<s>[INST] Find $x$ for $\overline{x}\oplus(\overline{7}\odot\overline{11}) = \overline{2} \odot (\overline{x}\oplus \overline{9})$ 
Find $x \in \mathbb{Z}$ such that the equality $\overline{x}\oplus(\overline{7}\odot\overline{11}) = \overline{2} \odot (\overline{x}\oplus \overline{9})$ is true in $\mathbb{Z_{12}}$

My reasoning:
Initially, defining an equivalence relation in $\mathbb{Z}$ with, such that $a\thicksim b \Leftrightarrow a - b$ is multiple of $n$ and $n>1$. So we have $a\equiv b \text{ (mod }n)$. For $x\in\mathbb{Z}$ and its equivalent class denoted as $\overline{x}$, we have
$$\overline{x} = \{a\in\mathbb{Z}: a\equiv z \text{ (mod }n)\}$$
So, the equivalent classes are
$\overline{0} = \{..., -2n, -n, 0, n, 2n, ...\}\\
\overline{1} = \{..., -2n+1, -n+1, 1, n+1, 2n+1, ...\}\\
\vdots\\
\overline{n-1} = \{..., -n-1, n-1, n-1, ...\}
$
And the congruent modulo $n$ in $\mathbb{Z}$ gives
$\mathbb{Z}_n = \mathbb{Z}/\sim = \{\overline{0}, \overline{1}, \overline{2}, ... , \overline{n-1}\}$
Once the equality given is true in $\mathbb{Z_{12}}$, such that $\mathbb{Z_{12}} = \{\overline{0}, \overline{1}, \overline{2}, ... , \overline{11}\}$, we know $x\in [0, 11]$. Also, once $n > 1$ the following properties are true:

$\overline{a}\oplus(\overline{b}\oplus \overline{c}) = (\overline{a}\oplus\overline{b})\oplus \overline{c} \quad (\text{Associativity of } \oplus)\\
\overline{a}\oplus\overline{b}= \overline{b}\oplus\overline{a} \quad (\text{Commutativity of } \oplus)\\
\overline{a}\odot(\overline{b}\odot\overline{c}) = (\overline{a}\odot\overline{b})\odot \overline{c} \quad (\text{Associativity of } \odot)\\
\overline{a}\odot\overline{b}= \overline{b}\odot\overline{a} \quad (\text{Commutativity of } \odot)\\
\overline{a}\odot(\overline{b}\oplus\overline{c}) = (\overline{a}\odot\overline{b})\oplus(\overline{a}\odot\overline{c}) \quad (\text{Distributivity of } \odot)\\
\text{For } \overline{a}, \overline{b}, \overline{c} \in\mathbb{Z}_n
$

Also,
$\overline{a}\oplus\overline{b} = \overline{a+b}$
and
$\overline{a}\odot\overline{b} = \overline{ab}$

Then, for the given question I tried the following

$\overline{x}\oplus(\overline{7}\odot\overline{11}) = \overline{2} \odot (\overline{x}\oplus \overline{9})$

As $\overline{7}\odot\overline{11} = \overline{77}$, we have $\overline{72} = \overline{0}$ for $\mathbb{Z}_{12}$, then $\overline{77} = \overline{5}$.
$\overline{x}\oplus\overline{5} = \overline{2} \odot (\overline{x}\oplus \overline{9}) \implies \overline{x}\oplus\overline{5} = (\overline{2}\odot\overline{x})\oplus(\overline{2}\odot\overline{9}) $
As $\overline{2}\odot\overline{9} = \overline{18}$, we have $\overline{12} = \overline{0}$ for $\mathbb{Z}_{12}$, then $\overline{18} = \overline{6}$.
$\overline{x}\oplus\overline{5} = (\overline{2}\odot\overline{x})\oplus(\overline{2}\odot\overline{9}) \implies \boxed{\overline{x}\oplus\overline{5} = (\overline{2}\odot\overline{x})\oplus\overline{6}}$
From the last step I couldn't finish it. Taking all the solutions for $x$ I could find that $x=11$, but I was not able to arrange the equality to get the solution.
For $\color{red}{\overline{x}\oplus\overline{5}} = \color{blue}{(\overline{2}\odot\overline{x})\oplus\overline{6}}$
$x = 1: \color{red}{\overline{6}} = \color{blue}{\overline{8}}\\
x = 2: \color{red}{\overline{7}} = \color{blue}{\overline{10}}\\
x = 3: \color{red}{\overline{8}} = \color{blue}{\overline{12} = \overline{2}}\\
x = 4: \color{red}{\overline{9}} = \color{blue}{\overline{14} = \overline{2}}\\
\vdots\\
x = 11: \color{red}{\overline{16} = \overline{4}} = \color{blue}{\overline{28} = \overline{4}}\\
$

How can I solve the question?
Sorry for the simple question, but I was not sure about how to proceed in the boxed step. In this context, what I could have done?

 [/INST] Guide:
Modulo operations is almost like regular algebraic manipulation with caution when it comes to division.
If the question is $x+5 = 2x + 6$, we would add $-x-6$ on both sides.
Now looking at
$$\overline{x} \oplus \overline{5} =(\overline2 \odot \overline{x}) \oplus \overline6$$
Try to perform $\oplus (\overline{11} \odot \overline{x}) \oplus \overline{6}$ on both sides. I will leave the remaining steps to you.
Another comment is personally I like to write $\pmod{12}$ as a reminder to what we are working on.
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