Note that every coefficient in the first equation is a multiple of $3$, so you can divide through by $3$ and replace $3x+6y=30$ with the equivalent equation $x+2y=10$. You now have the system
Now it’s very apparent that if you subtract the top equation from the bottom one, you’ll get an equation that does not contain $x$:
or, after simplification,
You can easily solve that for $y$ and then substitute the result back into either of the original equations and solve the resulting equation for $x$.
Even if the coefficients in the first equation had not all been multiples of $3$, I could have used this approach. Suppose that the first equation had been $3x+5y=32$; dividing through by $3$ would have left me with the system
and I can still subtract one equation from the other to get a single equation without $x$ in it. The arithmetic is a little messier, but the principle is exactly the same and can be applied to any system of this kind.