Someone please help!

Question

anonymous · Answer

With what?

anonymous · Answer

attachment

anonymous · Answer

Just wondering are you in connections?

anonymous · Answer

Yes I am

anonymous · Answer

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

anonymous · Answer

thanks

johnweldon1993 · Answer

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 · Answer

Hint would be to factor the denominators :)

anonymous · Answer

oh ok give me a sec

anonymous · Answer

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 · Answer

Correct

anonymous · Answer

it would be -4 and 1

johnweldon1993 · Answer

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

anonymous · Answer

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

johnweldon1993 · Answer

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

anonymous · Answer

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

johnweldon1993 · Answer

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 · Answer

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

anonymous · Answer

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

johnweldon1993 · Answer

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 · Answer

ok well how do I start to simplify?

johnweldon1993 · Answer

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 · Answer

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

anonymous · Answer

so like this?

johnweldon1993 · Answer

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

anonymous · Answer

ok so -3 and 5?

johnweldon1993 · Answer

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

anonymous · Answer

huh ok

anonymous · Answer

so its B?

johnweldon1993 · Answer

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

anonymous · Answer

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 · Answer

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

johnweldon1993 · Answer

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

anonymous · Answer

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

johnweldon1993 · Answer

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

anonymous · Answer

haha good xD