# rational exp - simplify

• Mar 20th 2011, 03:36 PM
reallylongnickname
rational exp - simplify
$\frac{2}{6x^2} + \frac{4}{9x^3}$

I need a common denominator?
• Mar 20th 2011, 03:50 PM
pickslides
Try $\displaystyle 6x^2 \times 9x^3$

Or find the lowest common mulitple of the 2. Maybe $\displaystyle 18x^3$
• Mar 20th 2011, 04:05 PM
reallylongnickname
$\frac{18}{54x^5}+\frac{24}{54x^5} = \frac{42}{54x^5}$

How's this look?
• Mar 20th 2011, 04:23 PM
pickslides
That's a good start but its not quite correct.

$\displaystyle \frac{2}{6x^2}+\frac{4}{9x^3}$

$\displaystyle \frac{2}{6x^2}\times \frac{9x^3}{9x^3}+\frac{4}{9x^3}\times \frac{6x^2}{6x^2}$

$\displaystyle \frac{\dots}{54x^5}+\frac{\dots}{54x^5}$
• Mar 20th 2011, 04:43 PM
reallylongnickname
$\frac{18x^3}{54x^5}+\frac{24x^2}{54x^5} = \frac{42x^5}{54x^5} = \frac{42}{54}$
• Mar 20th 2011, 04:46 PM
e^(i*pi)
You're not done cancelling yet. Recall your six times table

You're on the right track though
• Mar 20th 2011, 05:08 PM
reallylongnickname
Oh. $\frac{7}{9}$
• Mar 20th 2011, 05:11 PM
pickslides
Quote:

Originally Posted by reallylongnickname
$\frac{18x^3}{54x^5}+\frac{24x^2}{54x^5} = \frac{42x^5}{54x^5} = \frac{42}{54}$

Be careful with your like terms $\displaystyle 18x^3+24x^2 \neq 42x^5$
• Mar 20th 2011, 06:45 PM
reallylongnickname

$= \frac{18x^3 + 24x^2}{54x^5}$
• Mar 20th 2011, 06:53 PM
pickslides
Quote:

Originally Posted by reallylongnickname

$= \frac{18x^3 + 24x^2}{54x^5}$

That is better, I would conclude that $\displaystyle \frac{18x^3 + 24x^2}{54x^5}= \frac{6\times 3x^3 + 6\times 4x^2}{6\times 9x^5} = \frac{6( 3x^3 + 4x^2)}{6\times 9x^5}= \frac{ 3x^3 +4x^2}{ 9x^5}$
• Mar 20th 2011, 07:07 PM
Prove It
Quote:

Originally Posted by pickslides
That is better, I would conclude that $\displaystyle \frac{18x^3 + 24x^2}{54x^5}= \frac{6\times 3x^3 + 6\times 4x^2}{6\times 9x^5} = \frac{6( 3x^3 + 4x^2)}{6\times 9x^5}= \frac{ 3x^3 +4x^2}{ 9x^5}$

$\displaystyle = \frac{x^2(3x + 4)}{9x^5} = \frac{3x + 4}{9x^3}$...
• Mar 21st 2011, 04:26 AM
HallsofIvy
In fact, it would have been better to use $18x^3$ as your common denominator rather than $18x^5$.
• Mar 21st 2011, 05:21 AM
Prove It
Or even better still, note that $\displaystyle \frac{2}{6x^2} = \frac{1}{3x^2}$, then your LCD is $\displaystyle 9x^3$...