I have a system of modular equations all mod $p$. I know that if I have many different $p_i$ (not necessarily prime, just pairwise coprime), I can use Chinese Remainder Theorem, but I can't seem to figure out how to do it over a single modulus.
I have the system: $$ \begin{align} a*(576885) + k*(1323376) &\equiv 2001 \ (\text{mod } 1334537) \quad (1)\\ a*(576885) + k*(1161723) &\equiv 243 \ (\text{mod } 1334537) \quad (2) \\ \end{align} $$
I know that the system has a solution, I used a solver, but I would like to know how to get to the solution manually. Since $1334537$ is prime, can I use the usual methods? (i.e multiply $(2)$ by $-1$ and then add to $(1)$ and then use Extended Euclidean Algorithm to find the inverse of $1323376-1161723$)
Also, how would I go over a bigger system with more equations and variables? What tools can I use? Does linear algebra still hold?