Two Questions: (1) Is My Proof Method Legitimate, and (2) How Might Proof by Cases be Accomplished? Prove 5x-4 is even iff 3x+1 is odd.
I have two questions: First, for example, could we assume that 5x-4 is even such that
5x-4 = 2k, for some integer k
and then manipulate the above equation to produce 5x-4, which would be conducted as follows
5x-4 = (3x + 2x) + (-5 + 1) = 2k
3x + 1 = 2k - 2x + 5
       = 2(k-x+2)+1,
which shows that 3x+1 is odd, since (k-x+2) is an integer.
Would the above sequence of steps be acceptable?
My second question is, How might this proof be demonstrated by cases? I understand that there are easier methods of proof, but we were asked in class to do a proof by cases.
 A: The difference $(5x-4)-(3x+1)=2x-5$ is odd.
It's  sickening to hear that you are asked "to do a proof by cases", instead of invoking the overriding principle: Find the invariant! There are enough sad problems (e.g., the four color theorem) where chasing cases is indispensable so far.
A: Proof by cases might look something like this:


*

*Suppose $x$ is even. Then, $5x-4$ is even. Similarly, if $5x-4$ is even, then $x$ is even. Not then that $3x+1$ is odd.

*Suppose $x$ is odd. Then, $5x-4$ is odd. Similarly, if $5x-4$ is odd, then $x$ is odd. Note then that $3x+1$ is even.


These arguments can be formalized by writing $x = 2k$ for $x$ even, and $x = 2k+1$ if $x$ odd, but suffice it to say that we may permit the argument "an odd times an even is even; an odd times an odd is odd."
We can write these statements as follows:
$$
x\ \textrm{even}\ \Longleftrightarrow 5x-4\ \textrm{even} \implies 3x+1\ \textrm{odd} \\
x\ \textrm{odd}\ \Longleftrightarrow 5x-4\ \textrm{odd} \implies 3x+1\ \textrm{even}
$$
Apply the law of the excluded middle to these two cases to establish the bidirectional implication
$$x\ \textrm{even}\ \Longleftrightarrow 5x-4\ \textrm{even} \Longleftrightarrow 3x+1\ \textrm{odd}$$
A: Your steps look good for proving "$5x-4$ is even $\Rightarrow$ $3x + 1$ is odd", now you have to prove the other direction, that "$3x + 1$ is odd $\Rightarrow$ $5x-4$ is even". This may be what is meant by "cases*". (equivalently, you can prove the contrapositive, and show "$5x-4$ is not even (i.e. odd) $\Rightarrow$ $3x + 1$ is not odd (i.e. even)")
*: i.e. your case structure is something like:
Case 1: Assume $5x-4$ is even $\dots$
Case 2: Assume $5x - 4$ is odd $\dots$
A: This looks ok,but an easier way would be to set up a system of linear equations and try and solve the system. If it has a unique solution,then it should be the case that the solution of 3y+1 is odd and the solution of 5x-4 is even.Consider the following system: 
                       5x-4 = m 
                       3y+1 = n 
Now try and solve the system,preferably as a matrix. You should get an odd integer as the solution for 3y+1 and an even one for 5x-4.  
A: Hint $\,\ 3x\!+\!1\,$ odd $\iff 3x\,$ even $\iff \overbrace{3x+ \color{#c00}2(x\!-\!2)}^{\large 5x-4}\,$ even, $\ $ by  $\ $ even $\pm$ even $\,=\,$ even.
