Solving a Linear Congruence [duplicate]
HINT: $21\equiv-5\pmod{26}$, so $21\cdot5\equiv-25\equiv1\pmod{26}$, and $21\cdot5\cdot19\equiv1\cdot19\pmod{26}$.
If $19 \equiv 21 x \mod 26$ then we must have $19=21x+26y$ for some integer $y$. You could use the Euclidean Algorithm to solve this problem, finding integers $s,t$ such that $\gcd(21,26) = 21s+26t$
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 $19 \cong 42k + 21 \pmod{13}$.