If $x + y = 5xy$ , $y + z = 6yz$ , $z + x = 7zx$ . Find $x + y + z$ . 
If $x + y = 5xy$ , $y + z = 6yz$ , $z + x = 7zx$ . Find $x + y + z$ .

What I Tried :
I used some clever ways to get $x + y + z = 26xyz$ , but I suppose we have some solution as a number .
All all $3$ to get :-
$$2(x + y + z) = 5xy + 6yz + 7zx$$
Or,
$$ 2(x + y + z) = (xy + xy + xy + xy + xy) + (yz + yz + yz + yz + yz + yz) + (zx + zx + zx + zx + zx + zx + zx)$$
That is,
$$ 2(x + y + z) = (xy + zx) + (xy + zx) + (xy + zx) + (yz + zx) + (yz + zx) + (yz + zx) + (yz + zx) + (xy + yz) + (xy + yz)$$
Now see that $(xy + zx) = x(y + z) = 6xyz$ , similarly $(yz + zx) = 5xyz$ and $(xy + yz) = 7xyz$
So
$$2(x + y + z) = 3(6xyz) + 4(5xyz) + 2(7xyz)$$
$$\Rightarrow (x + y + z) = \frac{52xyz}{2} = 26xyz$$
I tried till this , then I have no idea . Can anyone help?
 A: Assuming $x,y,z \neq 0$, we have that,
$$x + y = 5xy \iff \frac1x+\frac1y =5$$
$$y + z = 6yz \iff \frac1y+\frac1z =6$$
$$z + x = 7zx \iff \frac1z+\frac1x =7$$
then solve for $1/x$, $1/y$, $1/z$.
The case $x=0 \lor y=0 \lor z=0$ is trivial.
A: Taking @user's hint a little further (and also assuming $x,y,z,\neq 0$), writing
\begin{align}
u &=\frac{1}{x}\\
v&=\frac{1}{y}\\
w&=\frac{1}{z}
\end{align}
Yields
\begin{align}
u+v &= 5\\
v+w &= 6\\
z+u&=7
\end{align}
thus
$$\begin{pmatrix}
1 & 1 & 0\\
0 & 1 & 1\\
1 & 0 & 1\\
\end{pmatrix}
\begin{pmatrix}
u\\
v\\
w\\
\end{pmatrix}=
\begin{pmatrix}
5\\
6\\
7
\end{pmatrix}
$$
Thus
$$
\begin{pmatrix}
u\\
v\\
w\\
\end{pmatrix} = \frac{1}{2}
\begin{pmatrix}
1 & -1 & 1\\
1 & 1 & -1\\
-1 & 1 & 1
\end{pmatrix}
\begin{pmatrix}
5\\
6\\
7
\end{pmatrix}$$
Whence
$$\begin{pmatrix}
u\\
v\\
w\\
\end{pmatrix}=\frac{1}{2}\begin{pmatrix}
6\\
4\\
8\\
\end{pmatrix}$$
And returning to the representation of these $u, v, w$ in terms of $x, y, z$ gives the desired result.
A: Multiply both sides of $x+y=5xy$, $y+z=6yz$ and $z+x=7zx$ by $z,x$ and $y$ respectively and add them together to obtain:
$xz+zy+xy+xz+zy+xy=2(xz+zy+xy)=18xyz$ so $xz+zy+xy=xy+z(x+y)=xy+5xyz=9xyz$.
So $xy=4xyz$ and hence $z=\frac{1}{4}$, etc.
