Find all solutions to linear congruences Find all the solutions of each of the linear congruences below:
\begin{align}
&(a) &10x &\equiv 5 \pmod{15},\\
&(b) &6x  &\equiv 7 \pmod{26},\\
&(c) &7x  &\equiv 8 \pmod{11}.
\end{align}
I'm not entirely sure how to get these solutions by hand.  I know how to prove there are solutions.
For example:
$(a) \quad\gcd(10,15)=5 $ and we know $5|5$.
From there I set $10x+15y=5$ and divide through by $5$. Leaving us with $2x+3y=1$.  I know some solutions for $x$ and $y$, such as $x=-1$ and $y=1$, but that's all I have thus far.
 A: It seems like you're familiar with the theorem in the comments above. For part (a), by inspection, you can see that $x\equiv 2$ is a solution. Since $g=\gcd(10,15)=5$ and $m/g=15/5=3$, you know there are $5$ total solutions $\pmod{15}$, and the others are found just be adding $3$ successively until you've found all $5$. 
For (b), $\gcd(6,26)=2$ but $2\nmid 7$, so how many solutions can there be?
Part (c) is nice because $7$ and $11$ are coprime, so $7$ is actually invertible here. Try to find $7^{-1}\pmod{11}$, and then multiply both sides of $7x\equiv 8\pmod{11}$ by it to find the unique solution for $x$ modulo $11$.
A: If $\rm\:M\:|\:AX-B\:$ then $\rm\:D\:|\:M,A\:\Rightarrow\:D\:|\:B,\:$ so cancelling the maximal such $\rm\:D = gcd(M,A)\:$ yields
$$\rm M\:|\:AX-B\iff m\:|\:aX-b\quad for\ \ \ m\, =\, \frac{M}D,\,\ a\, =\, \frac{A}D,\,\ b \,=\, \frac{B}D$$
The latter equation has unique solution $\rm\: X \equiv b/a\pmod{m},\:$ since $\rm\:gcd(m,a)=1\:$ implies that $\rm\:a\:$ is invertible mod $\rm\,m\,$ by Bezout's Identity.
Finally note $\rm\ x\equiv c\pmod m\iff x\equiv  c\!+\!m,\, c\!+\!2m,\ldots, c\!+\!Dm\pmod{Dm = M}$
A: The second equation has no solution, as $6 x + 26 k$ is always even, and can never be 7.
