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?

I wouldn't use fractions, instead use the usual division algorithm, note that every $7$ numbers, there is a multiple of $7$, ever $14$ a multiple of $14$, et cetera to motivate writing a number as

$$n=14q+r$$

with $0\le r < 14$ each time. Then say every number is also of the form

$$n=7q'+r'$$

with $0\le r'< 7$ and emphasize that clearly $r$ is unique. This is, of course, because you just count how many up you have to go from the nearest multiple of $7$, if you are $4$ more, then you are clearly not $3$ more.
If you like visuals you can demonstrate to the student with a simple list
$$\underbrace{\color{red}{0}}_{7\cdot 0},1,2,3,4,5,6,\underbrace{\color{red}{7}}_{7\cdot 1}, 8, 9, 10, 11, 12, 13, \underbrace{\color{red}{14}}_{7\cdot 2},\ldots$$
If the student knows enough about well-ordering, you can make this rigorous rather than simply intuitive since you can look at natural numbers of the form 

$$\{n-7k: k\in\Bbb Z\}$$

and just define $r$ to be the minimal element of this set.
From either approach, you can write
$$14q+10=7(2q+1)+3$$
so that $q'=2q+1$ and $r'=3$.

Addendum: If you want to emphasize how things are evenly space for the other remainders, you can make the same list with different highlighting, here I'll do $14$ and highlight the related $7$ information

$$\underbrace{\color{red}{0}}_{14\cdot 0},1,2,\underbrace{\color{orange}{3}}_{7\cdot 1+3},4,5,6,7,8,9,\underbrace{\color{blue}{10}}_{14\cdot 0+10=7\cdot 1+3},11,12,13,\underbrace{\color{red}{14}}_{14\cdot 1},$$
  $$15,16,\underbrace{\color{orange}{17}}_{7\cdot 2+3},18,19, 20,21,22,23,\underbrace{\color{blue}{24}}_{14\cdot 1+10=7\cdot 3+3},25,26,27,\underbrace{\color{red}{28}}_{14\cdot 2},$$
  $$29,30,\underbrace{\color{orange}{31}}_{7\cdot 4+3},32,33,34,35,36,37,\underbrace{\color{blue}{38}}_{14\cdot 2+10=7\cdot 5+3},\ldots$$

This illustrates exactly how the $7q'+3$ numbers are distributed, and it's easy to see how they overlap with the $14q+10$ numbers every other one.
