Solving a congruence system (which has other solutions) I am trying to solve this congruence system :
$$\begin{cases}
3x+7y\equiv 5[11]\\
8x+4y\equiv6[11]
\end{cases}
$$
I multiply first by $8$ and $3$ both equations resp.
$$\begin{cases}
24x+56y\equiv40[11]\equiv7 [11]\\
24x+12y\equiv18[11]\equiv7 [11]
\end{cases}
$$
We then substract the second equation from the first one.
$$44y\equiv0[11],$$
Since $gcd(11,44)=11$, then previous equation is equivalent to : $4y\equiv0[11]$, and then we find that :
$$y\equiv0[11],$$
and now I replace it in the following equation : $3x+7\times 0\equiv5[11]$ which is equivalent to :
$3x\equiv5[11]$, hence
$$x\equiv 9[11]$$
As a result $(9+11k,0+11k')$ with $k,k' \in \mathbb{Z}$ are solutions of the system.
Now my remark is the following : I noticed that also $(0+11k,7+11k')$ with $k,k' \in \mathbb{Z}$ and also $(7+11k,4+11k')$ with $k,k' \in \mathbb{Z}$ are also solutions of the system which I don't understand.
 A: Since upon elimination the two equations results in $44y \equiv 0 \pmod {11}$ which is true for all integers $y$, the two equations are not independent. As a result, we must express the values of $x$ in terms of $y$ using one of the equations. For example:
$$3x+7y\equiv 5 \pmod {11} \implies 12x + 28y \equiv 20 \pmod {11} \implies x\equiv 9+5y\pmod {11}$$
To verify that $(9+5y,y)$ gives all solutions, we substitute this to the second equation, and we have:
$$8x+4y \equiv 72+44y\equiv 6\pmod {11}$$
so both equations are satisfied. $(9,0), (0,7), (7,4)$ are particular solutions, modulo $11$.
A: A note on reducing $\quad ax\equiv b\pmod m\quad$ and a divider $\quad d$

*

*if $\gcd(m,d)=1$ then you can divide $a,b$ by $d$ and keep the same modulo

e.g. $\quad 6x\equiv 9\pmod{11}\iff 2x\equiv 3\pmod{11}$

*

*if $\gcd(m,d)\neq 1$ and $d$ divides $a,b$ then you can reduce the equation to a lower modulo

e.g. $\quad 6x\equiv 9\pmod{15}\iff 2x\equiv 3\pmod{5}$

*

*In all other cases, try multiplying by $a^{-1}$ or $b^{-1}$ depending on which is relatively prime with the modulo.

e.g. $\, 2^{-1}\equiv 6\pmod{11}\, $ thus $\, 2x\equiv 3\pmod{11}\iff 12x\equiv x\equiv 18\equiv 7\pmod {11}$

In the present case $44y\equiv 0\pmod{11}$, you get in first case for $d=4$ and second case for $d=11$.
$44y\equiv 0\pmod{11}\iff 11y\equiv 0\pmod{11}\iff y\equiv 0\pmod 1\quad$ which is always true for a random integer $y$.
Note that since $44$ is a multiple of $11$ the equation is also simply $0y\equiv 0\pmod{11}$, and this is just some $0=0$ equation, bringing no fuel to the problem (or rather that $y$ is a free variable).
A: The determinant of the matrix
$\left[ \begin{array}{cc}
   3 & 7 \\
   8 & 4
\end{array} \right]$
is $3 \cdot 4 - 8 \cdot 7 = 1 - 1 = 0 \pmod{11}$
Hence the two equations are linearly dependent.
A little work results in $10(3x+7y) = 30x + 70y = 8x + 4y \pmod{11}$
Hence there is no solution.
