# fractions again

• Jul 31st 2013, 07:37 AM
uperkurk
fractions again
$\displaystyle \left\frac{1}{x}+\frac{1}{3}=\right\frac{3}{2x}-1$

So the lowest common multiple is 6? As they both already have an x in them. So... $\displaystyle 6\times\left(\frac{1}{x}+\frac{1}{3}\right)=\left( \frac{3}{2x}-1\right)\times6$

So then I have $\displaystyle \frac{6}{6x}+2=\frac{18}{12x}-6$

$\displaystyle x+2=\frac{3}{2x}-6$

is this correct so far? What next?
• Jul 31st 2013, 08:04 AM
uperkurk
Re: fractions again
Nevermind I saw what I did wrong... fractions is stupidly difficult, so much messing around is needed to simple add or subtract a damn fraction...
• Jul 31st 2013, 08:46 AM
topsquark
Re: fractions again
Don't worry about it. There are any number of students out there who have a hard time with fractions. It seems to be a concept that is difficult to learn.

Why don't you post your solution? That way someone who is also having troubles can take a look at it.

-Dan
• Jul 31st 2013, 11:24 AM
uperkurk
Re: fractions again
Quote:

Originally Posted by topsquark
Don't worry about it. There are any number of students out there who have a hard time with fractions. It seems to be a concept that is difficult to learn.

Why don't you post your solution? That way someone who is also having troubles can take a look at it.

-Dan

It just takes me longer than most people to spot the "obvious" and sometimes I just get a complete mind blank and it's very frustrating. I even forget the most basic of things like $\displaystyle 2$ is the same thing as $\displaystyle \frac{2}{1}$

anyway, my solution:

$\displaystyle \frac{1}{x}+\frac{1}{3} = \frac{3}{2x}-1$

$\displaystyle 6x\times\left(\frac{1}{x}+\frac{1}{3}\right)=\left (\frac{3}{2x}-1\right)\times6x$

$\displaystyle \frac{6x}{x}+2x=\frac{18x}{2x}-6x$

after cancellation:

$\displaystyle 6+2x=9-6x$

$\displaystyle 6-9=-6x-2x$

$\displaystyle 8x=3$

$\displaystyle x=\frac{3}{8}$

This is what it's like for every question I have like this, I need to write down all the steps otherwise I just lose track of what is what, very annoying (Headbang)
• Jul 31st 2013, 01:55 PM
topsquark
Re: fractions again
Quote:

Originally Posted by uperkurk
This is what it's like for every question I have like this, I need to write down all the steps otherwise I just lose track of what is what, very annoying (Headbang)

That's pretty much how everyone has to learn this stuff. You're doing good.

-Dan