Still stuck...

Mercury3190

New member
Joined
Jan 30, 2006
Messages
9
Thanks for the help there but I don't know how to apply it to my problem, with pluses and minuses and that -1...
Could you explain a bit more please? :?
 
This one:

Hard to explain, but in words: [(1 over a) + (1 over b)] that whole thing divided by: [(a^2) over (b^2)] with a minus 1 next to that...
It looks like:

1 + 1
-- --
a b
----------
a^2
---- -1
b^2

I need to express it in simplest form, and when I asked my teacher I was told the answer was:

b over [(a^2) - (ab)], or: b
--------
(a^2) - ab

I got a little help earlier today, but it didn't advance me at all (I tried multiplying the top and bottom together to get rid of fractions over fractions - is that even how it's done?)

Now I need to show the work but I don't understand this at all...
Please show me the steps (as in step by step becuase I'm so stuck) to this process so I know from now on. :D
 
Unfortunately, the problem, the way you have posted it, is very unclear. I can't make heads or tails of it.
 
Hmmm

Well, if the box with a sample problem is there like it should be, put [(1/a)+(1/b)]/[[(a^2)/(b^2)]-1] in place of the sample. When it "simplifies" the equation, you'll see in a nicer format the original equation (as well as the answer - but not how to get that answer).
So are you able to see my exact problem?
If not, here goes:
It is like a double fraction (fraction(s) in denominator, fraction(s) in numerator as well).
On the top half it reads like this:
(1 over a) plus (1 over b)
Then there is another division line (separating the fractions).
On the bottom:
[(a squared) over (b squared)] minus 1 (could be writtn as (-1 over 1))
That's the whole problem.
I just need to simplify it - that site says the answer will be:
"b" over (a squared) minus ab.
It looks almost like a square shaped problem when you write it down.
Does that description help?[/img]
 
happy said:
Unfortunately, the problem, the way you have posted it, is very unclear. I can't make heads or tails of it.
It is clarified in the other thread.

Eliz.
 
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