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Try clearing the fractions by multiplying both sides by (b+1)(b-1)
so 1 * (b+1) and 1 * (b-1)? @myininaya
So (b-1)+(b+1)=2

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Other answers:

the b+1 will cancel in the first fraction the b-1 will cancel in the second fraction in the last fraction the b^2-1 will cancel because (b-1)(b+1) is the same as b^2-1
Recall b^2-1=(b-1)(b+1)
Ahh okay. So there are no solutions?
there is no solution very good. you knew that because 1 isn't in the domain of the original equation right?
It should be (b-1)+(b+1)=2
Ahh okay thanks. Changed it. Anything else?
I think it looks fine.

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