We seek to solve the sytem of equations:
$\left\{ \begin{array}{cc} \bar 3x + \bar2y &= \bar1 \\ \bar 5x + \bar1y &= \bar4 \end{array} \right.$
In $\mathbb{Z}_{19}$
The following properties of modular arithmetic are assumed:
$a\equiv c \pmod n$ and $b \equiv d \pmod n$ imply that:
$a\cdot b \equiv c\cdot d \pmod n$ and $a+b \equiv c+d \pmod n$
We have a system of equations mod 19. We can then simply add and subtract integer multiples of one equation to another and this will not change the congruence relationships, as such by doing linear combinations we get:
$\left\{ \begin{array}{cc} \bar 3x + \bar2y &= \bar1 \\ \bar 5x + \bar1y &= \bar4 \end{array} \right.$ $\implies$ $\left\{ \begin{array}{cc} -\bar {2}x + \bar1y &= -\bar3 \\ \bar 5x + \bar1y &= \bar4 \end{array} \right.$ $\implies$ $\left\{ \begin{array}{cc} -\bar {2}x + \bar1y &= -\bar3 \\ \bar 1x + \bar3y &= -\bar2 \end{array} \right.$ $\implies$
$\left\{ \begin{array}{cc} \bar 1x + \bar3y &= -\bar2 \\ -\bar {2}x + \bar1y &= -\bar3 \end{array} \right.$ $\implies$ $\left\{ \begin{array}{cc} \bar 1x + \bar3y &= -\bar2 \\ \bar7y &= -\bar7 \end{array} \right.$ $\implies$ $\left\{ \begin{array}{cc} \bar 1x + \bar3y &= -\bar2 \\ \bar 1y &= -\bar1 \end{array} \right.$ $\implies$
$\left\{ \begin{array}{cc} \bar 1x &= \bar1 \\ \bar 1y &= -\bar1 \end{array} \right.$
And thus we know that:
$\bar1x=\bar1 \iff x = 19k+1$ for $k\in \mathbb{Z}$
$\bar 1y = -\bar1 \iff y = 19k' -1 \iff y = 19k'' + 18$ for $k',k'' \in \mathbb{Z}$
Are these all the possible solutions? Is this the correct approach?