Congruence system with same modulus and same variable? I have this particular problem:
$$\begin{cases}
3k \equiv 2 \pmod 8 \dots(*) \\
7k \equiv 2 \pmod 8 \dots(**)
\end{cases}
$$
I know that the solution for this is $k = 8q + 6$. I can find this easily if I solve one of the equations alone.
Now, let's assume I subtract $(*)$ from $(**)$. I get $4k\equiv0[8]$ and $k = 2q$, which isn't coherent.
For example, if I take $q = 2$ then $3k=12$, which does not satisfy $(*)$ (nor $(**)$).
I can't figure out where I messed up. Please help me to understand this.

You didn't mess up anywhere, except for your interpretation of what you did. The result that $k=2q'$ (I'm using a different letter to avoid confusion with the $q$ from $k=8q+6$) is correct — the solutions $k=8q+6$ indeed satisfy this property that you found:
$$k=8q+6=2q', \quad \text{where} \quad q'=4q+3.$$
When you have two equations to begin with, and you combine them e.g. by subtracting, what you get is an implication but NOT an equivalent equation. In other words:


*

*each $k$ that satisfies the original system of equations also satisfies the new equation;

*but values of $k$ that satisfy the new equation do not have to satisfy the original system.


As an example, think of the usual system of equations that I'm sure you've seen before; say, something like:
$$\begin{cases} 2x+3y=11, \\ 3x+4y=12. \end{cases}$$
When you subtract the first equation from the second, you'll get
$$x+y=1.$$
Does it follow from the original system? Of course, it does. Is it equivalent to the original system? Definitely, NOT: the original system has a unique solution, while the new equation alone has infinitely many solutions (pairs $(x,y)$ that satisfy it). You would need to put it together with one of the original equations (for example, as in the substitution method) to solve the original system completely.
