I'm stuck with this exercise:
Find all $(a,b) \in \Bbb Z^2$ such that $b \equiv 2a \pmod 5$ and $28a+10b=26$
It's from my algebra class, we are looking into diophantic and congruence equations.
I started by looking to the $(a,b)$ that would solve the equation: $14a+5b=13$, those would be of the form $a=-13+5s$ and $b=39-14s$. Here is where I don't understand what I should do.
I've looked into $a$'s congruence mod 5 and got $5s \equiv 3 (5)$.
If $b$ should be two times $a$ then it would be: $10s\equiv1(5)$. Right?
So if I'm on the right path I still don't see how should I make to combine the first solution for b and this congruence requirement. What should I try? Thanks a lot.