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?
 A: For those who love tedious algebra, from
$$\begin{cases}
\frac1a+\frac1b+\frac1{a+x}=0\\
\frac1a+\frac1c+\frac1{a+y}=0\\
\frac1a+\frac1x+\frac1y=0
\end{cases}$$
We have:
$$a = \frac {-1}{\frac1x+\frac1y} = -\frac{xy}{x+y}$$
$$b = \frac {-1}{\frac 1a+ \frac 1{a+x}}=-\frac {a(a+x)}{2a+x}$$
$$c = \frac {-1}{\frac 1a+ \frac 1{a+y}}=-\frac {a(a+y)}{2a+y}$$
$$\begin{align}a+b+c &= a\left(1-\frac{a+x}{2a+x}-\frac {a+y}{2a+y}\right)
\\&=a\left(1-\frac{x-\frac{xy}{x+y}}{x-\frac{2xy}{x+y}}-\frac {y-\frac{xy}{x+y}}{y-\frac{2xy}{x+y}}\right)
\\&=a\left(1-\frac{x^2}{x^2-xy}-\frac {y^2}{y^2-xy}\right)
\\&=a\left(1-\frac x{x-y}-\frac y{y-x}\right)
\\&=a\left(1-\frac {x-y}{x-y}\right)\\&=0
\end{align}$$
The question remains, to find a more elegant solution.
A: 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 $$a+b+c=0.$$
A: Ok so by @player3236's hint , he said that taking $(x = y)$ will create a problem as then it would imply that $\frac{1}{a} + \frac{1}{a+x} = 0$ which is giving $\frac{1}{b} = 0$ and stopping me to move on.
For now, I am taking say $x = 1$ , $y = 2$ .
This gives :- $a = -\frac{2}{3}$ in the $3$rd equation .
Now in the $1$st equation, we have :-
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{a+x} = 0$$
$$\rightarrow \frac{-3}{2} + 3 + \frac{1}{b} = 0$$
$$\rightarrow \frac{1}{b} = \frac{-3}{2}$$
$$\rightarrow b = -\frac{2}{3}$$
Similarly solving for $c$ in the $2$nd equation we get :- $c =  \frac{4}{3}$ .
This gives $(a + b + c) = 0$.
Fortunately I got my answer to this question, but can I get some generalization of why $(a + b + c) = 0$ every time, i.e. , instead of putting $x = 1$ , $y = 2$ can we find $(a + b + c)$ .
