I've been trying this for a little while now,
if $ax+by = d, \ $ then $a'x+b'y=d$
where $x>0\ $ and $0 \leq b' \leq x\ $ and $a,b,a',b',x,y, \in \mathbb{Z}$
My first thought is:
$ \begin{align} ax+by &= d \\ ax &= d - by \\ \end{align} $ Implies that (since $x \neq 0 $ ), $\begin{align} x &| d - by \\ d & \equiv by \pmod{x} \end{align} $ I'm not sure that helps. I also thought maybe from, $ax = d - by,\ $ I could say,
$ \begin{align} d - by &= qx +r, \ 0 \leq r \leq x \\ d &= by + qx + r \end{align} $
But I can't see a way of pulling the information from $r$ which ideally I could somehow rewrite as $ry$ so that $b'= r$. Any hints? Thanks again. Again, not technically homework just a problem from a book.