anonymous
  • anonymous
j
Mathematics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.