Find $a+b+c$, given that $x+y\neq -1$ and  \begin{align*}
ax+by+c&=x+7,\\
a+bx+cy&=2x+6y,\\
ay+b+cx&=4x+y.
\end{align*}
Since these equations look symmetric, at least on the left-hand side, we add all three together.  This gives us  \begin{align*}
&a(x+1+y)+b(y+x+1)+c(1+y+x)\\
&\qquad\qquad=x+7+2x+6y+4x+y.
\end{align*} We then factor out $(1+x+y)$ on the left and simplify the right side to get \begin{align*}
(1+x+y)(a+b+c)&=7x+7y+7, \text{ so}\\
(1+x+y)(a+b+c)&=7(1+x+y).
\end{align*}

Since $x+y\neq -1$, we know $x+y+1\neq 0$, so we can divide both sides by that term and find that $a+b+c=\boxed{7}$.