Annoying maths question

[Mercutio]

Ok. This one bothering me.

The maths is actually "relatively" complex as far as I can tell.

Assuming:

x/(x+(50*y))=z

How do I calculate z based on a fixed value for x and y?

eg:
x/(x+(50*y))=z
x=200, y=1: 200/(200+(50*1))=0.8
x=400, y=2: 400/(400+(50*2))=0.8

How do I work out x with z fixed as 0.8 and y as 4? Just the example maths would do nicely.

 

[fornowagain]

If Z is fixed solve it for one of the other variables, e.g. x=200y y=4 then x=800

X/(X+(50y))=0.8

1.25X= X+50y

0.25X = 50y

X = 200y

 

[Mercutio]

What i'm after is being able to use fixed values of z (i.e the 0.8 mentioned) and varying values of y (whole numbers only) in excel and have it spit out the x value. It's just that with x being both sides of the divide I can't get my head around it.

 

[Mercutio]

If Z is fixed solve it for one of the other variables, e.g. x=200y y=4 then x=800

X/(X+(50y))=0.8

1.25X= X+50y

0.25X = 50y

X = 200y

Hmmm, between this and what a collegue at work (night shift) just scribled on paper I ALMOST get it.
It's very annoying, I last did complex maths equations about 15 years ago and even then it was at a push

Could anyone throw it into a fairly simple excel based single line formula i'd appreciate it muchly.

Use the above values as an example (x=200, y=4, z= 0.8).

 

[Reaper 392]

if you're looking to work out X for any fixed values of Y and Z then you can either plug the numbers in and do some rearrangement to make X = some number, or you could do some rearrangement to make X a function of Y and Z like this:

x/(x+(50*y))=z

multiply both sides by (x+(50*y)) to get rid of the fraction
x = z(x+50y)

multiply out the brackets
x = zx + 50zy

group the X's on one side by taking away zx from both sides
X - Zx = 50zy

factorise the left hand side to get the Xs and Zs separate
x(1-z) = 50zy

divide both sides by 1-z to get X = some function:
x = 50zy/(1-z)

then plug y=4 and Z=0.8 in and you'll get your answer for X
x = (50*4*0.8)/(1-0.8)
x = 160/0.2
x = 800

hopefully i havent done anything wrong here, but its some stupid hour of the morning so i've probably done something wrong somewhere

 

[Mercutio]

That's awesome.
I'd got as far as:

1x-zx=z+(50y) but couldn't work out how to make it simpler. Your's works perfectly, thanks very much

 

[scrlk]

Have a play about with Wolfram Alpha if you have the chance.

E.g. http://www.wolframalpha.com/input/?i=simplify+1x-zx=z+(50y)+

 

[JonJ678]

I'd cheat and run a root finding algorithm on it. Something along the lines of guess x, is z = 0.8? If not, guess another x and iterate.

Perhaps I'm spending too much time programming these days.