There is an error: $\rm\:w=7r\!+\!5\,\Rightarrow\,2w = 7(2r)\!+\!\color{#C00}{10} = 7(5k\!+\!1)\!+\!\color{#C00}{10} = 35k\!+\!\color{#C00}{17},\:$ not $\rm\:35k\!+\!\color{#0A0}{12}.$
Since $\rm\:35k\!+\!17 = 2w\:$ is even, $\rm\:k\:$ is odd, $\rm\:k = 2j\!+\!1,\,$ so $\rm\:w = (35(2j\!+\!1)\!+\!17)/2 = 35j+26.$
Remark $\ $ It is easier to do the division by $2$ before the substitution. Namely, we have $\rm\:2r = 5k\!+\!1\:$ so $\rm\:k\:$ is odd, $\rm\:k = 2j\!+\!1,\:$ thus $\rm\:r = (5k\!+\!1)/2 = (5(2j\!+\!1)\!+\!1)/2 = 5j\!+\!3.\:$ Therefore $\rm\:w = 7r\!+\!5 = 7(5j\!+\!3)\!+\!5 = 35j\!+\!26.$ Notice how the numbers remain smaller this way.
I emphasize again, it's much more intuitive if you learn about modular arithmetic (congruences). For many examples see my posts on Easy CRT (easy version of the Chinese Remainder Theorem)
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.
Best Answer
Hint : $\sum a_k\times10^k\equiv\sum a_k \pmod {9}$