You already know how to solve a linear equation.
x = 2y - 6
You already now how to substitute.
(2y - 6)^2+4y^2=36
Expanding and Simplifying are not new
4y^2 - 24y + 36 + 4y^2 = 36
8y^2 - 24y + 36 = 36
8y^2 - 24y = 0
y^2 - 3y = 0
If all else fails, we could here resort to the Quadratic Formula. In this case, since we lost the constant term, it factors easily.
y(y - 3) = 0
y = 0 or y = 3
My point here, is that you should not consider every problem you have not seen as a brand new experience. You have the tools. Use them.