Let $X = N(0,\frac{1}{\alpha})$, $Y = 2X + 8 + N_{y}$, and $N_{y}$ be a noise $N_{y} = N(0,1)$. Then, $P(y|x) = \frac{1}{\sqrt{2\pi}}exp\{ -\frac{1}{2}(y - 2x - 8)^{2} \}$ and $P(x) = \sqrt{\frac{\alpha}{2\pi}}exp\{-\frac{\alpha x^{2}}{2}\} $.
The mean vector is:
$$\mathbf{\mu} = \left( \begin{array}{c} \mu_{x}\\ \mu_{y}\end{array} \right)= \left( \begin{array}{c} 0\\ 8\end{array} \right).$$
The question is how to calculate the variance of Y.
I know that the correct answer is
$$\frac{4}{\alpha} + 1, $$
but don't know how to get from $$var(Y) = E[(Y-\mu_{y})^{2}] = E[(2X+N_{y})^{2}] $$
to
$$\frac{4}{\alpha} + 1. $$
Can anybody help? UPDATE: Thank you All for answers