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There is a video on youtube where a guy in the comments proves the following:

If $15l equiv 2 mod7$, then $l equiv 2 mod7$.

He does it like this:

15L = 2 (mod 7)

=> 15L = 7k + 2 for some k in the integers

Let k = 2T where T is an integer

=> 15L = 14T + 2

=> L = 14T - 14L + 2

=> L = 7(2T - 2L) + 2

Let H = (2T - 2L), then H is an integer.

=> L = 7H + 2

=> L = 2 (mod 7)

What bothers me is the following line "Let k = 2T where T is an integer". Why replace $k$ with a multiple of 2? We would get the same result if we do not even replace $k$ and leave it as it is for example:

L = 7k - 14L + 2

=> L = 7(k - 2L) + 2

=> L = 2 (mod 7)

Is my method correct as well or is there some deeper reasoning as to why would he replace $k$ with $2T$?

EDIT:
Here is the video if anyone is interested, the comment is made by the user RB:

https://www.youtube.com/watch?v=LInNgWMtFEs&lc=z23ts1qyloeagpzhm04t1aokgn15f4y4gqsns5m1d5p3rk0h00410.1543158282809495

I'd say you are right and the video is wrong. For instance, we could have $k=19$, which can't be written as $k=2T$ for an integer $T$.

By the way, a possibly simpler approach to the whole thing is to note that $15equiv 1 pmod{7}$.

So $2equiv 15lequiv l pmod{7}$.

Your concerns about the video are justified.

E.g. we have $15times 9=135=7times 19+2$ but there is no integer $k$ such that $15times 9=135=14k+2$.

Your method is okay.

On base of $7mid 14l$ you can also observe that: $$7mid 15l-2iff7mid 14l+l-2iff 7mid l-2$$

The argument is incorrect since $, 15,l = 7,k+2,$ does not imply $,2mid k,,$ (e.g. $ l,k = -5,-11$). Further, the argument uses unidirectional inferences where bidirectional inferences are required. Below is one correct way to do the proof in that manner.

$$begin{align}
15, l &equiv 2!pmod{! 7}\
iff exists, k!: 15,l &= 2+7,k\
iff exists, k!: l &= 2+7(k!-!2l)\
iff exists, j!: l &= 2+7,j\
iffqquadquad , l &equiv 2!pmod{! 7}
end{align}qquadqquad$$

It's simpler to use basic rules of modular arithmetic. By the Congruence Product Rule we deduce

$!bmod 7!:, color{#c00}{15equiv 1},Rightarrow, color{#c00}{15},lequiv color{#c00}1,lequiv l $ thus $ 2equiv 15,lequiv l$

If someone in their comments indeed says that, then your doubts are well justified — that step is plain wrong. In the setting of this question, $k$ does NOT have to be an even number, so we can NOT (in general) set it to be $2t$ for an integer $t$.

Quick example: if $l=9$, then $15l=15cdot9=135equiv2 mod7$, but then $k=133/7=19$ can't be represented as "$k=2t$ for an integer $t$".

Your solution, however, is perfectly correct!

Why not using

• 
  $color{blue}{15 equiv 1 mod 7}$?
  $$Rightarrow color{blue}{15}l equiv color{blue}{1}l equiv 2 mod 7$$