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?

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- Sep 5th 2011, 03:04 PMDavid GreenRearranging 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, 03:07 PMSironRe: Rearranging formulas
I would write:

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

$\displaystyle \Leftrightarrow a=5b-6c-2bc+c$

$\displaystyle \Leftrightarrow a-5b+5c=-2bc$

$\displaystyle \Leftrightarrow b=...$ - Sep 5th 2011, 03:16 PMDavid GreenRe: Rearranging formulas
- Sep 5th 2011, 03:18 PMSironRe: Rearranging formulas
Because:

-6c+c=-5c - Sep 5th 2011, 03:21 PMpickslidesRe: Rearranging formulas
- Sep 5th 2011, 03:28 PMDavid GreenRe: Rearranging formulas
- Sep 5th 2011, 03:33 PMpickslidesRe: Rearranging formulas
- Sep 5th 2011, 03:36 PMDavid GreenRe: Rearranging formulas
- Sep 5th 2011, 03:50 PMpickslidesRe: Rearranging formulas
I appreciate the honesty

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

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

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

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

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

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

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

__Spoiler__: - Sep 5th 2011, 03:54 PMSironRe: 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, 04:03 PMDavid GreenRe: Rearranging formulas
- Sep 6th 2011, 09:52 AMDavid GreenRe: Rearranging formulas
- Sep 6th 2011, 01:46 PMpickslidesRe: Rearranging formulas
You can't just remove the 'c'. This has made your working of the problem incorrect.

- Sep 6th 2011, 03:07 PMDavid GreenRe: Rearranging formulas
Hi Pickslides

Please advise what you did to loose a C in your first line?

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 ?

The final answer I am sure you had;

b = 4b - 5c / a - c

what changed?

Thanks

David - Sep 7th 2011, 01:44 PMpickslidesRe: Rearranging formulas
Post PM

From this step:

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

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

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

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

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

Now divide $\displaystyle 5-2c$ from both sides

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

All good?