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
 A: It's much easier to solve the linear system for the reciprocals then take the reciprocal to get the result. 
A: If the reciprocals are freaking you out, just let $t=\frac1x,u=\frac1y,v=\frac1z,$ so you have the system $$\begin{cases}t +u+v=0\\4t +3u+2v=5\\3t +2u+4v=-4\end{cases}$$ Once you've solved this, as long as none of $t,u,v$ is $0,$ you can simply let $x=\frac1t,y=\frac1u,z=\frac1v.$ If one or more of $t,u,v$ is $0,$ then the system has no solution.
A: how about this, let:
$$X=\frac1x,\,Y=\frac1y,\,Z=\frac1z$$
and it is much easier to solve the following:
$$\begin{pmatrix}
1&1&1\\
4&3&2\\
3&2&4
\end{pmatrix}
\begin{pmatrix}
X\\
Y\\
Z
\end{pmatrix}=
\begin{pmatrix}
0\\5\\-4
\end{pmatrix}
$$
A: Observe that the system's matrix of the reciprocal of the unknowns is:
$$A=\begin{pmatrix}1&1&1&0\\
4&3&2&5\\
3&2&4&-4\end{pmatrix}\stackrel{R_2-4R_1,\,R_3-3R_1}\longrightarrow\begin{pmatrix}1&1&1&0\\
0&-1&-2&5\\
0&-1&1&-4\end{pmatrix}\stackrel{R_3-R_2}\longrightarrow$$
$$\begin{pmatrix}1&1&1&0\\
0&-1&-2&5\\
0&0&3&-9\end{pmatrix}$$
Try to finish the exercise now, taking into account that the third column represents $\;\cfrac 1z\;$ , the second $\;\cfrac1y\;$ and the first one $\;\cfrac1x\;$
