<s>[INST] Determine all possible integers $x$ and $y$ s.t. $3x + 7y \equiv 14 \pmod{28}$ and $x + 3y \equiv 8 \pmod{28}$. Determine all possible ints $x$ and $y$ that satisfy $3x + 7y \equiv 14 \pmod {28}$ and $x + 3y \equiv 8 \pmod {28}$. The answer should be in terms of $x\equiv r$ and $y \equiv s$ where $r$ and $s$ are remainders.
Here is what I have tried. Am I headed in the right direction?
$8\cdot (3x + 7y) \equiv 14 \pmod {28}$ and $14 \cdot (x + 3y) \equiv 8 \pmod {28}$
$24x + 56y \equiv 112 \pmod{ 28}$ and $14x + 42y \equiv 112 \pmod {28}$
$24x + 56y \equiv 14x + 42y \pmod {28}$
This is where I start to get unsure,
$24x + 28y + 28y \equiv 14x + 14y + 28y \pmod {28}$
$24x \equiv 14x + 14y \pmod {28}$
$48x \equiv 28x + 28y \pmod {28}$
So then...?
$20x \equiv 1 \pmod {28}$?
Stuck here. Is this even the right idea?
 [/INST] We have
\begin{eqnarray*}
 3x+7y \equiv 14 \pmod{28} \\
x+3y \equiv 8 \pmod{28}.  
\end{eqnarray*}
Multiply the second equation by $3$ and then subtract the first
\begin{eqnarray*}
 3x+9y \equiv 24 \pmod{28} \\
2y \equiv 10 \pmod{28}.  
\end{eqnarray*}
So
\begin{eqnarray*}
 y &\equiv 5 \pmod{28} \\
x & \equiv 21 \pmod{28}  
\end{eqnarray*}
or\begin{eqnarray*}
 y &\equiv 19 \pmod{28} \\
x & \equiv 7 \pmod{28}.  
\end{eqnarray*}
</s>