How do you find one and what is it exactly? For example, 2x is congruent to 3 (mod 5), what is a solution, what would it look like, how would you write it?
When you write $\displaystyle 2x\equiv 3(\bmod 5)$ you are asking to find all $\displaystyle x\in \mathbb{Z}$ so that $\displaystyle 5$ divides $\displaystyle 2x-3$.
Notice that $\displaystyle 2x\equiv 3(\bmod 5)$ if and only if $\displaystyle 2x\equiv 8(\bmod 5)$ if and only if $\displaystyle x\equiv 4(\bmod 5)$. That is it! We found the solution. When we write $\displaystyle x\equiv 4(\bmod 5)$ it means $\displaystyle x = 5k+4$ for some $\displaystyle k\in \mathbb{Z}$. Therefore, all solutions to $\displaystyle 2x\equiv 3(\bmod 5)$ have the form $\displaystyle x=5k+4$.
Notice that $\displaystyle 2x\equiv 3(\bmod 5)$ if and only if $\displaystyle 2x\equiv 8(\bmod 5)$ if and only if $\displaystyle x\equiv 4(\bmod 5)$. That is it! We found the solution. When we write $\displaystyle x\equiv 4(\bmod 5)$ it means $\displaystyle x = 5k+4$ for some $\displaystyle k\in \mathbb{Z}$. Therefore, all solutions to $\displaystyle 2x\equiv 3(\bmod 5)$ have the form $\displaystyle x=5k+4$.[/quote]
I guess my problem is that I do not see why 2x must equal 8.
Notice that $\displaystyle 2x\equiv 3(\bmod 5)$ if and only if $\displaystyle 2x\equiv 8(\bmod 5)$ if and only if $\displaystyle x\equiv 4(\bmod 5)$. That is it! We found the solution. When we write $\displaystyle x\equiv 4(\bmod 5)$ it means $\displaystyle x = 5k+4$ for some $\displaystyle k\in \mathbb{Z}$. Therefore, all solutions to $\displaystyle 2x\equiv 3(\bmod 5)$ have the form $\displaystyle x=5k+4$.[/quote]
So if I understand this correctly, in the case 6x is congruent to 10 (mod 15), then there is no solution since 6 never divides evenly into 10+15k, where k is an integer.
The congruence $\displaystyle 6x\equiv 10(\bmod 15)$ never has a solution. So see this say there is such an $\displaystyle x$ then $\displaystyle 15 | (6x - 10)$ which means $\displaystyle 15k = 6x - 10$ for some $\displaystyle k\in \mathbb{Z}$. But then $\displaystyle 15k - 6x = 10 \implies 3(5k - 2x) = 10$. This is a problem because $\displaystyle 10$ has no factor of $\displaystyle 3$.
Once you understood what I wrote above try generalizing the result. Let $\displaystyle a,n$ be positive integers and consider $\displaystyle ax \equiv b(\bmod n)$. Let $\displaystyle d=\gcd(a,n)$. Prove that if the congruence is solvable then $\displaystyle d$ divides $\displaystyle b$.