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...
 A: 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
A: $$4x+5y \equiv 7 (\text{mod } 8),$$
$$3x+7y \equiv 2 (\text{mod } 8).$$
Equation $1$ subtract equation $2$
$$4x+5y=7,$$
$$x+6y=5 \Rightarrow x=5-6y.$$
Substitute the second equation into the first one $$4(5-6y)+5y=7,$$
that is
$$20-24y+5y=7 \Rightarrow 20-19y=7,$$
since we are in modulo $8$
$$4-3y=7 \Rightarrow -3y=3 \Rightarrow y=-1 = 7.$$
Then, substitute $y=7$ into equation $1$ $$4x+35=7 \Rightarrow4x=-28 \Rightarrow 4x=-4\ \text{or}\ 12 \Rightarrow x=-1\ \text{or}\ 3.$$
Inputting the $(x,y)$ pair $(-1,7)$ doesn't satisfy either of the two equations, so we may disregard this and are left with
$$x=3,\quad y=7.$$
Probably very sloppy on notation, but I'm learning how to solve such systems in a number theory course myself so just thought I'd put my attempt in there.
A: 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.
A: 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}
