Someone please help!

- anonymous

Someone please help!

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- anonymous

With what?

- anonymous

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- anonymous

Just wondering are you in connections?

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- anonymous

Yes I am

- anonymous

I am as well. I had trouble on a question like this ask @johnweldon1993 he'll probably be more help here.

- anonymous

thanks

- johnweldon1993

So we first start with the restrictions...they can come up at ANY point of our simplification process..
So restrictions here would be whatever 'x' values make the denominator = 0 (in either fraction)
what would those be?

- johnweldon1993

Hint would be to factor the denominators :)

- anonymous

oh ok give me a sec

- anonymous

ok so the first one factored would be (x+4)(x-1) and I have to find what x would be to make this equal 0?

- johnweldon1993

Correct

- anonymous

it would be -4 and 1

- johnweldon1993

Okay great...2 restrictions down so far
But...what about the other fraction now?

- anonymous

so you said I just have to do it for the denominator?

- johnweldon1993

mmhmm, we know dividing anything by 0 makes for an undefined answer...so whenever denominators = 0 we have a restriction

- anonymous

alright so for this one it would be -4 and -1

- johnweldon1993

That is correct...and yet I dont see any answer choices with a -1 as a restriction...odd...but oh well, aybe they just missed one
Okay so now we begin to simplify...how would we begin?

- anonymous

none of my answer look right with this... are you sure this is how you do it?

- anonymous

the restrictions should be -4, 1, -1 but those aren't in any answers

- johnweldon1993

Yup. Find restrictions...then simplify..and I know thats why I was confused as to why the -1 wasnt there
And we also will have more restrictions in a little bit

- anonymous

ok well how do I start to simplify?

- johnweldon1993

Well the first thing I would do....is turn this into multiplication...we would do that by flipping the second fraction
\[\large \frac{5-x}{x^2 + 3x - 4} \times \frac{x^2 + 5x + 4}{x^2 - 2x - 15}\]
make sense?

- anonymous

\[\frac{ 5-x }{ (x+4)(x-1) } \ \times \frac{ (x+4)(x+1) }{ (x+3)(x-5) } \]

- anonymous

so like this?

- johnweldon1993

Perfect...and NOW we simplified...we need to check for any more restrictions!!

- anonymous

ok so -3 and 5?

- johnweldon1993

Right...so literally in total...we SHOULD have 5 restrictions on the problem...but they only give us 4

- anonymous

huh ok

- anonymous

so its B?

- johnweldon1993

It would be ye...you simplified and got that right?

- anonymous

well its the only one with the correct restrictions. and yes I simplified to the point of knowing it couldn't be the other ones

- anonymous

thank you so much ^.^ a medal and a hug for you *hug* lol

- johnweldon1993

Well thank you but the hug was the best part! :P lol

- anonymous

haha well you're honestly the first I've ever given a hug for helping me :P ^.^

- johnweldon1993

Woo! I get the special hug!!! lol :P I'll treasure it always haha

- anonymous

haha good xD

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