Form the extended coefficients matrix and apply Gauss Reduction as much as possible.
To begin with, if $\,p=5\,$ then the first equation is $\,3y=4\Longrightarrow y=4\cdot 3^{-1}=4\cdot 2=3\,$ , and substituing in eq. 2 we get $\,3x+3=1\Longrightarrow 3x=-2=3\Longrightarrow x=1\,$ , so there's a unique solution: $\,(1,3)\,$
If $\,p\neq5\,$ we get
$\begin{pmatrix}5&3&4\\3&6&1\end{pmatrix}\stackrel{R_1/5}\longrightarrow \begin{pmatrix}1&3/5&4/5\\3&6&1\end{pmatrix}\stackrel{R_2-3R_1}\longrightarrow \begin{pmatrix}1&3/5&4/5\\0&21/5&-7/5\end{pmatrix}$
From here we get
$R_2\Longrightarrow \frac{21}{5}y=-\frac{7}{5}\Longrightarrow 21y=-7\stackrel{\text{if}\,\,p\neq 3}\Longrightarrow y=-\frac{7}{21}=-\frac{1}{3}$
and then
$R_1\Longrightarrow x+\frac{3}{5}y=\frac{4}{5}\Longrightarrow x=\frac{4}{5}+\frac{3}{5}\frac{1}{3}=1$
and the solution for $\,p\neq 3,5\,$ is $\,\displaystyle{\left(1\,,\,-\frac{1}{3}\right)}\,$
When $\,p=3\,$ the system is clearly inconsistent (watch thesecond equation!), whereas for $\,p=7\,$ , as the reduction process we carried on above shows at the end in the second row $\,(0\,,\,0\,,\,0)\,$ , we get one single (linearly independent) equation in two unknowns and thus there are several solutions: each pair of the form
$\left(x\,,\,y\right)\,\,\,,\,\,s.t.\,\,\,5x+3y=4\pmod 7$