4. Given $\left\{\begin{array}{l}\frac{x+9 y}{9 x+7 y}=\frac{m}{n}, \\ \frac{x+9 y}{9 x+8 y}=\frac{m+a n}{b m+c n} .\end{array}\right.$

If all real numbers $x, y, m, n$ that satisfy (1) also satisfy (2), then the value of $a+b+c$ is $\qquad$.
$4 \cdot \frac{41}{37}$. For (1), by the theorem of the ratio of sums and differences, we get
$$
\begin{aligned}
\frac{m+a n}{b m+c n} & =\frac{(x+9 y)+a(9 x+7 y)}{b(x+9 y)+c(9 x+7 y)} \\
& =\frac{(1+9 a) x+(9+7 a) y}{(b+9 c) x+(9 b+7 c) y} .
\end{aligned}
$$

From (2), we have $\frac{x+9 y}{9 x+8 y}=\frac{(1+9 a) x+(9+7 a) y}{(b+9 c) x+(9 b+7 c) y}$.
The above equation holds for all real numbers $x, y, m, n$ if and only if there exists a non-zero constant $k$ such that
$$
\left\{\begin{array}{l}
1+9 a=k, \\
9+7 a=9 k, \\
b+9 c=9 k, \\
9 b+7 c=8 k .
\end{array}\right.
$$

From the first two equations, we get $a=0, k=1$; thus, from the last two equations, we solve for $b$
$$
\begin{array}{l}
=\frac{9}{74}, c=\frac{73}{74} . \\
\quad \therefore a+b+c=\frac{82}{74}=\frac{41}{37} .
\end{array}
$$