$x\equiv 2\:(\text{mod }6)$ 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?