Subtracting the first equation from the second, the third equation from the first, and the second equation from the third, we find
\begin{align*}
(b - 1) x + (c - 7) y &= (2 - 1) x + (6 - 7) y - ( - 4 - 7) = x - y - 11, \\
(a - 8) x + (b - 7) y &= (8 - 2) x + (5 - 6) y + 3 - 4 = 6x - y - 1, \\
(a - 2) x + (c - 6) y &= (8 - 2) x + (5 - 7) y + 3 - 7 = 6x - 2y - 4.
\end{align*}We can also express the original system as a linear system in terms of $u$ and $v,$ where
\begin{align*}
u &= a - 1, \\
v &= b - 6.
\end{align*}Adding/modifying equations, we can express the system as
\begin{align*}
(b - 1) x + (c - 7) y &= x - y - 11, \\
u x + (b - 7) y &= 3x - y - 1, \\
u x + (c - 6) y &= 4 - 2x + y.
\end{align*}We can add together the equations $u x + (b - 7) y = 3x - y - 1$ and $u x + (c - 6) y = 4 - 2x + y$ to get $2ux + (b + c - 13) y = 7 - x.$ On the other hand, adding the equations $(b - 1) x + (c - 7) y = x - y - 11$ and $u x + (c - 6) y = 4 - 2x + y,$ we get $(b + u) x + 2cy = -7.$ This means
\[2ux + (b + c - 13) y = 2(b + u) x + 4cy = -14.\]Since $x + y \neq -1,$ we can write $y = \frac{-x - 14}{x + 1},$ so the equation $2(b + u) x + 4cy = -14$ can be expressed as
\[2(b + u) x + 4c \cdot \frac{-x - 14}{x + 1} = -14(1 + x).\]Since $u = a - 1,$ $b + u = b + a - 1$ and $b + a = (b + u) + 1.$ Thus,
\[2(b + u) x + 4c \cdot \frac{-x - 14}{x + 1} = -14(1 + x) = c(2x + 14).\]This simplifies to
\[2(b + u) = c.\]Then
\[(b - 1) x + (c - 7) y = x - y - 11,\]or $bx - x + cy - 7y = -y - 11,$ which becomes $(b + u) x + 2cy = - y - 11.$
Therefore, $a + b + c = 2(b + u) = \boxed{-11}.$
Real numbers $a$, $b$, $c$, $x$, and $y$ satisfy
\begin{align*}
ax+by+c&=x+7,\\
a+bx+cy&=2x+6y-4,\\
ay+b+cx&=8x+5y-3,
\end{align*}
and $x + y \neq -1$. Find $a+b+c$.
1 answer