I was reading this exercise: If $x=\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}$, then the value of $x$ is? Solution: When $a+b+c\not =0$, from the given equalities we have $$a=(b+c)x, \,\,\, b=(a+c)x\, \, \, c=(a+b)x.$$ By adding them up, we obtain $2(a+b+c)x=a+b+c$, so $x=\frac{1}{2}$. Since $b+c\not =0$, $a+c\not =0$, $a+b\not =0$, if one $a,b,c$ is $0$, then, from the given equalities, the other two are zeros also, a contradiction. Therefore $abc\not =0$
When $a+b+c=0$, then $b+c=-a$, $a+c=-b$, $a+b=-c$, so $$x=\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=-1$$.
I have doubts regarding this solution, do you think it is correct? I mean this in the sense that the exercise does not prevent me or says as a condition that $a\not =b \not =c\not =a$.