Solve the simultaneous congruences: $2n + 6m\equiv 2 \pmod{10}$ and $n + m\equiv -2 \pmod{10}$ To solve the simultaneous congruences
$$2n + 6m\equiv 2 \pmod{10}$$
and
$$n + m\equiv -2 \pmod{10}$$
I tried to add the two congruences together so I got:
$$3n + 7m\equiv 0 \pmod{10}$$
But I am not sure if that's right and if it is, what to do next to calculate the two separate variables. If the question is like
$n\equiv x \pmod y$ then it's simple enough to calculate

What you did can easily be justified by
$$
-2n - 6m \equiv -2 \pmod{10},
$$
which gets you to $n + m \equiv -2n -6m \pmod{10}$, which is what you got. So 
$$3n \equiv -7m \equiv 3m \pmod{10}$$ and since $3$ is coprime to $10$, $3$ has a multiplicative inverse mod $10$ and so we can multiply out the $3$ on both sides. So $n \equiv m \pmod{10}$. Can you solve it now? 
In case you are interested the multiplicative inverse of $3$ in this case is $7$.
