When $n$ is divided by $14$, the remainder is $10$. What is the remainder when $n$ is divided by $7$? I need to explain this to someone who hasn't taken a math course for 5 years. She is good with her algebra.
This was my attempt:

Here's how this question works. To motivate what I'll be doing,
  consider  \begin{equation*} \dfrac{5}{3} = 1 + \dfrac{2}{3}\text{.}
\end{equation*} This is because when 5 is divided by 3, 3 goes into 5
  once (hence the $1$ term) and there is a remainder of $2$ (hence the
  $\dfrac{2}{3}$ term). Note the following: every division problem can
  be decomposed into an integer (the $1$ in this case) plus a fraction,
  with the denominator being what you divide by (the $3$ in this
  case).
So, when $n$ is divided by 14, the remainder is 10. This can be
  written as  \begin{equation*} \dfrac{n}{14} = a + \dfrac{10}{14}
\end{equation*} where $a$ is an integer. 
We want to find the remainder when $n$ is divided by 7, which I'll
  call $r$. So \begin{equation*} \dfrac{n}{7} = b + \dfrac{r}{7}\text{,}
\end{equation*}  where $b$ is an integer. 
Here's the key point to notice: notice that  \begin{equation*}
\dfrac{n}{7} = \dfrac{2n}{14} = 2\left(\dfrac{n}{14}\right)\text{.}
\end{equation*} This is because $\dfrac{1}{7} = \dfrac{2}{14}$. 
Thus, \begin{equation*} \dfrac{n}{7} = 2\left(\dfrac{n}{14}\right) =
2\left(a + \dfrac{10}{14}\right) = 2a + 2\left(\dfrac{10}{14}\right) =
2a + \dfrac{10}{7} = 2a + \dfrac{7}{7} + \dfrac{3}{7} = (2a+1) +
\dfrac{3}{7}\text{.} \end{equation*} So, since $a$ is an integer, $2a
+ 1$ is an integer, which is our $b$ from the original equation. Thus, $r = 3$.

To her, this method was not very intuitive. She did understand the explanation. Are there any suggestions for how I can explain this in another way?

To motivate this to students who are just beginning to learn about remainders, it often helps to use real-world examples, e.g. if changing $\,n\,$ pennies (1 cent coins) into dimes (10 cent coins) leaves 8 cents, then changing these $\,n\,$ pennies into nickels (5 cent coins) leaves 3 cents, because we can first change them to dimes, then change each dime to 2 nickels, then change the remaining 8 pennies to 1 nickel and 3 pennies. Algebraically
$$\  n = 10q + 8\, =\, 5(2q) + 5 + 3\, =\, 5(2q+1) + 3$$ 
In modular language,  $\ {\rm mod}\,\ \color{}5\!:\ \begin{array}{r}\color{#c00}{10\equiv 0}\\ \color{#0a0}{8\equiv 3}\end{array}\,\ \Rightarrow\ \begin{array}{r}\color{#c00}{10}q+\color{#0a0}8\\ \,\equiv\ \color{#c00}0\,+\,\color{#0a0}3\end{array}\ $ by $ $  Congruence Sum, Product Rules.
Most students can easily visualize the money-changing. The goal is to help them translate that intuition into rigorous algebra, or, more powerfully, congruence algebra - as above.
