[MissKitten]

Okay, I'm in precalc/trig, so you'd think I'd know how to do this, considering it's algebra II. Were reviewing Alg II this semester, so we have an exponents review thing that were doing. The only problem is, I kinda forgot how to do this. Can you guys point me towards the right direction in solving the following problems? Thanks!

(3+(a^-1))^-1

((a^-1)--(b^-1))^-1

Sorry for the confusing exponents!

[Thalia]

Not completely sure what you are asking for. Do these equal 0 or something? Do you want to solve for a or b?

Do these look right?

(3+a^-1)^-1

(a^-1+b^-1)^-1

If you want to get rid of the exponents, you can do the first one like this: The -1 exponent around the entire expression means 1 over the expression. So you have 1/(3+a^-1). The -1 exponent with the a means you have to flip that one. So you get 1/[3+(1/a)].

Doing the same thing to the 2nd one gives you 1/[(1/a)+(1/b)].

If you find a common denominator for the bottom part, you can clean them up a bit. Not convinced that was what you were looking for so please clarify what you are trying to do.

[MissKitten]

Ah, sorry about that. Yeah, we need to simplify as much as possible, and completely get rid of the exponents.

[Thalia]

OK. So for the first one, you can multiply the 3 by a/a and combine. That should give you 1/[(3a+1)/a]. a gets flipped up to the top. That should give you a/(3a+1).

For the second one, you should manipulate the bottom of the fraction so that the denominator is ab. That gets flipped up to the top. ab/(a+b). Does that help?

EDIT: Err. That would be after doing what I put in the previous post.

Edited January 17, 2012 by Thalia

[MissKitten]

Yeah, that does. Thanks so much!

[Thalia]

No problem.