If $\frac{1}{a} + \frac{1}{b} + \frac{1}{a+x} = 0$ ; $\frac{1}{a} + \frac{1}{c} + \frac{1}{a+y}$ ; $\frac{1}{a} + \frac{1}{x} + \frac{1}{y} = 0$. 
If $a \neq 0$ , $b \neq 0$ , $c \neq 0$ and if :- $\frac{1}{a} + \frac{1}{b} + \frac{1}{a+x} = 0$ ; $\frac{1}{a} + \frac{1}{c} + \frac{1}{a+y}=0$ ; $\frac{1}{a} + \frac{1}{x} + \frac{1}{y}  = 0$ , find $(a+b+c)$ .

What I Tried :- No information is given about $x$ and $y$ . So I thought of putting $x = y = 1$ , and this silly thing came out in the end .
Now, as $x = y = 1$ , I have $a = \frac{1}{-2}$ from the $3$rd equation .
So from the $1$st equation I get :- $$\frac{1}{a} + \frac{1}{b} + \frac{1}{a+x} = 0$$
$$ \rightarrow -2 + \frac{1}{b} + 2 = 0$$
$$ \rightarrow \frac{1}{b} = 0$$
This definitely looks absurd (also it's given that $b \neq 0$), so I guess putting $x = y = 1$ was a big mistake .
I don't have any other cool ideas for now as I see that doing it algebraically is going to include a lot of simplification and stuffs, and since there are $5$ variable there must be some shortcut of this .
Can anyone help?

From the first and the second equality we obtain:
$$x=-\frac{a(a+2b)}{a+b}$$ and $$y=-\frac{a(a+2c)}{a+c}.$$
Thus, $$\frac{1}{a}-\frac{a+b}{a(a+2b)}-\frac{a+c}{a(a+2c)}=0$$ or
$$(a+2b)(a+2c)=(a+b)(a+2c)+(a+c)(a+2b)$$ or
$$a(a+b+c)=0$$ or $$a+b+c=0.$$
