Well if $(2a + 11)$ is a multiple of $9$, then
$$(2a + 11) = 0\ (mod \ 9)$$
so $$(2a + 2) = 0\ (mod\ 9)$$
and since $(2a + 2)$ is even, it follows that
$$(a + 1) = 0\ (mod\ 9)$$
Well if $(2a + 11)$ is a multiple of $9$, then
$$(2a + 11) = 0\ (mod \ 9)$$
so $$(2a + 2) = 0\ (mod\ 9)$$
and since $(2a + 2)$ is even, it follows that
$$(a + 1) = 0\ (mod\ 9)$$