Solve the linear Diophantine equation $15x− 6y= 3$.
Using your solutions to this equation, solve $15x ≡ 3 \pmod 6$.
We start with
$$ 15x -6y = 3 $$
By Proposition 1.17, this has integer solutions if $g=\gcd(15, -6) = 3$ divides 3, which it does.
By inspection, a solution is $x=1, y=2$. So the general solution is, by Proposition 1.18,
$$ x = 1 -2t , \quad y =2 - 5t $$
for some integer $t$.
The linear congruence $15x \equiv 3 \pmod 6$ is equivalent to
$$ 15x-3 = 6y $$
for some integer $y$.
We've shown that $x=1-2t$ for some integer $t$. This is equivalent to
$$ x \equiv 1 \pmod 2$$
This is equivalent to a set of $g=3$ solutions in modulo 6.
$x \equiv 1 \pmod 6$
$x \equiv 3 \pmod 6$
$x \equiv 5 \pmod 6$
(All of these are congruent to 1 in modulo 2).
Alternative Solution
The linear congruence $15x \equiv 3 \pmod 6$ is equivalent to
$$ 15x-3 = 6y $$
for some integer $y$.
We've shown that $x=1-2t$ for some integer $t$. That is
$$ x \equiv 1-2t \pmod 6 $$
where $t=0, 1, 2$ because there are $g=3$ incongruent solutions. The solution are therefore
$x \equiv 1 \pmod 6$
$x \equiv -1 \pmod 6$
$x \equiv -3 \pmod 6$
Substituting with the equivalent non-negative residues
$x \equiv 1 \pmod 6$
$x \equiv 3 \pmod 6$
$x \equiv 5 \pmod 6$