Anomalous Cancellation on 2-Digit Numbers

Theorem
There are exactly four anomalously cancelling vulgar fractions having two-digit numerator and denominator when expressed in base $10$ notation:

Proof
Let $\dfrac {\sqbrk {ax} } {\sqbrk {xb} }$ be an anomalously cancelling vulgar fraction.

Then we have:

From the last equation:

Therefore $\dfrac {9 a b} {10 a - b}$ must be an integer.

In particular, if $10 a - b$ is divisible by a prime greater than $9$, $\dfrac {9 a b} {10 a - b}$ cannot be an integer,

since none of $9,a,b$ can be a multiple of that prime.

Moreover, from $x \le 9$,

So we have to check whether $x = \dfrac {9 a b} {10 a - b}$ is an integer for $\dfrac b {10 - b} \le a < b$, when $b$ ranges from $2$ to $9$.

The only solutions for $\tuple {a, x, b}$ are $\tuple {1, 6, 4}, \tuple {1, 9, 5}, \tuple {2, 6, 5}, \tuple {4, 9, 8}$,

which corresponds to the fractions $\dfrac {16} {64}, \dfrac {19} {95}, \dfrac {26} {65}, \dfrac {49} {98}$.