Why is solving $ax + by = c$ for $x$ and $y$ equivalent to solving $ax \equiv c$ (mod $b$), wherein $a, b, c, x, y \in\mathbb Z$?

Question

I am learning about modular equations. In my math textbook it says "For the diophantine equation $ax + by = c$, $a, b, c \neq 0$. To find which integers $x$ and $y$ satisfy the equation, the equality can be reformulated to a congruence: Determine $x$ so that $ax \equiv c$ (mod $b$)."

I don't see why it is possible to rewrite $ax + by = c$ to $ax \equiv c$ (mod $b$) when solving the diophantine equation. Why is this possible?

Answer 1

If $ax \equiv c \pmod b$, it must mean that $ax-c$ is divisible by $b$, so that: $$ax - c = kb \implies ax-kb = c$$ And if $ax+by=c$, then obviously $ax \equiv c \pmod b$, so the equations are equivalent.

Answer 2

This answer serves as a summary and elaboration of Elie Louis answer.

$$\eqalign{ & ax \equiv c\left( {\bmod b} \right) \Leftrightarrow ax\bmod b = c\bmod b \cr & ax = pb + r{\rm{ }} \wedge {\rm{ }}c = qb + r \cr & ax - c = pb + r - qb - r = pb - qb = b(p - q) = kb,{\rm{ }}k = p - q \Rightarrow b|(ax - c) \cr & ax - kb = c \Leftrightarrow ax + by,{\rm{ }}y = - k \cr & a,b,c,k,p,q,x,y \in {{\bf{Z}}^*} \cr} $$