$x\equiv 2\:(\text{mod }6)$ and $x \equiv 3\:(\text{mod }9)$
attempted solution:
$x = 2, 8, 14, 20,$
$x = 2+6m$
$x = 3, 12, 21, 30, 39$
x = $3+9m$
$2+6m = 3+9m$ $-1 = 3m$ $-1/3 = m$
$m $is not an integer, therefore there is no common solutions?
$x\equiv 2\:(\text{mod }6)$ and $x \equiv 3\:(\text{mod }9)$
attempted solution:
$x = 2, 8, 14, 20,$
$x = 2+6m$
$x = 3, 12, 21, 30, 39$
x = $3+9m$
$2+6m = 3+9m$ $-1 = 3m$ $-1/3 = m$
$m $is not an integer, therefore there is no common solutions?
I see no reason why both constants should be the same. It should be more like
$$2+6m=3+9n$$
Doesn't look as helpful, but if you rearrange it like this
$$2=3+9n-6m$$
you will see that the right side is divisible by 3, but the left side is not. Therefore, there are no solutions.
Hint $\rm \ \ \begin{eqnarray}\rm x &\equiv&\,\rm a\,\ (mod\ m)\\ \rm x &\equiv&\rm \,b\,\ (mod\ n)\end{eqnarray}\Rightarrow\: a\!+\!jm = x = b\!+\!kn\:\Rightarrow\:gcd(m,n)\mid jm\!-\!kn = b\!-\!a $
Hence $\rm\ b\!-\!a = \pm1\:\Rightarrow\:gcd(m,n)=1.\ $ Since this fails in your system, it has no solution.
Conversely, a solution exists if $\rm\ gcd(m,n)\mid b\!-\!a,\:$ see the Chinese Remainder Theorem (CRT).