Using elementary matrix row and column operations on the system of two diophantine equations, namely, $N=an+b$ and $N=cn+d$, where $n\in\mathbb{N}^0$, it can be shown that the intersection of these two arithmetic progressions is another arithmetic progression $N=(ac)n+c\delta+d$ where $\delta\in\mathbb{N}:a|\left(c\delta+d-b\right)$. 


For example the intersection of $N=5n+3$ and $N=7n-2$ by the above formula is $N=35n+33$


Is there a way to transform $\delta$ such that the condition of divisibility is eliminated?



Intersection of two arithmetic progressions
You do not need a divisibility criterion, the intersection of two such arithmetic progressions can be found using the chinese remainder theorem.


In your example notice that any such $N$ in the intersection satisfies:


$N \equiv 3 \bmod 5$


and


$N \equiv 5 \bmod 7$ 


Solving gives $N \equiv 33 \bmod 35$.


