elementary number theory - Modular Arithmetic.......

I know that a duplicate of this question is posted here ,but i am still
posting this because the post contain only hint.I solved this using myself ,and

i want to verify it.

Question

Suppose that $a$ and $b$ are integers, $a ≡ 4 (mod 13)$, and
$b ≡ 9 (mod 13)$.
Find the integer $c$ with $0 ≤ c ≤ 12$ such that

1. $c ≡ 9\,\,a (\text {mod} \,\,13)$

My attempt

Given

$\Rightarrow$ $a ≡ 4 (mod 13)$

we can write it using symmetric as,

$\Rightarrow$ $4 ≡ a (mod 13)$

$\Rightarrow$ $b ≡ 9 (mod 13)$.

We can write above as

$\Rightarrow$ $4*b ≡ 9a (mod 13)$.

and our question is

$c ≡ 9\,\,a (\text {mod} \,\,13)$

I am stuck here , any way to move forward?

Answer

Just compute, using $a\equiv 4\bmod 13$, that
$$9a\equiv 9\cdot 4=36\equiv 10\bmod 13.$$
hence we have $c=10$. We do not need to knwo $b$, by the way.