• Apr 30th 2012, 08:32 PM
Azusa
I'm not quite sure what the exact name for this type of problem.

[Instructions] Perform the indicated operations on the rational expressions and simplify.

Attachment 23739

I do not need the answer, but instead, a guide on how to solve the problem.
I am aware that I need to move the negative exponents on a denominator, but I'm not sure if it will go on the bottom -OR- make a fraction on the numerator and put the negative exponent(and its variable) on the denominator of the fraction on the numerator (and so on). I hope that makes sense. Can anybody enlighten me, please (Preferably, with numbers, as opposed to sentences - just like what I did. x_x)?

Once the negative exponents have been taken care of, it would be great to have the rest of the problem solved as well.

Any help is appreciated. Thank You,
Azusa.
• Apr 30th 2012, 09:10 PM
Prove It
What do you mean by perform the indicated operations? Your terms are not alike, so you can't add them, and trying to divide will make things very messy...

Do you mean you want to write this with positive powers?
• Apr 30th 2012, 09:23 PM
Azusa
To be honest, I do not know. But it is on the same section where I just pretty much simplified everything.
Didn't run into much problems until this one.
The answers are as follows (it may help):
Attachment 23740

Azusa.
• Apr 30th 2012, 09:34 PM
Prove It
That means you're probably expected to write everything with positive powers.

\displaystyle \begin{align*} \frac{3x^{-2} + 5y^{-3}}{2x^{-3} - 3y^{-2}} &= \frac{\frac{3}{x^2} + \frac{5}{y^3}}{\frac{2}{x^3} - \frac{3}{y^2}} \\ &= \frac{\frac{3y^3 + 5x^2}{x^2y^3}}{\frac{2y^2 - 3x^3}{x^3y^2}} \\ &= \frac{x^3y^2\left(3y^3 + 5x^2\right)}{x^2y^3\left(2y^2 - 3x^3\right)} \\ &= \frac{x\left(3y^3 + 5x^2\right)}{y\left(2y^2 - 3x^3\right)} \end{align*}
• Apr 30th 2012, 09:55 PM
Azusa
Thank You. Makes more sense now, but I have one last question..

Why did Attachment 23741, turn into Attachment 23742?

I get the denominator part; you just combined them. But how about the numerator? Just cross multiply? Why did you cross multiply?

EDIT2: Also, I do not get how you went from Attachment 23743, to Attachment 23744

Again, Thank You For Your Help,
Azusa.
• Apr 30th 2012, 11:15 PM
Azusa
bump.
• May 1st 2012, 02:56 AM
Prove It
To add fractions you need to get a common denominator.
• May 1st 2012, 03:05 AM
Azusa
I get it now. Thank you so much for your help!
• May 3rd 2012, 05:20 AM
HallsofIvy
Quote:

Originally Posted by Azusa
I'm not quite sure what the exact name for this type of problem.

[Instructions] Perform the indicated operations on the rational expressions and simplify.

Attachment 23739

I do not need the answer, but instead, a guide on how to solve the problem.
I am aware that I need to move the negative exponents on a denominator, but I'm not sure if it will go on the bottom -OR- make a fraction on the numerator and put the negative exponent(and its variable) on the denominator of the fraction on the numerator (and so on). I hope that makes sense. Can anybody enlighten me, please (Preferably, with numbers, as opposed to sentences - just like what I did. x_x)?

Once the negative exponents have been taken care of, it would be great to have the rest of the problem solved as well.

Any help is appreciated. Thank You,
Azusa.

The first thing I would do is multiply both numerator and denominator by $x^3y^3$ to get
$\frac{3xy^3+ 5x^3}{2y^3- 3x^3y}$