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.
Since you don't know modular arithmetic we can instead give a direct divisibility proof.
$$2^{60}\!-2^{45}\!-2^{30}\!-2^{15}\!-1\, =\, \underbrace{2^{30}(2^{30}\!-1)-2^{5}(2^{40}\!-1)-2^{5}(2^{10}\!-1)}_{\Large\color{#c00}{ 96}\, n}\!-2^5\!-2^5\!-1$$
$\color{#c00}{96} = 2^5\cdot 3\,$ divides all RHS summands of form $\,2^J (2^{2K}\!-1)$ since $\,J\ge 5\,$ so $\,2^5$ divides $2^J,\,$ and $\,3\,$ divides $\,2^{2K}\!-1\ $ (put $\ a=2\ $ in $\,a+1\,$ divides $\,a^2-1\,$ divides $\,(a^2)^K-1).$
So it has form $\,\color{#c00}{96}\,n - 2^5\!-2^5\!-1 = 96n-65 = 96(n\!-\!1)+96-65 = 96(n\!-\!1)+\color{#0a0}{31}$
Best Answer
First note that $r(x)=ax+b$ for some $a,b\in\mathbb Z$
For $x=2$ you get $r(2)=-5$ (not $r(x)=-5$) $ \Rightarrow 2a+b=-5$. Do the same for $x=6$ to find r(6).Then find $a,b$.