## anonymous 5 years ago I am really unsure of how to do this, can anyone please help me! b + 17 - 1 = b – 2 ------ ------ ----- b2 – 1 b + 1 b – 1

1. anonymous

To double check, is this: $\frac{b+17}{b^2-1} - \frac{1}{b+1} = \frac{b-2}{b-1}?$

2. anonymous

yes.

3. anonymous

Well, the first thing you need to do with this kind of equation is to get rid of the denominator. So what can you multiply both sides of the equation by that will cancel out the denominator?

4. anonymous

well I think that I need to factor the (b^2 -1) to (b+1)(b-1) but I am not sure

5. anonymous

That's a good first step. So to get rid of that denominator, you'd ideally multiply $$\frac{b+17}{b^2-1}$$ by $$(b+1)(b-1)$$. But to do that, you'd have to multiply the other fraction on that side of the equation ($$b+1$$) and the other side of the equation ($$\frac{b-2}{b-1}$$) by that. What would happen if you did that?

6. anonymous

it would make the (b-1) to be something like b^3 - 1 ???

7. anonymous

So: \begin{align} \frac{b+17}{b^2-1}\cdot(b+1)(b-1) &= b+17\\ \frac{1}{b+1}\cdot(b+1)(b-1) &= ??\\ \frac{b-2}{b-1}\cdot(b+1)(b-1) &= ?? \end{align} What should replace the question marks?

8. anonymous

I feel completly stupid right now and maybe it is just because I have been trying to figure this problem out for about an hour now but... the first set would = b^2 -1 / b+1 the second set would be = b^3+2 / b-1 I think but that doesnt make sense to me

9. anonymous

Haha. So you're right, but you made life more difficult by not cancelling out things! On the first set, you can cancel out the $$b+1$$ from the numerator and denominator, and on the second set, you can cancel out the $$b-1$$ terms. If you do that, what do you get?

10. anonymous

(b-1) (b-2)(b+1)

11. anonymous

Exactly! So if you multiply the entire equation by $$(b+1)(b-1)$$, you would get: $(b+17) - (b-1) = (b-2)(b+1)$ At this point you can start using FOIL to multiply things out, then add like terms to get a quadratic equation. Does that make sense?

12. anonymous

I think so....

13. anonymous

So what's the quadratic equation you get at the end?

14. anonymous

I am probably wrong ( way wrong) but I am coming up with -b^2-16b+17 = b^2-b-2

15. anonymous

Well, the right side of the equation is correct, but you seem to be multiplying the two terms on the left side when you should just be subtracting the second from the first.

16. anonymous

okay so... 18 =b^2-b-2

17. anonymous

Awesome! And if you move everything to one side of the equation, you get: $b^2-b-20 = 0$ Does that look more like an equation you can solve?

18. anonymous

possibly... the answer that I am coming up with would be b=$(\sqrt{20+b})$ which isnt exactly seem like it is solved...

19. anonymous

Do you know how to factor quadratic equations?

20. anonymous

wow now I really feel like an idiot... it has been one of those days!

21. anonymous

so b= 5 or -4

22. anonymous

Exactly! The one thing you have to keep in mind is that once you get a solution you need to also state that b can't be any value that would make the original denominators equal 0. What values are those?

23. anonymous

0, 1, and -1 ?

24. anonymous

Well, if $$b=0$$, then $$b+1 = 1$$, $$b-1=-1$$, and $$b^2-1=-1$$, so that wouldn't be a problem, right? So the only values $$b$$ can't equal are 1 and -1.

25. anonymous

thank you so much, I really appreciate it!

26. anonymous