Through many operations in an exercise I've reached this point
$17≡4a+19b(mod$ $27)$
$8≡11a(mod$ $27)$
I want to find both a and b to have the answer I need, I've been trying to use the Chinese rest theorem but found it doesn't work in my case, I'm starting converting the second one into
$11a+27a≡1(mod$ $27)$
Despite this I'm not sure if is correct and I still don't know how to clear this $a$ in the second equation in order to replace it in the first.
Any help will be really appreciated