Unable to use Chinese Remainder Theorem on a question. So I had recently learned about Chinese Remainder theorem, and I got a Question to solve:
Find X if-
$X\equiv 1\bmod 2\\X\equiv 2 \bmod 3\\X\equiv 3 \bmod 4\tag*{}$
So I went on to find the solution modulus, $N=24$ & $N_1=12\\ N_2 = 8\\ N_3 = 6\tag*{}$
And now I just had to find the three inverse modulus, but for the first one
$12y_1\equiv 1 \bmod 2\tag*{}$ there's no $y_1$ satisfying the equation. I was stuck on this part. Please Highlight what I am doing wrong.

$x \equiv 1 (\mod 2) \Rightarrow x-1$ divisible by $2$ $\Rightarrow x-1+2=x+1$ is divisible by $2$.
$x \equiv 2 (\mod 3) \Rightarrow x-2$ divisible by $3$ $\Rightarrow x-2+3=x+1$ is divisible by $3$.
$x \equiv 3 (\mod 4) \Rightarrow x-3$ divisible by $4$ $\Rightarrow x-3+4=x+1$ is divisible by $4$.
Because $x+1$ is divisible by $2;3;4$, it is also divisible by the least common multiple of them, which is $12$ (but not neccesarily the product of them or $24$, because these numbers are not relatively prime)
We can imply that $x \equiv -1 $(mod $12)$ or $x \equiv 11 $(mod $12)$, which means $x$ can be expressed in the form $x=12k+11$ for non-negative integers $k$.
