Mathematical way to solve integer numbers $217 = (20x+3)r+x$ Is there any mathematical way to find the integer numbers that solve the following equation:
$$217 = (20x+3)r+x$$

$$217 = (20x+3)r+x\\
217=20xr+3r+x$$
Now
$$(5x+\frac{3}{4})(4r+\frac{1}{5})=20xr+3r+x+\frac{3}{20}$$
Therefore
$$217+\frac{3}{20}=(5x+\frac{3}{4})(4r+\frac{1}{5})$$
Multiplying both sides by $20$ we get
$$4343=(20x+3)(20r+1)$$
There are only few ways of writing $4343$ as a product of two integers, in each case solve. Note that the product has to be of terms of the form $1 \pmod{20}$ and $3 \pmod{20}$.
P.S. Here is a faster way to get to the product:
$$217 = (20x+3)r+x$$
Note that $x=\frac{(20x+3)-3}{20}$. Therefore
$$217 = (20x+3)r+\frac{(20x+3)-3}{20}=(20x+3)(r+\frac{1}{20})-\frac{3}{20}$$
Move $\frac{3}{20}$ on the other size and multiply by $20$.
