"reversing" non-linear equation system I'm not a mathematician and I'm facing a problem with those equations that I found in a book of history of colorscience.
The equations were created by MacAdam to transform the classical colorimetric diagram of the CIE into something better.
The CIE diagram plots chromaticity with 2 coordinates x,y
MacAdams plots the transformed chromaticities D,M
While it's easy to calculate D,M with given x,y (see codes below),
I find it impossible so far to do the opposite, i.e., to find the reverse formulae that will compute x,y with given D,M
Do you have any idea of how this could be done?
Thanks
The transformation from $x,y$ to $D,M$ is as follows:
$$a = \frac{10x}{2.4x+34y+1}\\
    b = \frac{10y}{2.4x+34y+1}\\
    D = 3751a^2-10a^4-520b^2+13295b^3+32327ab-25491a^2b-41672ab^2+10a^3b-5227\sqrt a+2952a^\frac14\\    
c = \frac{10x}{42y-x+1}\\
    d = \frac{10y}{42y-x+1}\\
    M = 404d - 185d^2+52d^3+69c(1-d^2)-3c^2d+30cd^3$$

As said in your post, comments and answers, you have two equations equations $D$  and $M$ for two unknwons $x$ and $y$. Provided a "reasonable" guess of the solution which could be obtained as suggested by Axel Kemper), you could use Newton-Raphson method (even with numerical derivatives since your equations are quite awful).   
Another solution is to consider the Euclidian norm of the system and minimize ($D^2+M^2$) with respect to $x$ and $y$. This will bring you to the first local minimum; but, if, at this point, the norm is $0$, then you have a solution to the problem.  
By the way, is it established that the solution is unique (that is to say that to one pair of $D,M$ corresponds only one pair of ($x,y$) ?
