I was given $$ 6x+14y=4 \space \mod 5 $$
I took this approach: $$ 6x+14y-5z=4, \space \text{ for some } z $$
Let $$ w=\frac{6}{(6,14)}x+\frac{14}{(6,14)}y $$
Then, $$ (6,14)w+5z=4 \quad , \quad 6w+5z=4 $$
where $w=-6$ and $z=8$
Hence, the general solution is
$$ w=-6+5s \quad , \quad z=8-6s $$
Substitute for $w$ yields
$$ \frac{6}{(6,14)}x+\frac{14}{(6,14)}y =-6+5s \quad,\quad 6x+14y=-6+5s $$ where $x=\frac{5}{6}s$ and $y=-\frac{3}{7}$ which is a particular solution
Hence, the general solution would be
$$ x=\frac{5}{6}s+14t \quad , \quad y=-\frac{3}{7}-8t $$
How can I check that this general solution holds true?
${\rm mod}\ 5\!:\ 4\equiv 6x+14y\equiv x-y\iff y\equiv x+1\ $
So the general solution is $\, (x,y) \equiv (x,x\!+\!1) \equiv (0,1) + x(1,1)$