Prove divisibility law $\,b\mid a,c\,\Rightarrow\, b\mid ka + lc$ for all $k,l\in\Bbb Z$ We have to prove $b|a$ and $b|c \Rightarrow b|ka+lc$ for all $k,l \in \mathbb{Z}$.
I thought it would be enough to say that $b$ can be expressed both as $b=ka$ and $b=lc$. Now we can reason that since $ka+lc=2b$ and $b|2b$, it directly follows that $b|ka+lc$?
But I'm not sure if that works for any value of $k$ and $l$ (namely $k$ and $l$ are defined through quotient between $a$ and $c$, respectively).
What am I missing?

An alternative presentation of the solution (perhaps slightly less elementary than the already proposed answers) is to work in the quotient ring. 
You can write that in $\mathbb{Z}/b\mathbb{Z}$
\begin{align*}
 \overline{ka+lc}&=\bar{k}\bar{a}+\bar{l}\bar{c}\\
&=\bar{0}+\bar{0}\\
&=\bar{0}
\end{align*}
where $\bar{a}=\bar{c}=\bar{0}$ because $b | a$ and $b | c$.
