Understanding simplification of constants in derivation of variance of regression coefficient In looking over TooTone's answer in Derive Variance of regression coefficient in simple linear regression, there's a step in line 3 below where $(\beta_0 + \beta_1x_i + u_i )$ is simplified to $u_i$ on the basis that only $u_i$ is a random variable.  I'm having trouble understanding why the $\beta_0 + \beta_1x_i$ terms can be ignored as a result of this?  Is anyone able to offer more detail on either a reason or an intermediate step that I'm not following?
\begin{align}
\text{Var}(\hat{\beta_1})
& = \text{Var} \left(\frac{\sum_i (x_i - \bar{x})(y_i - \bar{y})}{\sum_i (x_i - \bar{x})^2} \right) \\
&= \text{Var} \left(\frac{\sum_i (x_i - \bar{x})(\beta_0 + \beta_1x_i + u_i )}{\sum_i (x_i - \bar{x})^2} \right), \;\;\;\text{substituting in the above} \\
&= \text{Var} \left(\frac{\sum_i (x_i - \bar{x})u_i}{\sum_i (x_i - \bar{x})^2} \right), \;\;\;\text{noting only $u_i$ is a random variable} \\
&=  \frac{\sum_i (x_i - \bar{x})^2\text{Var}(u_i)}{\left(\sum_i (x_i - \bar{x})^2\right)^2} , \;\;\;\text{independence of } u_i \text{ and, Var}(kX)=k^2\text{Var}(X) \\
&= \frac{\sigma^2}{\sum_i (x_i - \bar{x})^2} \\
\end{align}
 A: If $X$ and $Y$ are independent random variables, then
$$
\textrm{Var}(X+Y)=\textrm{Var} X+\textrm{Var} Y.\qquad (1)
$$
This is because
$$
\mathbb E\left[(X+Y)^2\right]=\mathbb E[X^2]+2\mathbb E[XY]+\mathbb E[Y^2]\overset{indep!}=\mathbb E[X^2]+2\mathbb EX\mathbb EY+\mathbb E[Y^2]=\left(\mathbb EX+\mathbb EY\right)^2.
$$
Similarly if $a$ is deterministic and $X$ is random, then $$\textrm{Var}(aX)=a^2\textrm{Var}X,\qquad (2)$$ since
$$
\mathbb E[(aX)^2]=\mathbb E[a^2X^2]=a^2\mathbb E[X^2].
$$
Finally, if $a$ is deterministic and $X$ is random, then
$$\textrm{Var}(a+X)=\textrm{Var}X,\qquad (3)$$
since
$$
\mathbb E\left[a+X-\mathbb E(a+X)\right]^2=\mathbb E\left[X-\mathbb EX\right]^2.
$$
We can use these three properties to answer your question. You have a sequence of independent random variables $(u_i)$ and sequences of constants $(a_i),(b_i)$ where
$$a_i=\frac{x_i-\overline{x_i}}{\sum_i(x_i-\overline{x_i})^2},\qquad b_i=\beta_0+\beta_1 x_i$$
and you wish to simplify the expression
$$
\textrm{Var}\left(\sum_i a_i(b_i+u_i)\right).
$$
Using the third property,
$$
\textrm{Var}\left(\sum_i a_i(b_i+u_i)\right)=\textrm{Var}\left(\sum_i a_iu_i\right).
$$
Using the first property and independence,
$$
\textrm{Var}\left(\sum_i a_iu_i\right)=\sum_i \textrm{Var}(a_iu_i).
$$
Using the second property,
$$
\sum_i \textrm{Var}(a_iu_i)=\sum_i a_i^2\textrm{Var}u_i.
$$
Altogether, this fills in the missing reasoning.
A: The property used here is that $\mathbb{Var}(\text{const}+X) = \mathbb{Var}(X)$.  If you expand the terms out formally, from the second step to the third, you get:
$$\begin{equation} \begin{aligned}
\mathbb{Var} (\hat{\beta}_1)
&= \mathbb{Var} \left(\frac{\sum_i (x_i - \bar{x})(\beta_0 + \beta_1 x_i + u_i )}{\sum_i (x_i - \bar{x})^2} \right) \\[6pt]
&= \mathbb{Var} \left(\frac{\sum_i (x_i - \bar{x})(\beta_0 + \beta_1 x_i) }{\sum_i (x_i - \bar{x})^2} + \frac{\sum_i (x_i - \bar{x}) u_i}{\sum_i (x_i - \bar{x})^2} \right) \\[6pt]
&= \mathbb{Var} \left( \text{const} + \frac{\sum_i (x_i - \bar{x}) u_i}{\sum_i (x_i - \bar{x})^2} \right) \\[6pt]
&= \mathbb{Var} \left( \frac{\sum_i (x_i - \bar{x}) u_i}{\sum_i (x_i - \bar{x})^2} \right). \\[6pt]
\end{aligned} \end{equation}$$
