Re: Equation problem

*To*: mathgroup at smc.vnet.net*Subject*: [mg109162] Re: Equation problem*From*: "Sjoerd C. de Vries" <sjoerd.c.devries at gmail.com>*Date*: Wed, 14 Apr 2010 23:12:30 -0400 (EDT)*References*: <hq39ug$na6$1@smc.vnet.net>

What you want is mathematically impossible. x and y are not linearly related. Anyway, to bring you as close as possible to what you want you could try to use the correct Mathematica syntax for your equation: y == ((1 + Sqrt[5.71 10^8/x])*x)/6.3810^9 double = to make it a real equation, otherwise it's a Set operation. Square brackets for functions and a capital letter for any built-in function. Powers of 10 written as powers of 10 and not of e. Then you can use Solve to solve this equation for x: Solve[y == ((1 + Sqrt[5.71 10^8/x])*x)/6.3810^9, x] {{x -> 2.27747*10^-28 (1.25358*10^36 + 7.70097*10^34 y - 4.39404*10^35 Sqrt[8.13912 + 1. y])}, {x -> 2.27747*10^-28 (1.25358*10^36 + 7.70097*10^34 y + 4.39404*10^35 Sqrt[8.13912 + 1. y])}} So, two solutions none of which have the shape you'd like. But as said above that can't be done, not in Mathematica not in anything else. Cheers -- Sjoerd On Apr 14, 4:40 am, Asttro <imor... at gmail.com> wrote: > I just started using Mathematica and have no idea how to do the following= ... > > I have the starting equation: > > y=((1+sqrt(5.71e8/x))*x)/6.38e9 > > What I want is to obtain x=a+b*y, where a and b are parameters. > > Is this possible in Mathematica?