1. anonymous

With what?

2. anonymous

3. anonymous

Just wondering are you in connections?

4. anonymous

Yes I am

5. anonymous

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

6. anonymous

thanks

7. 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?

8. johnweldon1993

Hint would be to factor the denominators :)

9. anonymous

oh ok give me a sec

10. 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?

11. johnweldon1993

Correct

12. anonymous

it would be -4 and 1

13. johnweldon1993

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

14. anonymous

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

15. johnweldon1993

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

16. anonymous

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

17. 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?

18. anonymous

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

19. anonymous

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

20. 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

21. anonymous

ok well how do I start to simplify?

22. 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?

23. anonymous

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

24. anonymous

so like this?

25. johnweldon1993

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

26. anonymous

ok so -3 and 5?

27. johnweldon1993

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

28. anonymous

huh ok

29. anonymous

so its B?

30. johnweldon1993

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

31. 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

32. anonymous

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

33. johnweldon1993

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

34. anonymous

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

35. johnweldon1993

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

36. anonymous

haha good xD