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!
On
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
If one of $a$, $b$ and $c$ are zero, then they all are. So assume none of them are, and divide your three equations by $ab$ etc. to get $$5=\frac1a+\frac1b,$$ $$7=\frac1b+\frac1c$$ and $$6=\frac1a+\frac1c.$$ So three linear equations for $1/a$, $1/b$ and $1/c$. There ought to be a unique solution to this; I'll let you find it!