Proving $d\ell= ad+b$ for integers that $d$ must divide $b$.

Question

Proving $d\ell= ad+b$ for integers that $d$ must divide $b$.

If $d\ell$ is a product we can write it as $\underbrace{d+d+d+\ldots}_{\ell \text{ times}}$ and $ad$ as a sum of $\underbrace{d+d+d+\ldots}_{a \text{ times}}$ which means $b$ must be a sum of $d$'s, but how can I prove that to be true?

Answer 1

Move $ad$ to the LHS and factorize $d$ out of both terms on the LHS. Since $d$ divides LHS then it must divide $b$ the only term left on the RHS as well.