<s>[INST] 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?
 [/INST] 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.
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