# Thread: Rearrange in terms of z

1. ## 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?

2. ## Re: Rearrange in terms of z

Code:
           cdf        hef
a(z-b) = - ... - cg - ... - hi
f+z        f+z
Note: multiplication implied (like, hi means h*i)

3. ## Re: Rearrange in terms of z

Yes absolutely, that's more elegant.

4. ## 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)

5. ## Re: Rearrange in terms of z

I would start by mulitplying both sides through by $f+z$ . Then you have no nasty fractions. After this multiply out any brackets and then group terms of $z$ and $z^2$ together, what do you get?

6. ## Re: Rearrange in terms of z

Yep, I follow, how does one group those terms?

7. ## Re: Rearrange in terms of z

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 !!

8. ## Re: Rearrange in terms of z

Originally Posted by TomByrne
Yep, I follow, how does one group those terms?
Expand all the brackets first and Wilmer will show you!

9. ## 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

10. ## Re: Rearrange in terms of z

az = -cdf/(f+z)-cg-hef/(f+z)-hi+ab

11. ## Re: Rearrange in terms of z

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?
AGREE??!!

12. ## 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).

13. ## 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?

14. ## Re: Rearrange in terms of z

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)

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

Any questions?!

What you think, Pickslide? You proud of me?

15. ## Re: Rearrange in terms of z

Thanks, great. What are the square brackets in this situation?

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