Solve the linear congruence $28x \equiv 63 \pmod {105} $
Now $gcd(28,105)=7$ and $7|63$ so it has solution The congruence is equivalent to $4x \equiv 9 \pmod {15}$ Now $gcd(4,15)=1$ So $4u+15v=1$ Here $u=4$, $v=-1$ Implies $4(4)+15.(-1)=1$ Implies $4(4) \equiv 1 \pmod {15}$ Now $4(36) \equiv 9 \pmod {15}$ Now there are how many solution of $28x \equiv 63 \pmod {105} $ .what they are..
As you've already noted, you're trying to solve $15|4x-9$, which is equivalent to $15|4x-24$ and hence $15|x-6$.