Hi all,

a chap in work is doing some maths and is stuck on this. I said I would have a look but I cannot remember how to do these. Maths was too many moons ago for me.

Anyone know how to do it or what type of problem it is so I can have a go.

Cheers,

y=((a+x)/(5-bx))

[Edited on 18/3/16 by emsfactory]

[Edited on 18/3/16 by emsfactory]

Looks like a quadratic equation, so easy I'll let someone else do it

y=((a+x)/(5-bx))
5y-ybx=a+x
5y-ybx-a=x
5y-a=x+ybx
5y-a=x(1+yb)
x=(5y-a)/(1+yb)

does that work? been a while (30 years...)

I love a bit of algebra.

A quick scribble suggests
X = (5y-a) /(yb+1)

Damn too slow! Thats how i did it as well.

[Edited on 18/3/16 by bi22le]

Yeah thats where we got too. I am thinking there is more to it though. Some formula to be applied to make it nicer.

would need y a and b values to.input.if he wants a numerical value of x

x=(5y-a)/(1+yb)

This is as far as you can take it unless you start with derivatives

then it becomes

(d)/(dx)((a+x)/(5-b x)) = (a b+5)/(b x-5)^2

Presumably a and b are constants, x and y are variables.

You have too many unknowns to solve this equation, but:

when y = 0, x = -a
when x = 0, y= a/5

Given values for the constants a and b, you could then draw a graph of the equation - it's linear, so should be a straight line.

quote:

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Originally posted by jonrotheray
Presumably a and b are constants, x and y are variables.

You have too many unknowns to solve this equation, but:

when y = 0, x = -a
when x = 0, y= a/5

Given values for the constants a and b, you could then draw a graph of the equation - it's linear, so should be a straight line.

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Nope thats it. You cant simplify it.