# Thread: fractions with negative exponents

1. Just a quick one what would happen it that was in the denominator?

Good night.

2. Originally Posted by anthonye
Just a quick one what would happen it that was in the denominator?

Good night.
then you cannot split it apart. sleep well.

3. Then how would you solve it please post example.

Thanks

4. Can someone please show an example.

5. In that situation it generally can't be simplified very much. Here's my random example.

$\frac{x^{-1}y^{-3}z^{2}}{x^4y^{-2}+7}$

$=\displaystyle\frac{\frac{1}{x}\times\frac{1}{y^3} \times z^2}{x^4\frac{1}{y^2}+7}$

Now consider the numerator:
$=\displaystyle{\frac{1}{x}\times\frac{1}{y^3}\time s z^2}$

$=\displaystyle\frac{z^2}{xy^3}$

The denominator:

$\displaystyle x^4\frac{1}{y^2}+7$

$=\displaystyle\frac{x^4}{y^2}+7$

$=\displaystyle\frac{x^4+7y^2}{y^2}$

So I could rewrite my example as:

$\displaystyle\frac{\frac{z^3}{xy^3}}{\frac{x^4+7y^ 2}{y^2}}$
$
=\displaystyle\frac{z^3}{xy^3}\times\frac{y^2}{x^4 +7y^2}$

And you'd carry on simplifying from there. I don't know how standard this method is, though I would assume it can be applied to most, if not all, situations.

6. Unfortunately
going from 6 terms to 7 terms
wouldn't often be classed as a "simplification".

7. Originally Posted by Quacky
In that situation it generally can't be simplified very much. Here's my random example.

$\frac{x^{-1}y^{-3}z^{2}}{x^4y^{-2}+7}$

$=\displaystyle\frac{\frac{1}{x}\times\frac{1}{y^3} \times z^2}{x^4\frac{1}{y^2}+7}$

Now consider the numerator:
$=\displaystyle{\frac{1}{x}\times\frac{1}{y^3}\time s z^2}$

$=\displaystyle\frac{z^2}{xy^3}$

The denominator:

$\displaystyle x^4\frac{1}{y^2}+7$

$=\displaystyle\frac{x^4}{y^2}+7$

$=\displaystyle\frac{x^4+7y^2}{y^2}$

So I could rewrite my example as:

$\displaystyle\frac{\frac{z^3}{xy^3}}{\frac{x^4+7y^ 2}{y^2}}$
$
=\displaystyle\frac{z^3}{xy^3}\times\frac{y^2}{x^4 +7y^2}$

And you'd carry on simplifying from there. I don't know how standard this method is, though I would assume it can be applied to most, if not all, situations.
Unfortunately
going from 6 terms to 7 terms
wouldn't often be classed as a "simplification".
Haha, that's because I didn't finish. I hope you're teasing me. If not, then how are you defining 'term'? I don't see how you've identified 7.

$=\displaystyle\frac{z^3y^2}{x^5y^3+7xy^5}$

$=\displaystyle\frac{z^3}{x^5y+7xy^3}$

If I've understood you correctly, then by your definition there are still 6 "terms" here, and whilst it might not be mathematically any simpler, I know which I'd rather use! I would argue that there are just three terms in my final answer and three terms in my initial expression, but regardless, whilst the amount of terms stays the same, there are no negative indices in my final answer. Also, by your definition of a term, Skeeter's previous example also fails to be "simplified," unless I'm misinterpreting what you mean. There was a thread in the chatroom about this, actually - 'term' is a very ambiguous word. But I'm waffling: my point is that whilst it might be mathematically indifferent, I acknowledged that there was little that could be done in the way of simplification, and perhaps my method is not classed as 'simplification' as you would define it, but you can ask anyone which they'd rather work with, and I think you can probably predict the most common answer.

Edit: Furthermore, the OP was asking how you'd solve such a fraction. I think that by writing it in this form, solving for any of the variables is a lot simpler than it was in the original mess.

8. Originally Posted by Quacky
In that situation it generally can't be simplified very much. Here's my random example.

$\dfrac{x^{-1}y^{-3}z^{2}}{x^4y^{-2}+7}$
$\dfrac{x^{-1}y^{-3}z^{2}}{x^4y^{-2}+7} \cdot \dfrac{xy^3}{xy^3} = \dfrac{z^2}{x^5y+7xy^3}$

9. Originally Posted by skeeter
$\dfrac{x^{-1}y^{-3}z^{2}}{x^4y^{-2}+7} \cdot \dfrac{xy^3}{xy^3} = \dfrac{z^2}{x^5y+7xy^3}$
. I knew that.

Anyway, my point was that it couldn't be simplified very much, but that got rid of the negative exponents. Is that not what the OP was asking?

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