# Hah, okay. I've got one hour to turn this in...

• September 15th 2008, 07:57 PM
Hah, okay. I've got one hour to turn this in...
And I'm terrible at this stuff..If someone wouldn't mind walking me through these, I'd be super grateful.

1.)
https://webwork.math.wisc.edu/webwor...b48def6941.png

2.)https://webwork.math.wisc.edu/webwor...e8997006d1.png

3.)https://webwork.math.wisc.edu/webwor...35ff8db3b1.png

With this one, I know i need to add up the nominator and denominator by finding a common denominator for each..then to take:

[(t^2-s^2)/(s^(2)t^(2))]/[(t+s)/(st)]

But then I'm stuck.

• September 15th 2008, 08:52 PM
Jameson
#1)What are you supposed to do here?

#2) Factor $s^2-9$. You should get $(s+3)(s-3)$ and you get some nice cancellation.

#3)Multiply the whole thing by (st)/(st)
• September 15th 2008, 09:14 PM
o_O
Or notice you can apply the difference of squares formula to the numerator and notice that something cancels ;):
$\frac{\displaystyle \frac{1}{s^2} - \frac{1}{t^2}}{\displaystyle\frac{1}{s} + \frac{1}{t}} = \frac{\displaystyle\left(\frac{1}{s} - \frac{1}{t}\right)\left(\frac{1}{s} + \frac{1}{t}\right)}{\displaystyle\frac{1}{s} + \frac{1}{t}}$
• September 15th 2008, 09:27 PM
Jameson
Quote:

Originally Posted by o_O
Or notice you can apply the difference of squares formula to the numerator and notice that something cancels ;):
$\frac{\displaystyle \frac{1}{s^2} - \frac{1}{t^2}}{\displaystyle\frac{1}{s} + \frac{1}{t}} = \frac{\displaystyle\left(\frac{1}{s} - \frac{1}{t}\right)\left(\frac{1}{s} + \frac{1}{t}\right)}{\displaystyle\frac{1}{s} + \frac{1}{t}}$

Much more elegant.
• September 15th 2008, 11:47 PM
Moo
Hello,
Quote:

$\frac{6ab-8b^2}{15a^2-20ab}=\frac{2b(3a-4b)}{5a(3a-4b)}$
$=\frac{2b}{5a}$