Modular Arithmatic - Solving congruences I'm sure this is pretty basic but I'm struggling to understand how to go about solving this problem for my homework.  The question states "Solve the following congruences for x".  The first problem is $2x+1\equiv 4\pmod 5$.

There are many ways to solve the problem. The conceptually simplest, but most tedious, is to test one by one the possibilities $x\equiv 0\pmod{5}$, $x\equiv 1\pmod{5}$, and so on up to $x\equiv 4\pmod{5}$. Quickly we find that $x\equiv 4\pmod{5}$. (This approach would become quite unpleasant if $5$ were replaced by $97$.) 
It is simpler to use some algebra. So rewrite as $2x\equiv 3\pmod{5}$. Since $3\equiv 8\pmod{5}$, it is convenient to rewrite the congruence as $2x\equiv 8\pmod{5}$. Then since $2$ and $5$ are relatively prime, we can divide by $2$, getting $x\equiv 4\pmod{5}$.
A fancier version is to start from $2x\equiv 3\pmod{5}$. Now multiply both sides by $3$ (the modular inverse of $2$). We get $6x\equiv 9\pmod{5}$. But $6\equiv 1\pmod{5}$ and $9\equiv 4\pmod{5}$, so we conclude that $x\equiv 4\pmod{5}$. 
Remark: We used congruence notation throughout, since it is very important to get accustomed to it. But $2x\equiv 3\pmod{5}$ means that $5$ divides $2x-3$. So we want to solve $2x-3=5k$, that is, $2x=3+5k$. So we want to find a $k$ such that $3+5k$ is divisible by $2$. It is clear that $k=1$ works, giving $2x=8$ so $x=4$. Any number congruent to $4$ modulo $5$ will also work, giving answer $x\equiv 4\pmod{5}$. 
