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...

Write this linear system in matrix form:
$$\begin{pmatrix}3&7\\4&5\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}\equiv\begin{pmatrix}2\\7\end{pmatrix}$$
and multiply both sides by the adjugate matrix, which is  invertible mod. $8$:
\begin{align}
&\begin{pmatrix}5&1\\4&3\end{pmatrix}\begin{pmatrix}3&7\\4&5\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}\equiv\begin{pmatrix}5&1\\4&3\end{pmatrix}\begin{pmatrix}2\\7\end{pmatrix}\\[1ex]
\iff &\begin{pmatrix}3&0\\0&3\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}\equiv\begin{pmatrix}1\\5\end{pmatrix}\\[1ex]
\iff 3&\begin{pmatrix}3&0\\0&3\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}\equiv3\begin{pmatrix}1\\5\end{pmatrix}\\[1ex]\iff & \begin{pmatrix}1&0\\0&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}\color{red}x\\\color{red}y\end{pmatrix}\equiv\begin{pmatrix}\color{red}3\\\color{red}7\end{pmatrix}.
\end{align}
