How to prove if $\frac{20b}{19b-20}=a$ is possible or not, where $a$ and $b$ are positive integers? This is a follow up question to the one posted here. (The method posted here works for all cases where there isn't a coefficient in front of $b$ in the denominator)
Here is the problem:
$$\frac{20b}{19b-20}=a$$
$a$ and $b$ have to be positive integers. Prove whether the equation is possible or not possible. By the way, the answer is that it isn't possible.
What I've attempted
My first method was to check if $20b$ are divisible by $1$ or $5$. This didn't work, as you will see. We know $19b-20$ will never be $1$. I tried to check whether $19b - 20$ will be a multiple of $5$, however this approach was wrong, as it may be a multiple of $5$, but that does not mean that it is a factor of $20b$.
My second attempt was playing with the equation algebraically. I got to $20(b+a)=19(ba)$. This shows that $ba$ has to be even, however it does not prove whether the equation is possible or not.
I don't know how to approach these problems that require you to disprove the whether an expression is divisible, especially if the variables are both present in both the denominator and the numerator. Could someone please provide some assistance in proving/disproving this? Thanks for your time.

If $b=1$, then we get $a=-20$ which is not a positive integer.
In the following, $b\ge 2$.
We have
$$a=\frac{20b}{19b-20}=\frac{19b-20+b+20}{19b-20}=1+\frac{b+20}{19b-20}$$
So, $\frac{b+20}{19b-20}$ has to be a positive integer.
Then, we have to have
$$\frac{b+20}{19b-20}\ge 1\implies b+20\ge 19b-20\implies 18b\le 40\implies b\le 2$$
If $b=2$, then we get $a=\frac{20}{9}$ which is not an integer.
