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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
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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0To double check, is this: \[ \frac{b+17}{b^21}  \frac{1}{b+1} = \frac{b2}{b1}? \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well I think that I need to factor the (b^2 1) to (b+1)(b1) but I am not sure

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That's a good first step. So to get rid of that denominator, you'd ideally multiply \(\frac{b+17}{b^21}\) by \((b+1)(b1)\). 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 (\(\frac{b2}{b1}\)) by that. What would happen if you did that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it would make the (b1) to be something like b^3  1 ???

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So: \[ \begin{align} \frac{b+17}{b^21}\cdot(b+1)(b1) &= b+17\\ \frac{1}{b+1}\cdot(b+1)(b1) &= ??\\ \frac{b2}{b1}\cdot(b+1)(b1) &= ?? \end{align} \] What should replace the question marks?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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 / b1 I think but that doesnt make sense to me

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Haha. 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 \(b1\) terms. If you do that, what do you get?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Exactly! So if you multiply the entire equation by \((b+1)(b1)\), you would get: \[ (b+17)  (b1) = (b2)(b+1) \] 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So what's the quadratic equation you get at the end?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am probably wrong ( way wrong) but I am coming up with b^216b+17 = b^2b2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay so... 18 =b^2b2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Awesome! And if you move everything to one side of the equation, you get: \[ b^2b20 = 0 \] Does that look more like an equation you can solve?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0possibly... the answer that I am coming up with would be b=\[(\sqrt{20+b}) \] which isnt exactly seem like it is solved...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do you know how to factor quadratic equations?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wow now I really feel like an idiot... it has been one of those days!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Exactly! 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, if \(b=0\), then \(b+1 = 1\), \(b1=1\), and \(b^21=1\), so that wouldn't be a problem, right? So the only values \(b\) can't equal are 1 and 1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thank you so much, I really appreciate it!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No problem! Glad it helped!
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