I solved today the equations:
$y=4\pmod 3$ and $y=3\pmod 4$ where $y$ is an integer.
(Probably infinite number of solutions) by writing down:
$3k_1=y-4$
$4k_2=y-3$
$\Rightarrow 3k_1=-1+4k_2$
and one solution (in integers) is for $k_2=5,k_1=1 \rightarrow y=19$
Is there a 'smarter' solving this kind of equations? Is there a sufficient/necessary conditions for a solution to exist?