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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

Mathematics
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To double check, is this: \[ \frac{b+17}{b^2-1} - \frac{1}{b+1} = \frac{b-2}{b-1}? \]
yes.
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?

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well I think that I need to factor the (b^2 -1) to (b+1)(b-1) but I am not sure
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?
it would make the (b-1) to be something like b^3 - 1 ???
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?
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
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?
(b-1) (b-2)(b+1)
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?
I think so....
So what's the quadratic equation you get at the end?
I am probably wrong ( way wrong) but I am coming up with -b^2-16b+17 = b^2-b-2
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.
okay so... 18 =b^2-b-2
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?
possibly... the answer that I am coming up with would be b=\[(\sqrt{20+b}) \] which isnt exactly seem like it is solved...
Do you know how to factor quadratic equations?
wow now I really feel like an idiot... it has been one of those days!
so b= 5 or -4
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?
0, 1, and -1 ?
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.
thank you so much, I really appreciate it!
No problem! Glad it helped!

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