<s>[INST] System of congruent equations Given the following system, how do you prove that there only is one solution congruent 8?
$$3x+7y \equiv 2 (\text{mod } 8)$$
$$4x+5y \equiv 7 (\text{mod } 8)$$
My main idea have been to solve the equation, then to come to the solution that there only is one solution, however I get flawed results to say the least. Here is my method:
$$\left\{
\begin{array}{ll}
3x+7y \equiv 2 (\text{mod } 8)
\\4x+5y \equiv 7 (\text{mod } 8)
\end{array}
\right.
\Leftrightarrow
\left\{
\begin{array}{ll}
6x+14y \equiv 4 (\text{mod } 8)
\\8x+10y \equiv 14 (\text{mod } 8)
\end{array}
\right.
\Leftrightarrow
\left\{
\begin{array}{ll}
6x+6y \equiv 4 (\text{mod } 8)
\\2y \equiv 6 (\text{mod } 8)
\end{array}
\right.
$$
$$
\Leftrightarrow
\left\{
\begin{array}{ll}
6x+18 \equiv 4 (\text{mod } 8)
\\2y \equiv 6 (\text{mod } 8)
\end{array}
\right.
\Leftrightarrow
\left\{
\begin{array}{ll}
-2x\equiv -14 (\text{mod } 8)
\\2y \equiv 6 (\text{mod } 8)
\end{array}
\right.
\Leftrightarrow
\left\{
\begin{array}{ll}
2x\equiv 6 (\text{mod } 8)
\\2y \equiv 6 (\text{mod } 8)
\end{array}
\right.
$$
Which then gives:
$$
\left\{
\begin{array}{ll}
x=4t+3
\\y=4s+3
\end{array}
\right.$$
$t$ and $s$ being arbitary whole numbers
So solution given congruent 8 is: 
$$x\equiv 3 (\text{mod } 8) \text{ or } x\equiv 7 (\text{mod } 8)$$
$$y\equiv 3 (\text{mod } 8) \text{ or } y\equiv 7 (\text{mod } 8)$$
However when you put that into the original equation, the answer is wrong. So where did i do wrong...
 [/INST] Don't multiply equations by $2$: as $2$ is not coprime to $8$ you're not getting equivalent equations.
One way: given:
$$\begin{align}3x+7y\equiv 2 \pmod 8\\4x+5y\equiv 7\pmod 8\end{align}$$
Take away the 1st equation from the 2nd:
$$\begin{align}3x+7y\equiv 2 \pmod 8\\x-2y\equiv 5\pmod 8\end{align}$$
Take away 3 times the 2nd equation from the 1st one:
$$\begin{align}13y\equiv -13 \pmod 8\\x-2y\equiv 5\pmod 8\end{align}$$
Divide the first equation by $13$, which is coprime to $8$:
$$\begin{align}y\equiv -1 \pmod 8\\x-2y\equiv 5\pmod 8\end{align}$$
Add the first equation, multiplied by 2, to the 2nd:
$$\begin{align}y\equiv -1\equiv 7 \pmod 8\\x\equiv 3\pmod 8\end{align}$$
Bonus reading about what I've actually been (partially) doing to the equations: https://en.wikipedia.org/wiki/Smith_normal_form
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