Solve the following in $\mathbb{Z}_{15}$
$[2][x] + [14][y] = [11] $
$[8][x] +[7][y] =[12]$
solving for $x$ I got
$[x] +[14][y]=[9]$ by multiplying the second equation by 2
So taking the difference of this with the first equation gives me
$[x] = [2]$
solving for y I got
$[8][x] + [11][y] = 14$ by multiplying the first equation by 4
This gives me
$[4][y] = [2] \Rightarrow 4y \equiv 2 \pmod{15}$
By inspection, $y=8$ is a particular solution, so
$[y]=[8]$
Is that all the solutions for $(x,y)$ I have to find?
I'm just a bit unsure about this because I know that these congruence class problems may have more solutions than normal.
Also it would be nice if someone has a tip or a trick for estimating how many solutions these problems can have
See, $[x]$ stands for a number of the form $x + 15k$, where $k \in \mathbb Z$ and $0 \leq x \leq 14$.
So the above equation can be rewritten as: \begin{gather} (2 + 15k_1)(x+15k_2) + (14+15k_3)(y + 15k_4) = 11 + 15k_5 \\ (8 + 15k_6)(x+15k_7) + (7+15k_8)(y + 15k_9) =12 + 15k_0 \end{gather}
Now, expand on both sides, and you can check that they simplify to: \begin{gather} 2x + 14y = 11 + 15m \\ 8x+7y=12 + 15n \end{gather}
where $m,n$ are constants depending on the $k_i$, $0 \leq i \leq 9$.
Hence, solving this system of equations, we get: $$ x=\frac{30n-15m+13}{14}, y = \frac{60m-15n+32}{49} $$
Now, $x$ and $y$ are integers whenever $m,n$ exist such that the above equations are integrally satisfied.
Now, we twerk the equations slightly, to illustrate something: \begin{gather} 15(2n-m) = 14x-13 \\ 15(4m-n) = 49y-32 \end{gather}
Now, note that: exactly one value of $x$ between $0$ and $14$ is such that $14x-13$ is a multiple of $15$, namely $x=2$. Also there is only one value of $y$ between $0$ and $14$ such that $49y-32$ divides $15$, namely $y=8$.
Hence, we have to solve the equations \begin{gather} 2n-m = \frac{14*2-13}{15} =1 \\ 4m-n = \frac{49*8-32}{15} = 24 \end{gather}
Solving, we see that $m=7,n=4$.
Hence, the only solution of this system is $x=2,y=8$.
In the end, you can see what it boiled down to:the fact that $15 | 14x-13$ and $15 | 49x-32$ had unique solutions was the important point in the logic. If they had more than one solution, then the solutions of the set of equations would not be unique.
Observe the procedure carefully: I took the equations, removed the congruences, added two constants $15m$ and $15n$ to each equation, solved it normally, and checked when the solutions are integers, that's all. It's very simple, and you can do it for other problems as well.