jmc

Since these equations look symmetric, at least on the left-hand side, we add all three together. This gives us $$\begin{array}{rl}& a(x+1+y)+b(y+x+1)+c(1+y+x)\\ & =x+7+2x+6y+4x+y.\end{array}$$ We then factor out $(1+x+y)$ on the left and simplify the right side to get $$\begin{array}{rl}(1+x+y)(a+b+c)& =7x+7y+7,\text{ so}\\ (1+x+y)(a+b+c)& =7(1+x+y).\end{array}$$

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