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.
 A: I think you're getting confused by the $3x$ part. The $3x$ plays no role in the problem.
Suppose you're given any number that is even. I'll call it y. Now we want to show $y+5$ is odd.
Therefore by definition
$y=2k$ for some integer $k$
Now
$y+5 = 2k+5$
Now we just need to show that $2k+5$ is 2 times an integer plus 1
$2k+5 = 2(k+2)+1$
So $2k+5$ is odd because it can be written in the form 2*integer +1 where the integer here is $k+2$. So $y+5$ is odd since $y+5 = 2k+5$
So if any number is even. Then that number plus 5 is odd. It doesn't matter if the original number is 3x or 8z or 3x^2-5x+x^3 etc...
A: 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.

A: Ok so we know that $3x$ is even, that means we can write $3x=2n$ for a suitable $n$, since even means that the number is divisible by two without remainder. But then we have $3x+5=2n+5=2n+(4+1)=2n+2\cdot 2+1=2(n+2)+1$ which clearly is odd.
A: Remember here that $n$ represents ANY natural number. You got to the answer but you didn’t even realize it. That’s probably because you are thinking syntactically rather than semantically. What I mean is the literal string of symbols $2(n+2)$ didn’t register to you as even because it is not the same as the string $2n$. But $n+2$ is a natural number just like $n$ is. So the the strings $2(n+2)$ and $2n$ both represent even numbers, and so $2(n+2) + 1$ is odd just as you have shown in your last line.
A: 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 $h=(n+2)$)
A: You are dealing with a problem where you are given too much information. Here is another way of doing it (we'll use $m$ for an integer).
If $3x$ is even then $x$ must be even, so we can put $x=2m$. [This is a consequence of the fact that $2$ is a prime, or can be proved in various ways]
Then $3x+5=6m+5=6m+4+1=2(3m+2)+1$.
Now we can put $3m=n$, an integer, to get $2(n+2)+1$.
For an alternative proof, we could put $3m+2=n$ and then we get $2n+1$.
Note that the last two sentences are alternatives to one another. They both use $n$, but $n$ is defined differently in the two cases.
What I'm trying to do here is to unpack how the different expressions for the same thing relate to each other. If you get your head round that you will be flying.
