Incongruent solutions to $7x \equiv 3$ (mod $15$) I'm supposed to find all the incongruent solutions to the congruency $7x \equiv 3$ (mod $15$)
\begin{align*}
7x &\equiv 3 \mod{15} \\
7x - 3 &= 15k \hspace{1in} (k \in \mathbb{Z}) \\
7x &= 15k+3\\
x &= \dfrac{15k+3}{7}\\
\end{align*}
Since $x$ must be an integer, we must find a pattern for $k$ that grants this.  We know that $\frac{k+3}{7}$ must be equal to some integer, say $m$.  Solving for $k$, we have $k=4+7m$.
Substituting this into our value for $x$, we get:
\begin{align*}
x & = \dfrac{15(4+7m) + 3}{7}.\\
&= \dfrac{63}{7} + \frac{105m}{7}.\\
&= 9+15m.
\end{align*}
So, $x = 9+15m, m\in \mathbb{Z}.$
So, is this what I was looking for?  I'm not exactly sure what is meant by incongruent solutions.

$$x = 9+15m, m\in \mathbb{Z}.$$
is the same as 
$$x \equiv  9 \pmod{15} \,.$$
Equations of the type $ax \equiv b \pmod{n}$ sometimes have no solution, sometimes have one solution $\pmod{n}$ and sometimes they have two or more (incongruent solutions) $\pmod{n}$. 
In this case, there is only one, the one you found. A much easier way to find it is to multiply your equation by $2$ and use 
$$14x \equiv -x \pmod{15}$$
