<s>[INST] Solve the systems of equations for real numbers $x$, $y$, $z$ Solve the system of equations in real solutions:
$x+2y+z$ $=$ $5(x+y)(y+z)$,
$y+2z+x$ $=$ $7(y+z)(z+x)$,
$z+2x+y$ $=$ $6(z+x)(x+y)$,
I substituded $a=x+y$, $b=y+z$ and $c=z+x$ and got:
$5ab-a-b$ $=$ $0$, 
$7bc-b-c$ $=$ $0$,
$6ac-a-c$ $=$ $0$, and got to the conclusion where all equals $0$, but can't prove it
Thank you for your help!
 [/INST] Considering $$5ab-a-b=0\tag 1$$ $$7bc-b-c=0\tag 2$$ $$6ac-a-c=0\tag 3$$ extract $b=\frac{a}{5 a-1}$ from $(1)$ (assuming $a\neq \frac 15$), $c=\frac{a}{6 a-1}$ from $(3)$ (assuming $a\neq \frac 16$). Replace in $(2)$ and simplify to get $$\frac{2 (1-2 a) a}{30 a^2-11 a+1}=0$$
Lord Shark's solution is better
</s>