Rearranging formulas

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• Sep 5th 2011, 04:04 PM
David Green
Rearranging formulas
a = 5b - 2c(3+b) + c make b the subject

a = 5b - 6c + b + c

b = 5b + 6c^2 / a

can I not do that?
• Sep 5th 2011, 04:07 PM
Siron
Re: Rearranging formulas
I would write:
$a=5b-2c(3+b)+c$
$\Leftrightarrow a=5b-6c-2bc+c$
$\Leftrightarrow a-5b+5c=-2bc$
$\Leftrightarrow b=...$
• Sep 5th 2011, 04:16 PM
David Green
Re: Rearranging formulas
Quote:

Originally Posted by Siron
I would write:
$a=5b-2c(3+b)+c$
$\Leftrightarrow a=5b-6c-2bc+c$
$\Leftrightarrow a-5b+5c=-2bc$
$\Leftrightarrow b=...$

Interesting, but a - 5b + 5c = - 2bc

why did - 6c become 5c?

I understand that change the sides change the sign, but why change 6 to 5?

I have got it now, - 2bc + c change which change -6c to 5c and -2bc + c became - 2bc(Cool)
• Sep 5th 2011, 04:18 PM
Siron
Re: Rearranging formulas
Because:
-6c+c=-5c
• Sep 5th 2011, 04:21 PM
pickslides
Re: Rearranging formulas
Quote:

Originally Posted by David Green

why did - 6c become 5c?

I understand that change the sides change the sign, but why change 6 to 5?

$a = 5b - 2c(3+b) + c$

$a = 5b - 6c-2bc + c$

$a = 5b - 6c+c-2bc$

$a = 5b - 5c-2bc$
• Sep 5th 2011, 04:28 PM
David Green
Re: Rearranging formulas
Quote:

Originally Posted by pickslides

$a = 5b - 2c(3+b) + c$

$a = 5b - 6c-2bc + c$

$a = 5b - 6c+c-2bc$

$a = 5b - 5c-2bc$

I have still got to get b on its own, and there is 2 b's?

So is the rule now to factor the b's out?
• Sep 5th 2011, 04:33 PM
pickslides
Re: Rearranging formulas
Quote:

Originally Posted by David Green
I have still got to get b on its own, and there is 2 b's?

So is the rule now to factor the b's out?

Yep, what do you get?
• Sep 5th 2011, 04:36 PM
David Green
Re: Rearranging formulas
Quote:

Originally Posted by pickslides
Yep, what do you get?

Not much yet because I am not sure about factoring mixed terms?

Will have to do some more research on this my course book has poor example of it?
• Sep 5th 2011, 04:50 PM
pickslides
Re: Rearranging formulas
I appreciate the honesty

$a = 5b - 2c(3+b) + c$

$a = 5b - 6c-2bc + c$

$a = 5b - 6c+c-2bc$

$a = 5b - 5c-2bc$

$a = 5b -2bc - 5c$

$a = 5\times b -2c\times b - 5c$

$a = b(5 -2c) - 5c$

Spoiler:
$a+5c = b(5 -2c)$

$b=\frac{a+5c}{5 -2c}$
• Sep 5th 2011, 04:54 PM
Siron
Re: Rearranging formulas
Sorry David, I didn't see that the first term had also a facor b so what I wrote doesn't make sense, pickslides solution is the good one.
• Sep 5th 2011, 05:03 PM
David Green
Re: Rearranging formulas
Quote:

Originally Posted by Siron
Sorry David, I didn't see that the first term had also a facor b so what I wrote doesn't make sense, pickslides solution is the good one.

Thanks for all your help Pickslides and Siron, in alphabetical order?

Very much appreciated, I will work on it later after work

Thanks

David
• Sep 6th 2011, 10:52 AM
David Green
Re: Rearranging formulas
Quote:

Originally Posted by pickslides
I appreciate the honesty

$a = 5b - 2c(3+b) + c$

$a = 5b - 6c-2bc + c$

Hi Pickslides

I have just realised what the spolier was you installed, however I have been looking at this today and came up with this;

a = 5b - 2c(3+b) + c First I loose the bracket
a = 5b - 6c - 2bc + c Not sure why but in a previous explanation Siron removed a C, so I have followed the same line of reasoning here.
a = 5b - 5c - 2b + c Now the whole point of this is to get "b" on it's own, and I still have 5b and another - 2b, so;
a = 5b - 2b - 5c + c
a = 3b - 5c + c Now I will subtract "C" from both sides
a - c = 3b - 5c From this point forwards I can't get "b" on it's own, so I need to go back to line "3" from the top and start again;

a = 5b - 5c - 2b + c Now this time I will try and subtract 1b
a = 4b - 5c - b + c Now I can subtract "c" from the RHS
a - c = 4b - 5c - b Now I can subtract "b" from the RHS
a - c + b = 4b - 5c Now I can divide both sides by "a - c"

b = 4b - 5c / a - c

How have I done working that lot out, since I knew nothing really last night?

Thanks again for all your help to all who helped and advised

Thanks

David

$a = 5b - 6c+c-2bc$

$a = 5b - 5c-2bc$

$a = 5b -2bc - 5c$

$a = 5\times b -2c\times b - 5c$

$a = b(5 -2c) - 5c$

Spoiler:
$a+5c = b(5 -2c)$

$b=\frac{a+5c}{5 -2c}$

Thanks

David
• Sep 6th 2011, 02:46 PM
pickslides
Re: Rearranging formulas
You can't just remove the 'c'. This has made your working of the problem incorrect.
• Sep 6th 2011, 04:07 PM
David Green
Re: Rearranging formulas
Quote:

Originally Posted by David Green
Thanks

David

Hi Pickslides

The original equation was;

a = 5b - 2c (3 + b) + C

In my first line of working out I got;

a = 5b - 6c - 2bc + c

in your first line of working out you got;

a = 5b - 6c -2bc

what did you do with the latter C ?

b = 4b - 5c / a - c

what changed?

Thanks

David
• Sep 7th 2011, 02:44 PM
pickslides
Re: Rearranging formulas
Post PM

From this step:

$a = 5b -2bc - 5c$

We need to isolate $b$ by factoring, recall $ab+ab = a(b+c)$ so

$a = 5\times b -2c\times b - 5c$

$a = b(5 -2c) - 5c$

$a+5c = b(5 -2c)$

Now divide $5-2c$ from both sides

$b=\frac{a+5c}{5 -2c}$

All good?
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