<s>[INST] Solving a linear system of reciprocals. Solve for $\begin{cases}\frac{1}{x} +\frac{1}{y}+\frac{1}{z}=0\\\frac{4}{x} +\frac{3}{y}+\frac{2}{z}=5\\\frac{3}{x} +\frac{2}{y}+\frac{4}{z}=-4\end{cases}$
I turn the equations into $\begin{cases}yz+xz+xy=0\\4yz+3xz+2xy=5xyz\\3yz+2xz+4xy=-4xyz\end{cases}$
Not sure if I am doing fine
 [/INST] how about this, let:
$$X=\frac1x,\,Y=\frac1y,\,Z=\frac1z$$
and it is much easier to solve the following:
$$\begin{pmatrix}
1&1&1\\
4&3&2\\
3&2&4
\end{pmatrix}
\begin{pmatrix}
X\\
Y\\
Z
\end{pmatrix}=
\begin{pmatrix}
0\\5\\-4
\end{pmatrix}
$$
</s>