elementary number theory - Solving a Linear Congruence

I've been trying to solve the following linear congruence with not much success:
19 congruent to $19\equiv 21x\pmod{26}$

If anyone could point me to the solution I'd be grateful, thanks in advance

Answer

Hint: $26 = 2 \cdot 13$ and the Chinese remainder theorem. Modulo $2$ we have to solve $1 \cong x \pmod 2$, that is $x = 2k + 1$ for some $k$, now solve $19 \cong 42k + 21 \pmod{13}$.