Find all solutions to this system of congruences

Question

$$x \equiv 11 \pmod{84} $$ $$ x \equiv 23 \pmod{36}$$

I have the bulk of the work done for this;
$x=11+84j$
$x=23+36k$

$\Rightarrow 11+84j \equiv 23 \pmod{36}$
$\Rightarrow 84j \equiv 12 \pmod{36}$
$\Rightarrow 12j \equiv 12 \pmod{36}$
$\Rightarrow j \equiv 1 \pmod{36}$
$\Rightarrow j = 1 + 36n$

Thus this system is true for any $x$ of the form $x=11+84(1+36n)=95+3024n$.

However, I know from messing around when trying to solve this that all $x$ of this form are not the entirety of the $x$ which satisfy the system; that would be $x=95+252n$.
I've also observed that $252$ is the lowest common multiple of $36$ and $84$, but I don't know how to tie this together to give a concrete method to finding all the solutions.

Answer 1

Correct is: $\ 12j\equiv 12\pmod {36}\iff 36\mid 12(j-1)\overset{\rm cancel\ 12}\iff \color{#c00}3\mid j-1\iff \color{#0a0}{j = 1+3n}.$

Therefore $\ x = 11+84\,\color{#0a0}j = 11+84(\color{#0a0}{1+3n}) = 95 + 252n,\,$ as claimed.

Answer 2

$x \equiv 11 \pmod{84}$, so $x = 84k + 11$, and $x \equiv 23 \pmod{36}$, so $x = 36n + 23$. So $84k + 11 = 36n + 23$, and $84k - 36n = 12$. So $7k - 3n = 1$. So $3n = 7k -1$, and $n = \frac{7k-1}{3} = 2k + \frac{k-1}{3}$. Thus $3 | k-1$, and $k = 3t + 1$, and $n = 2(3t+1) + t = 7t +2$. So $x = 36n + 23 = 36(7t + 2) + 23 = 252t + 95$ with $t \in \mathbb{Z}$