You solve modular systems of equations mostly just like normal systems of equations -- the main difference is that you have to be careful about things like division and square roots.
Guessing* at how you would solve the system normally, I'm going to say that you solved the second equation for $m$ and plugged it into the first equation. So, you have
$ m \equiv -2 - n \pmod{10}$
and so
$ 2n + 6m \equiv 2n + 6(-2 - n) = 2n - 12 - 6n = -4n - 12 \pmod{10}$
and thus
$-4n - 12 \equiv 2 \pmod{10}$
or
$-4n \equiv 14 \pmod{10}$
You would normally like to divide by $-4$ here. I'm going to assume you're familiar with the algorithm to find modular inverses of numbers -- e.g. if you want to divide by $3$, you could solve for its inverse and find $3 \cdot 7 \equiv 1 \pmod{10}$ so that multiplication by $7$ has the effect of dividing by $3$.
The problem here is that $4$ is not relatively prime to $10$ -- both are divisible by 2. You've probably seen the method to deal with this fact but have just forgotten, so I'll restate it here. If you can divide everything by the gcd, then you do that, and get
$-2n \equiv 7 \pmod 5$
If we were unable to divide everything by the gcd (e.g. if the right hand side was 13 instead of 14), then the equation has no solutions.
Anyways, you should now be able to solve $-2n \equiv 7 \pmod 5$, and thus finish solving the problem.
*: It's usually a good idea to explain how you think about going about a problem, so that the answers you get can be tailored to the way you think. This can often give better results than someone answering with a method geared towards how they think.