Hi,

I need to rearrange this function in terms of z, I'm struggling a little bit (asterisks are multiplication):

a*(z-b) = -c*d*(f/(f+z))-(c*g)-h*e*(f/(f+z))-(h*i)

Any help?

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- September 14th 2011, 07:59 PMTomByrneRearrange in terms of z
Hi,

I need to rearrange this function in terms of z, I'm struggling a little bit (asterisks are multiplication):

a*(z-b) = -c*d*(f/(f+z))-(c*g)-h*e*(f/(f+z))-(h*i)

Any help? - September 14th 2011, 08:33 PMWilmerRe: Rearrange in terms of z
So, is this your equation:

Code:`cdf hef`

a(z-b) = - ... - cg - ... - hi

f+z f+z

- September 14th 2011, 08:37 PMTomByrneRe: Rearrange in terms of z
Yes absolutely, that's more elegant.

- September 14th 2011, 08:41 PMWilmerRe: Rearrange in terms of z
Hokay Tom: start by multiplying through by f+z:

a(z-b)(f+z) = -cdf - cg(f+z) - hef - hi(f+z)

Do you follow that? - September 14th 2011, 08:42 PMpickslidesRe: Rearrange in terms of z
I would start by mulitplying both sides through by . Then you have no nasty fractions. After this multiply out any brackets and then group terms of and together, what do you get?

- September 14th 2011, 08:43 PMTomByrneRe: Rearrange in terms of z
Yep, I follow, how does one group those terms?

- September 14th 2011, 08:52 PMWilmerRe: Rearrange in terms of z
- September 14th 2011, 09:07 PMpickslidesRe: Rearrange in terms of z
- September 14th 2011, 09:08 PMTomByrneRe: Rearrange in terms of z
You're absolutely correct, I've screwed something up in the original formula, here is the correct version:

a*z-(a*b) = -c*d*(f/(f+z))-(c*g)-h*e*(f/(f+z))-(h*i)

sorry - September 14th 2011, 09:12 PMTomByrneRe: Rearrange in terms of z
az = -cdf/(f+z)-cg-hef/(f+z)-hi+ab

- September 14th 2011, 09:20 PMWilmerRe: Rearrange in terms of z
- September 14th 2011, 09:25 PMTomByrneRe: Rearrange in terms of z
I try to avoid drinking at work, although considering work is currently a laptop in a van, I suppose a drink wouldn't go astray (although it may not aid my abysmal maths skills).

- September 14th 2011, 09:36 PMTomByrneRe: Rearrange in terms of z
Well, all the same, I'm no closer to a solution. Can you help me or have I squandered my opportunity?

- September 14th 2011, 09:41 PMWilmerRe: Rearrange in terms of z
Now that you're sober...OK, that's correct.

Problem/complication is, when multiplying through by f+z, we get az(f+z) = azf + az^2:

so we have a ye olde quadratic; going through with this mess, simplifying and combining, we'll get:

ax^2 + x(af - ab + cg + hi) + f(cd + cg + he + hi - ab) = 0 : recognize the quadratic?

Best way (for me anyway) to handle such a huge mudder is:

Let u = af - ab + cg + hi

Let v = f(cd + cg + he + hi - ab)

Then, using quadratic formula:

x = [-u +- SQRT(u^2 - 4av)] / (2a)

Any questions?!

What you think, Pickslide? You proud of me? - September 14th 2011, 09:45 PMTomByrneRe: Rearrange in terms of z
Thanks, great. What are the square brackets in this situation?