1. ## Common denominator trick?

It's on the tip of my tongue and I can't remember a trick I remember learning to solve these.

$
\frac{3}{(x+2b)(x-b)} + \frac{2}{(3b-x)(x+2b)} + \frac{1}{(x-3b)(b-x)}
$

For some reason i feel like there was a trick where you switch the signs and cross multiplied or something like that to get the denominators the same...but i cant remember

or do i just have to do it the long way and multiply a million times

Thanks

2. Just looking at the last term:

$\frac{1}{(x-3b)(b-x)}$

= $\frac{1}{(-1)(3b-x)(x-b)}$

The trick was to factorise -1 out of the brackets.

If the first term's denominator has (x+2b) instead of (x+2) (is it a typo?), it would have been such a nice fraction...

3. Originally Posted by Gusbob
Just looking at the last term:

$\frac{1}{(x-3b)(b-x)}$

= $- \frac{1}{(-1)(3b-x)(x-b)}$

The trick was to factorise -1 out of the brackets.

If the first term's denominator has (x+2b) instead of (x+2) (is it a typo?), it would have been such a nice fraction...
yes its a typo

thanks :P

so where does that (-1) go now?

$
\frac{7b-x}{(x+2b)(x-b)(3b-x)} + \frac {1}{(-1)(3b-x)(x-b)}
$

i can see how this would turn into

$
\frac{(7b-x) + (x+2b)}{(x+2b)(x-b)(3b-x)}
$

but i dunno where i put that -1

4. $\frac{3}{(x+2b)(x-b)} + \frac{2}{(3b-x)(x+2b)} + \frac{1}{(x-3b)(b-x)}$

$= \dfrac{3(x-3b)}{(x+2b)(x-b)(x-3b)} - \dfrac{2(x-b)}{(x-3b)(x+2b)(x-b)} - \dfrac{x+2b}{(x-3b)(x-b)(x+2b)}$

$= \frac{3(x-3b)-2(x-b)-(x+2b)}{(x+2b)(x-b)(x-3b)}$

$= \frac{(3x -2x -x)+ (-9b +2b-2b)}{(x+2b)(x-b)(x-3b)}$

$= -\frac{9b}{(x+2b)(x-b)(x-3b)}$

5. Originally Posted by silencecloak
yes its a typo

thanks :P

so where does that (-1) go now?

$
\frac{7b-x}{(x+2b)(x-b)(3b-x)} + \frac {1}{(-1)(3b-x)(x-b)}
$

i can see how this would turn into

$
\frac{(7b-x) + (x+2b)}{(x+2b)(x-b)(3b-x)}
$

but i dunno where i put that -1
Just so you know, I accidentally put a minus sign in front of my original reply, its not meant to be there.

I would suggest you do this:

$\frac{7b-x}{(x+2b)(x-b)(3b-x)} - \frac {1}{(3b-x)(x-b)}$

6. Originally Posted by Isomorphism
$\frac{3}{(x+2b)(x-b)} + \frac{2}{(3b-x)(x+2b)} + \frac{1}{(x-3b)(b-x)}$

$= \dfrac{3(x-3b)}{(x+2b)(x-b)(x-3b)} - \dfrac{2(x-b)}{(x-3b)(x+2b)(x-b)} - \dfrac{x+2b}{(x-3b)(x-b)(x+2b)}$

$= \frac{3(x-3b)-2(x-b)-(x+2b)}{(x+2b)(x-b)(x-3b)}$

$= \frac{(3x -2x -x)+ (-9b +2b-2b)}{(x+2b)(x-b)(x-3b)}$

$= -\frac{9b}{(x+2b)(x-b)(x-3b)}$
how can you just take the $3b-x$ and $b-x$
and just flip the signs?

7. Originally Posted by silencecloak
how can you just take the $3b-x$ and $b-x$
and just flip the signs?
Isnt $-(3b-x) = x-3b$?

I will show what I did for the middle one...

$\frac{2}{(3b-x)(x+2b)} = \frac{2}{(3b-x)(x+2b)} {\color{red}\times \frac{-(x-b)}{-(x-b)}}$

$\frac{2}{(3b-x)(x+2b)} {\color{red}\times \frac{-(x-b)}{-(x-b)}} = \frac{-2(x-b)}{-(3b-x)(x+2b)(x-b)}$

Now use $-(3b-x) = x-3b$

$\frac{-2(x-b)}{-(3b-x)(x+2b)(x-b)} = \frac{-2(x-b)}{(x-3b)(x+2b)(x-b)}$

Understood?

8. Originally Posted by Isomorphism
Isnt $-(3b-x) = x-3b$?

I will show what I did for the middle one...

$\frac{2}{(3b-x)(x+2b)} = \frac{2}{(3b-x)(x+2b)} {\color{red}\times \frac{-(x-b)}{-(x-b)}}$

$\frac{2}{(3b-x)(x+2b)} {\color{red}\times \frac{-(x-b)}{-(x-b)}} = \frac{-2(x-b)}{-(3b-x)(x+2b)(x-b)}$

Now use $-(3b-x) = x-3b$

$\frac{-2(x-b)}{-(3b-x)(x+2b)(x-b)} = \frac{-2(x-b)}{(x-3b)(x+2b)(x-b)}$

Understood?
Yes, but isn't
$\frac{(3x -2x -x)+ (-9b +2b-2b)}{(x+2b)(x-b)(x-3b)}
$

-9b not 9b ?

or does that negative come back from somewhere

9. Originally Posted by silencecloak
Yes, but isn't
$\frac{(3x -2x -x)+ (-9b +2b-2b)}{(x+2b)(x-b)(x-3b)}
$

-9b not 9b ?

or does that negative come back from somewhere
Its there, I marked it in red here:
$= {\color{red} -}\frac{9b}{(x+2b)(x-b)(x-3b)}$

10. Originally Posted by Isomorphism
Its there, I marked it in red here:
$= {\color{red} -}\frac{9b}{(x+2b)(x-b)(x-3b)}$
Ah, my apologies.

Thanks for stickin with me.

Been a couple years since I have done all this and I have to pass a test to get into calculus!

Hmm one more thing

Would it be wrong to just go $3(3b-x) + 2(x-b) + (x+2b)$

all over the LCD, this comes out to 9b perfectly, but I have a feeling this is an isolated case where this works.

11. Originally Posted by silencecloak
Would it be wrong to just go $3(3b-x) + 2(x-b) + (x+2b)$

all over the LCD, this comes out to 9b perfectly, but I have a feeling this is an isolated case where this works.
Whether stuff cancels out is purely problem dependent... And what you said is not wrong