I have the congruence equation:
$$6x+y \equiv 19 \pmod{26}$$
One way to solve it is to start from:
$$y \equiv 19-6x \space \pmod{26}$$
and try all the $y\in \left \{ 0,\dots,25 \right \}$.
However my book states that there exist only $12$ possibilities for $x$, so I think the congruence equation
$6x+y \equiv 19 \pmod{26}$ can be simplified.
However I didn't succeed in doing it, can you help me?
Best Answer
$6x+y-19$ is to be divisible by $26$
hence $6x+y\equiv19\pmod2$
$\implies y\equiv1\pmod2$
If $y=2z+1, 6x+2z+1\equiv19\pmod{26}$
$\iff z\equiv9-3x\pmod{13}$