# Simplifying/Reducing

• Sep 15th 2011, 11:00 AM
yossarian87
Simplifying/Reducing
Could somebody please show me how to simplify this?

((1/4)x^(-3/4)*y^(3/4)/(3/4)x^(1/4)*y^(-1/4))

That is, in words, one-fourth "x" raised to the negative three-fourths, "y" raised to the three-fourths, over/divided by three-fourths "x" raised to the one-fourth, "y" raised to the negative one-fourth.

The answer is (y/3x), or in words, "y" over/divided by 3"x".

I just need to see how to get to that and preferably the quickest way.

Thank you so much for the help!
• Sep 15th 2011, 11:20 AM
mabentley
re: Simplifying/Reducing
I've converted to Latex for now so that it's easier to read:

$\frac{\bigg(\frac{1}{4}x^{-\frac{3}{4}}y^{\frac{3}{4}}\bigg)}{\bigg(\frac{3}{ 4}x^{\frac{1}{4}}y^{-\frac{1}{4}}\bigg)}$
• Sep 15th 2011, 11:25 AM
yossarian87
re: Simplifying/Reducing
Thanks!
• Sep 15th 2011, 11:33 AM
mabentley
re: Simplifying/Reducing
Quote:

Originally Posted by mabentley
I've converted to Latex for now so that it's easier to read:

$\frac{\bigg(\frac{1}{4}x^{-\frac{3}{4}}y^{\frac{3}{4}}\bigg)}{\bigg(\frac{3}{ 4}x^{\frac{1}{4}}y^{-\frac{1}{4}}\bigg)}$

When you divide with powers you subtract:

$\frac{1/4}{3/4}=\frac{4}{12}=\frac{1}{3}$

$x^{-\frac{3}{4}}-x^\frac{1}{4}=x^\frac{4}{4}=x^{-1}=\frac{1}{x}$

$y^\frac{3}{4}-y^{-\frac{1}{4}}=y^\frac{4}{4}=y^{1}=y$

Multiplying through:

$=\frac{y}{3x}$
• Sep 15th 2011, 11:43 AM
yossarian87
re: Simplifying/Reducing
Thank you very much!
• Sep 15th 2011, 11:49 AM
mabentley
re: Simplifying/Reducing
You are welcome, feel free to hit the thanks button on the bottom left of your reply (Wink)
• Sep 15th 2011, 12:03 PM
e^(i*pi)
Re: Simplifying/Reducing
Quote:

Originally Posted by mabentley
When you divide with powers you subtract:
$x^{-\frac{3}{4}}-x^\frac{1}{4}=x^\frac{4}{4}=x^{-1}=\frac{1}{x}$

$y^\frac{3}{4}-y^{-\frac{1}{4}}=y^\frac{4}{4}=y^{1}=y$

That's not true. It's only the powers that are subtracted.

. $\dfrac{x^{-3/4}}{x^{1/4}} = x^{-3/4 - 1/4} = x^{-1}$
• Sep 15th 2011, 02:03 PM
mabentley
Re: Simplifying/Reducing
Quote:

Originally Posted by e^(i*pi)
That's not true. It's only the powers that are subtracted.

. $\dfrac{x^{-3/4}}{x^{1/4}} = x^{-3/4 - 1/4} = x^{-1}$

Sorry the $x^{\frac{4}{4}}$ should have read $x^{-\frac{4}{4}}$. I can see that my wording could also have been clearer. Thanks.