Find a+b+c, given that

Eliminating x and y, we get a + b + c = 10.

Find a+b+c, given that $x+y\neq -1$ and $\begin{align*} ax+by+c&=x+7,\\ a+bx+cy&=2x+6y,\\ ay+b+cx&=4x+y. \end{align*}$

$\begin{array}{|lrcll|} \hline (1)& ax+by+c&=&x+7 \\ (2)& a+bx+cy&=&2x+6y \\ (3)& ay+b+cx&=&4x+y \\ \hline (1)+(2)+(3): & (ax+by+c)+(a+bx+cy)+(ay+b+cx) &=& (x+7)+(2x+6y)+(4x+y) \\ & (ax+bx+cx)+(ay+by+cy) +(a+b+c) &=& 7x+7y+7 \\ & x(a+b+c)+y(a+b+c) +(a+b+c) &=& 7(x+y+1) \\ & (a+b+c)(x+y+1) &=& 7(x+y+1) \\ & (a+b+c) &=& \dfrac{7(x+y+1)}{(x+y+1)} \quad | \quad x+y\neq -1 \\ & \mathbf{a+b+c} &=& \mathbf{7} \\ \hline \end{array}$