Can someone point me in the right direction

1. Mar 16, 2005

powp

Hello

I have to simplify the following expression can anybody help me get started??

$$\frac{\frac{-3xy^{-2}}{x-1}}{10(\frac{xy}{z^3})^{-2}}$$

Thanks

P

2. Mar 16, 2005

Crosson

The first thing is to take care of the negative exponent, then divide the fractions (by multiplying by the reciprical).

3. Mar 30, 2005

powp

Sorry I miss typed the problem(I think hard to tell)
$$\frac{\frac{-3xy^{-2}}{z^{-1}}}{10(\frac{xy}{z^3})^{-2}}$$

So this is what I have so far

$$\frac{\frac{-3xz^{1}}{y^{2}}}{10(\frac{z^3}{xy})^{2}}$$

$$\frac{\frac{-3xz^{1}}{y^{2}}}{10(\frac{z^6}{x^2y^2})}$$

$$\frac{\frac{-3xz^{1}}{y^{2}}}{(\frac{10z^6}{10x^2y^2})}$$

$${\frac{-3xz^{1}}{y^{2}}}X{\frac{10x^2y^2}{10z^6}$$

Am I on the right track?

4. Mar 30, 2005

b0mb0nika

yeah except one thing:
when you multiply the faction by 10, it only goes on the top, not on the bottom. So then when you invert it, its going to be on the botton
(x^2)(y^2) / (10 z^6)
the rest is fine now u just need to cancel the y^2 and the z

5. Mar 30, 2005

powp

so it is ?

$${\frac{-3xz^{1}}{y^{2}}}X{\frac{x^2y^2}{10z^6}$$

how do the z cancel each other out?

6. Mar 30, 2005

dextercioby

They don't

$$\frac{-3}{10}\frac{x^{3}}{z^{5}}$$

Daniel.

7. Mar 30, 2005

powp

does it work out like this??

$${\frac{-3x^3y^2z^1}{10y^{2}z^6}}$$

$${\frac{-3x^3y^{2-2}z^{1-6}}{10}}$$

$${\frac{-3x^3y^{0}z^{-5}}{10}}$$

$${\frac{-3x^3}{10z^{5}}}$$

8. Mar 30, 2005

dextercioby

Exactly like that.

Daniel.

9. Mar 30, 2005

powp

Thanks for you help