I have the following system of equations, \begin{align} \mathrm{(I)}&&\quad \mathbf{A} f_1(\mathbf{x})&=\mathbf{b}_1,\\ \mathrm{(II)}&&\quad\mathbf{A} f_2(\mathbf{x})&=\mathbf{b}_2,\\ \mathrm{(III)}&&\quad\mathbf{A} f_3(\mathbf{x})&=\mathbf{b}_3, \end{align} where I only need to solve any two equations (for the same $\mathbf{x}$) and $\mathbf{A} \in \mathbb{R}^{9\times N}$, $\mathbf{x} \in \mathbb{R}^{N}$ , $\mathbf{b}_i \in \mathbb{R}^{9}$ and $N \in \mathbb{N}$ (probably necessary to be at least bigger or equal to $18$). $\mathbf{A} f_i(\mathbf{x})$ is a matrix-vector product in the following sense: The functions $f_i(x)$ are given by \begin{align} f_1(x) &= \frac{1}{2}\left(x^2-x+\frac{1}{4}\right),\\ f_2(x) &= -x^2+\frac{3}{4},\\ f_3(x) &= \frac{1}{2}\left(x^2+x+\frac{1}{4}\right) \end{align} and act component-wise on vectors, e.g.: $$f_i(\mathbf{x})=\begin{pmatrix} f_i(x) \\ f_i(y) \\ f_i(z) \end{pmatrix}.$$ The system can therefore also be written as (just to provide some options) \begin{align} \mathrm{(I')}&&\quad \mathbf{A} (\mathbf{x}^2-\mathbf{x}) = \mathbf{A} \mathbf{x}^2-\mathbf{A}\mathbf{x}&=\mathbf{b}_1',\\ \mathrm{(II')}&&\mathbf{A} \mathbf{x}^2&=\mathbf{b}_2',\\ \mathrm{(III')}&&\quad\mathbf{A} (\mathbf{x}^2+\mathbf{x}) = \mathbf{A} \mathbf{x}^2+\mathbf{A}\mathbf{x}&=\mathbf{b}_3', \end{align} where then, e.g., $$\mathbf{x}^2 = \begin{pmatrix} x^2 \\ y^2 \\ z^2 \end{pmatrix}.$$ I need to solve any two of the equations (I) - (III). I do not care which two exactly. My problem is that the functions $f_i$ are linearly independent, which makes any system of two equations non-linear. I tried the whole day and lots of things. For example \begin{align} f_1(x) = f_3(x) -x \Rightarrow \end{align} \begin{align} \mathrm{(I)} &&\quad\mathbf{A} f_3(\mathbf{x}) - \mathbf{A}\mathbf{x} &= \mathbf{b}_1\\ \mathrm{(III)} &&\quad\mathbf{A} f_3(\mathbf{x}) &= \mathbf{b}_3 \end{align} which nearly gives a system of linear equations (in $f_3(\mathbf{x})$) if it wouldn't be for that pesky $-\mathbf{A}\mathbf{x}$.
I need to find an efficient solution which I can do numerically in a fast way (that's why I tried to get a system of linear equations). Is there any nice mathematical relation I am missing? I looked into regular chains but got scared. My system is not that complicated, or is it?
I'd also be grateful if anybody knows a way to determine if there is a solution at all. This is not a priority, as, if there is an efficient algorithm, I'd just let the computer try and possibly fail while moving on to a different set of matrices.
EDIT: Thanks to the comment by @AlexRavsky I looked into the Moore-Penrose inverse $\mathbf{A}^+ \in \mathbb{R}^{N\times 9}$ and can now rewrite the problem. The solutions to \begin{align} \mathrm{(I')}&&\mathbf{A} \mathbf{x}^2-\mathbf{A}\mathbf{x}&=\mathbf{b}_1'\quad\mathrm{and}\\ \mathrm{(II')}&&\mathbf{A} \mathbf{x}^2&=\mathbf{b}_2' \end{align} are therefore given by \begin{align} \mathrm{(I')}&&\mathbf{x}&=\mathbf{A}^+(\mathbf{b}_2'-\mathbf{b}_1')+\left[\mathbf{I}-\mathbf{A}^+\mathbf{A}\right]\mathbf{k}_1 \quad \mathrm{and}\\ \mathrm{(II')}&&\mathbf{x}^2&=\mathbf{A}^+\mathbf{b}_2'+\left[\mathbf{I}-\mathbf{A}^+\mathbf{A}\right]\mathbf{k}_2, \end{align} where $\mathbf{I}\in \mathbb{R}^{N\times N}$ is the identity matrix and $\mathbf{k}_1, \mathbf{k}_2 \in \mathbb{R}^N$ can be chosen arbitrarily. Keep in mind that in $\mathbf{x}^2$ the square function acts element-wise. These are $2 N$ equations for my $N$ unknowns. Now I need help solving these equations for $\mathbf{x}$.