJJ @Chelsey2015 @ChiakiNagoya @camman I'll award a medal!

OpenStudy (anonymous):

@Luigi0210 @zepdrix please help

zepdrix (zepdrix):

\[\large\rm \frac{x^2 + x - 12}{x^2 - x - 20}\div\frac{3x^2 -24x + 45}{12x^2 - 48x - 60}\]We gotta simplify huh? :) So I guess we'll need to do some good ole factoring.

OpenStudy (anonymous):

1 ——————————— 36 • (x - 5)^3• (x + 1)

OpenStudy (anonymous):

Well, dont we first flip the second fraction and switch the sign to multiplication? I just cant remember if we change the signs within the fraction when we flip it..

OpenStudy (anonymous):

thats what i got but could be wrong

zepdrix (zepdrix):

Yes, flip is a good first step!

OpenStudy (anonymous):

@yinkim52001 Thanks so much for trying to help! But I want to learn how to do this as I am struggling with the problem (:

zepdrix (zepdrix):

\[\large\rm \frac{x^2 + x - 12}{x^2 - x - 20}\cdot\frac{12x^2 - 48x - 60}{3x^2 -24x + 45}\]

OpenStudy (anonymous):

So the signs dont change within the fraction?

zepdrix (zepdrix):

correct. no change. example:\[\large\rm \frac{\left(\frac{a}{b}\right)}{\left(\frac{c}{d}\right)}=\frac{a}{b}\cdot\frac{d}{c}\]This is how we apply this "flip" idea ^ See how nothing changed to negative? :o

OpenStudy (anonymous):

Yes! So what is the next step? I factor both denominators and both numerators?

zepdrix (zepdrix):

Yesss. Stuck on any of those steps? :)

OpenStudy (anonymous):

Not yet! Lol I have to try it out. Will you please give me a minute?

zepdrix (zepdrix):

k :)

OpenStudy (anonymous):

Okay! So I have (x-3)(x+4) / (x-4)(x+5) times 12(x^2 - 4x -5) / 3(x^2 - 8x + 15)

OpenStudy (anonymous):

I'm not sure how to further factor the right fraction because the top has a prime number (5) and the denominator is giving me an issue too..

zepdrix (zepdrix):

\[\large\rm \frac{(x-3)(x+4)}{\color{red}{(x-4)(x+5)}}\cdot\frac{12(x^2-4x-5)}{3(x^2-8x+15)}\]Woops! :O Double check the red a sec. Rest looks good though.

OpenStudy (anonymous):

I put that ;p

zepdrix (zepdrix):

ya it's incorrect! :O fix it silly!

zepdrix (zepdrix):

The middle term is negative, so the LARGER number should get the negative sign, ya? :)

OpenStudy (anonymous):

I guess it would have to be (x + 4) and (x - 5) haha

zepdrix (zepdrix):

\[\large\rm \frac{(x-3)(x+4)}{(x+4)(x-5)}\cdot\frac{12(x^2-4x-5)}{3(x^2-8x+15)}\]Ooooo kay great!

OpenStudy (anonymous):

Yes (:

zepdrix (zepdrix):

You are correct about the 5 being prime. Which means, IF IT CAN FACTOR, the only options we have will be 1 and 5, ya?

OpenStudy (anonymous):

Ohhh! Right, I didn't even think about that for some reason ;p Let me try to factor further!

OpenStudy (anonymous):

Okay so that would factor out to (x - 5) (x + 1), but I tried factoring out the denominator and I couldn't figure it out because neither (x + 5)(x - 3) nor (x + 3)(x - 5) were foiling out to be (x^2 - 8x + 15)..

OpenStudy (anonymous):

WOW!

OpenStudy (anonymous):

Just realized they could noth be negative... Okat my bad

OpenStudy (anonymous):

both* okay*

zepdrix (zepdrix):

ya there ya go :) same sign since the 15 is positive

OpenStudy (anonymous):

Right! So let me see what I can do this far & I'll come back with my answer or if I get stuck ;p

zepdrix (zepdrix):

k brb gonna make a sammich c:

OpenStudy (anonymous):

No problem ;p

OpenStudy (anonymous):

Got it!!! It's 4(x+1) / x - 5 Thank you so so much! <3

zepdrix (zepdrix):

yayyy good job \c:/

OpenStudy (anonymous):

Thaaank you (:

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