In a certain study group, the equation

$$
\frac{x+7}{x-8}-6=\frac{5 x-55}{7-x}
$$

is solved as follows:

$$
\begin{gathered}
\frac{x+7-6(x-8)}{x-8}=\frac{5 x-55}{7-x} \\
\frac{-5 x+55}{x-8}=\frac{5 x-55}{7-x} \\
\frac{5 x-55}{8-x}=\frac{5 x-55}{7-x}
\end{gathered}
$$

When they reached this point, a member of the study group, $A$, remarks: »Since the numerators are equal, the denominators must also be equal. This can only be true if $8=7$, which is absurd!«

To which $B$ corrects: »It doesn't follow that $8=7$, but rather that $8-x=7-x$, which means $0=-1$, leading to a contradiction. Therefore, we are dealing with an impossibility, meaning there is no value of $x$ that satisfies our equation!«

In response, $C$ says: »That's not true; on the contrary! The equality of the denominators, along with the equality of the numerators, $5 x-55=5 x-55$, means $0=0$, indicating that we are dealing with an identity, and thus every value of $x$ satisfies our equation!«

To which $D$ adds triumphantly: »It's not true that every value of $x$ satisfies our equation, because the values $x=8$ and $x=7$ must be excluded!«

Who is right?
$\mathrm{Az}$

$$
\frac{5 x-55}{8-x}=\frac{5 x-55}{7-x}
$$

does not imply that the denominators are equal, but rather that the values of the identical numerators must each be 0. Therefore, neither $A$ nor $B$ is correct.

The denominators can never be equal, which is why the numerators must each vanish separately, and thus our equality cannot be an identity, even if we exclude the values $x=7$ and $x=8$. Therefore, neither $C$ nor $D$ is correct.

From our equation, the root is obtained from

$$
5 x-55=0
$$

which gives us

$$
x=11
$$