Showing that if $3x$ is even then $3x+5$ is odd I'm learning the absolute basics of how to do proofs, and am really struggling.

If 3x is even then 3x+5 is odd.

This is the solution:

I get that even numbers are 2n and odd numbers are 2n+1. For the life of me, I CANNOT get it into that form shown below. I feel so dumb. I tried looking up other answers before posting, but   nothing I found is this basic.
Work:
-Assumptions-
3x = 2n
3x+5 = 2k+1
-Trying to make sense of 3x-
3x+5 = 2k+1
3x = 2k-4
-Plugging in 2k-4 for 3x-
2k-4 = 2n
2k = 2n+4
k = n+2
-Plugging in n+2 for k-
3x+5 = 2(n+2)+1
...This is where I gave up. I don't know where I'm going with this anymore.

We know that $3x=2n$. So:
$$\color{blue}{3x}+5=\color{blue}{2n}+5$$
Because the "blue quantities" are equal. Now:
$$3x+5=2n+5=2n+4+1$$
In this step I just wrote $5$ as $4+1$:
$$ 3x+5=2n+5=2n+4+1=2n+2\cdot 2+1$$
In this step I just wrote $4$ as $2 \cdot 2$.
$$ 3x+5=2n+5=2n+4+1=\color{green}{2n+2\cdot 2}+1=\color{green}{2(n+2)}+1$$
The last step is valid because the green quantities are equal. In the end:
$$3x+5=2(n+2)+1$$
This means that $3x+5$ is odd because is of the form $2h+1$ with $h$ integer(in particular $h=(n+2)$)
