<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] Another method, that's specific to this question:
If x is even, then 4x is 0, so y = 3. Then 3x+21=2. The LHS is odd but the RHS is even, so "x is even" leads to a contradiction. So x is odd, 4x = 4, y =7, and x =3.
</s>