<s>[INST] 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.
 [/INST] Your problem is that $3x + 5 = 2k +1$ is not an assumption.  It is the conclusion you need to prove.
Your one and only assumption is that $3x = 2n$ for some integer $n$.
so you start with
$3x = 2n$.
.... then you do a bunch of steps ....
.... steps .....
.... and get in the end ........
Conclusion: $3x + 5 = 2(????????) + 1$ where $??????$ is some integer you come up with in you steps.
Let's see what happens when we try.  Let's take it nice and slow:
=======
$3x = 2n$.
$3x +5 = 2n + 5$
....hmmm,  we want $2(??????) + \color{red}1$ in the end so let's pull out the $+\color{red}1$ first.....
$3x + 5 = 2n+5 = 2n + (4 + \color{red}1)=(2n+4) +\color{red}1$
.... hmmm, okay that's the $+1$ now we want $2(\color{red}{??????}) + 1$.  To get the So we need to factor then $2$ out of $2n+4$ and see what we have left.... that will bee the $\color{red}{??????}$
$3x + 5 = (\color{red}{2n+4}) + 1$
$3x + 5= 2(\color{red}{n + 2}) + 1$
.... and that's it......
Conclusion:  $3x+5 = 2(\color{red}{n + 2})+1$.
The $??????$ we wanted turns out to be $\color{red}{n+2}$ an we have
$3x + 5 = (3x+4) + 1 = (2n+4) + 1  = 2(\color{red}{n+2}) + 1$.
And because $\color{red}{n+2}$ is an integer if we let $k = n+2$ be that integer $3x+5 = 2k + 1$ and so... $3x + 5$ is odd.
=======
Although if you want to work backwards
Conclusion:  $3x + 5 = 2k +1$ ..... and we want to solve for $k$ to show it is possible...
$3x + 5 -1 =2k + 1-1$
$3x +4 = 2k$
$k = \frac {3x + 4}2 = \frac {3x}2 + 2$.
.... but is $\frac {3x}2 + 2$ an integer?????
Well, $3x$ is even.  So there is an integer $n$ so that $3x = 2n$ so
$k = \frac {3x}2 +2 = \frac {2n}2 + 2 = n+2$.
So $k=n+2$ is the integer we want to conclude $3x+5 =2k +1$.
If we did it this way our proof would go:

$3x$ is even so there is an integer $n$ so that $3x = 2n$.  Let $k = n+2$; that is an integer.


$2k + 1 = 2(n+2)+1 = 2n + 5 = 3x + 5$.


So $3x+5 = 2k +1$ and that is odd.

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