How to solve a nonlinear system of three equations involving rational functions?

Question

How do I get $a$, $b$, and $c$?

Given

$$X=\frac{a+\frac{1}2b}{a+b+c}$$

$$Y=\frac{b(\frac{\sqrt3}{2})}{a+b+c}$$

$$Z=\frac{76a+150b+29c}{255}$$

in other words How do i get $a$, $b$, and $c$ on the left side using these three formulas?

is it possible? A= or B= or C= on the left. For a total of 3 equations.

Please do not provide a Matrix answer, I need this formula to use in a robot I haven't programed for matrix. Color matching algorithm.

I have a feeling there are many answers for any given X y and Z, but all answers should be between 0 and 1.

Answer 1

It's easy, write in a matrix form:

$$\begin{array}{l}\left( {x - 1} \right)a + \left( {x - \frac{1}{2}} \right)b + xc = 0\\ya + \left( {y - \frac{{\sqrt 3 }}{2}} \right)b + yc = 0\\76a + 150b + 29c = 255z\end{array}$$

or

$$\left[ {\begin{array}{*{20}{c}}{x - 1}&{x - \frac{1}{2}}&x\\y&{y - \frac{{\sqrt 3 }}{2}}&y\\{76}&{150}&{29}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}a\\b\\c\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\{255z}\end{array}} \right]$$

so, by inversing the coefficients matrix, you will have the solution to the system of equations:

$$\left[ {\begin{array}{*{20}{c}}a\\b\\c\end{array}} \right] = {\left[ {\begin{array}{*{20}{c}}{x - 1}&{x - \frac{1}{2}}&x\\y&{y - \frac{{\sqrt 3 }}{2}}&y\\{76}&{150}&{29}\end{array}} \right]^{ - 1}}\left[ {\begin{array}{*{20}{c}}0\\0\\{255z}\end{array}} \right]$$

by doing this; you will have a, b and c in terms of x, y, z; so, you have the solutions as follows:

$$\left[ {\begin{array}{*{20}{c}}a\\b\\c\end{array}} \right] = \frac{{255z}}{{29\sqrt 3 + 47\sqrt 3 x + 195y}}\left[ {\begin{array}{*{20}{c}}{\sqrt 3 x - y}\\{2y}\\{\sqrt 3 \left( {1 - x} \right) - y}\end{array}} \right]$$

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Verification:

You have the following values by putting them in the question or this answer formulas:

a=0.501960784
b=0.501960784
c=0

x=0.75
y=0.433012702
z=0.444875048