Question

How to solve 1/(b+1) + 1/(b-1) = 2/(b^2-1) ?

Answer 1

\[\frac{ 1 }{ b+1 } + \frac{ 1 }{ b-1 } = \frac{ 2 }{ b ^{2} -1}\]

Answer 2

you could multiply both sides (all terms) by b^2 -1
remember that b^2-1 can be written as (b-1)(b+1)
so some things will cancel

what do you get ?

Answer 3

Wouldn't it just be 1(b-1)/(b+1)(b-1) + 1(b+1)/(b+1)(b-1) = 2/(b+1)(b-1) ?

Answer 4

ok you found the common denominator. you can do that also

Answer 5

now add the numerators

Answer 6

1b-1+1b+1 would equal 2b ?

Answer 7

yes you get
\[ \frac{2b}{b^2-1} = \frac{2}{b^2-1} \]

Answer 8

What do you do after that? That's where I'm really confused.

Answer 9

If we multiply both sides by b^2-1 we get
\[ \cancel{b^2-1}\cdot \frac{2b}{\cancel{b^2-1}} = \cancel{b^2-1}\cdot \frac{2}{\cancel{b^2-1}} \]

Answer 10

The answer is b=1!

Answer 11

ordinarily yes, b would be 1
but in this case, we plug b=1 into the original equation, we get a divide by 0!
so this is one of those unfortunate equations that has no solution.

Answer 12

Oh okay, I understand now. Thank you so much!!