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.

We can solve this congruence equation in elementary way also. 
We shall write $[15]$ to denote the word mod 15. Fine?
Note that \begin{align*}
&7x\equiv 3[15]\\
-&15x +7x\equiv 3-15[15] ~~\text{because}~~ 15x\equiv 0\equiv 15[15]\\
-&8x\equiv -12[15]\\
&2x\equiv 3\left[\frac{15}{\gcd(15, -4)}\right]~~\text{since}~~ax\equiv ay[m]\Rightarrow x\equiv y\left[\frac{m}{\gcd(a, m)}\right]\\
&2x\equiv 3[15]~~\text{since}~~\gcd(15,-4)=\gcd(15, 4)=1\\
&2x\equiv 3+15[15]~~\text{since} 15\equiv 0[15]\\
&2x\equiv 18[15]\\
&x\equiv 9\left[\frac{15}{\gcd(15, 2)}\right]\\
&x\equiv 9[15]
\end{align*}
Thus the solution is given by $x\equiv 9[15]$.
