Fractional Division (ab/(a-b)^2 all over 1/(a-b)

[spwittbold]

Can't seem to quite grasp how the book is coming to it's solution:

Given:

[ab/(a-b)^2) divded by [1/a-b) as the problem.

Multiply by recipricol instead, so......

ab / (a-b)^2 x [(a-b) / 1]

First thing I did was expand the denomoniator of the 1st equation as follows

[ab/(a-b)(a-b)] x [(a-b)/1]

I thought first I could eliminate one set of a-b in the denominaotr of equation one and the numerator of the 2nd equation to end up with this:

ab/(a-b) x 1/1 ....


but the solution given in the book is (a-b)/ba....i don't get it?

 

[ksdhart2]

Okay, just so we're all on the same page, I'll summarize what I think your steps are showing. You started with:

\(\displaystyle \dfrac{ab}{(a-b)^2} \div \dfrac{1}{a-b}\)

You then transformed this into a multiplication problem by using the reciprocal of the second term:

\(\displaystyle \dfrac{ab}{(a-b)^2} \cdot \dfrac{a-b}{1}=\dfrac{ab}{(a-b)^2} \cdot (a-b)\)

Then you expanded out the denominator and cancelled one of the (a-b) terms, leaving:

\(\displaystyle \dfrac{ab}{a-b}\)

If the above is correct, then I concur with Denis that you have the right answer. I believe your textbook to be mistaken.

 

[mmm4444bot]

Given:

[ab/(a-b)^2) divded by [1/(a-b))

I'm not sure why you're mixing brackets with parentheses, but the missing grouping symbols (added in red above) are very important. Without them, the expression means something else.

[ab/(a-b)^2] / [1/(a-b)]

the solution given in the book is (a-b)/ba

If you posted the problem correctly, the book solution is a mistake. :cool: