# Rearrange in terms of z

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• Sep 14th 2011, 06:59 PM
TomByrne
Rearrange 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?
• Sep 14th 2011, 07:33 PM
Wilmer
Re: Rearrange in terms of z
So, is this your equation:
Code:

```          cdf        hef a(z-b) = - ... - cg - ... - hi           f+z        f+z```
Note: multiplication implied (like, hi means h*i)
• Sep 14th 2011, 07:37 PM
TomByrne
Re: Rearrange in terms of z
Yes absolutely, that's more elegant.
• Sep 14th 2011, 07:41 PM
Wilmer
Re: 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?
• Sep 14th 2011, 07:42 PM
pickslides
Re: Rearrange in terms of z
I would start by mulitplying both sides through by \$\displaystyle f+z\$ . Then you have no nasty fractions. After this multiply out any brackets and then group terms of \$\displaystyle z\$ and \$\displaystyle z^2\$ together, what do you get?
• Sep 14th 2011, 07:43 PM
TomByrne
Re: Rearrange in terms of z
Yep, I follow, how does one group those terms?
• Sep 14th 2011, 07:52 PM
Wilmer
Re: Rearrange in terms of z
Quote:

Originally Posted by TomByrne
Yep

GREAT! Now for the big question: are you SURE the term on left is a(z-b);
like is that "damn z" really there?
Asking because we'll be getting "z^2" ... which will spoil my evening !!
• Sep 14th 2011, 08:07 PM
pickslides
Re: Rearrange in terms of z
Quote:

Originally Posted by TomByrne
Yep, I follow, how does one group those terms?

Expand all the brackets first and Wilmer will show you!
• Sep 14th 2011, 08:08 PM
TomByrne
Re: 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
• Sep 14th 2011, 08:12 PM
TomByrne
Re: Rearrange in terms of z
az = -cdf/(f+z)-cg-hef/(f+z)-hi+ab
• Sep 14th 2011, 08:20 PM
Wilmer
Re: Rearrange in terms of z
Quote:

Originally Posted by TomByrne
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)

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

sorry

Huh? That's same thing Tom: a(z-b) = az - ab
You been drinking? (Giggle)
AGREE??!!
• Sep 14th 2011, 08:25 PM
TomByrne
Re: 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).
• Sep 14th 2011, 08:36 PM
TomByrne
Re: 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?
• Sep 14th 2011, 08:41 PM
Wilmer
Re: Rearrange in terms of z
Quote:

Originally Posted by TomByrne
az = -cdf/(f+z) - cg - hef/(f+z) - hi + ab

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?
• Sep 14th 2011, 08:45 PM
TomByrne
Re: Rearrange in terms of z
Thanks, great. What are the square brackets in this situation?
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