Common denominator trick?

**silencecloak** · May 7th 2008, 10:10 PM

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

**Gusbob** · May 7th 2008, 10:42 PM

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...

**silencecloak** · May 7th 2008, 10:44 PM

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

**Isomorphism** · May 7th 2008, 10:52 PM

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

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

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

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

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

**Gusbob** · May 7th 2008, 10:55 PM

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:

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

**silencecloak** · May 7th 2008, 10:56 PM

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?

**Isomorphism** · May 7th 2008, 11:08 PM

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?

**silencecloak** · May 7th 2008, 11:17 PM

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

**Isomorphism** · May 7th 2008, 11:29 PM

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

**silencecloak** · May 7th 2008, 11:31 PM

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.

**Isomorphism** · May 7th 2008, 11:57 PM

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

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