Morphing $\beta_1$ into a different form (OLS question in Statistics)

Question

I am currently studying Simple Linear Regression and I have successfully proven to myself how $\beta_1$ and $\beta_0$ are derived.

However, I have been stuck on a seemingly simple problem (I'm sure it is very simple but alas I am not at all mathematically/statistically inclined although I wish to be) and I simply cannot figure it out.

There are really 3 problems that are very similar and I have attached pictures of where I get stuck. Unfortunately, both my textbook Mathematical Statistics by Wackerly and my course notes skip a lot of steps which is very aggravating for me. I would greatly appreciate help with steps shown as I can learn the best when everything is made explicit. Thanks so much!

Please note that in each problem, to make things neater I have simply called

$\sum_{i=1}^n$ equal to $\sum$ since those parameters are the same in every summation.

Problem $#1$:

Problem $#2$

Problem $#3$

I did not make 3 separate questions as these are very similar to each other but I just cannot figure it out! I would go to office hours but today is a Saturday and it is not possible.

Thank you so much in advance! :)

Answer 1

To answer your first question:

\begin{align}\sum\limits_{i=1}^{n}\left(x_i - \bar{x}\right)\left(y_i - \bar{y}\right) &= \sum\limits_{i=1}^{n}\left(x_iy_i-x_i\bar{y}-y_i\bar{x}+\bar{x}\bar{y}\right) \\ &= \sum\limits_{i=1}^{n} x_iy_i - \sum\limits_{i=1}^{n} x_i\bar{y}-\sum\limits_{i=1}^{n} y_i\bar{x}+\sum\limits_{i=1}^{n}\bar{x}\bar{y} \\ &= \sum\limits_{i=1}^{n} x_iy_i - \bar{y}\sum\limits_{i=1}^{n} x_i - \bar{x}\sum\limits_{i=1}^{n} y_i + \bar{x}\bar{y}\sum\limits_{i=1}^{n} 1 \\ &= \sum\limits_{i=1}^{n}x_iy_i-\bar{y}\left(n\bar{x}\right) - \bar{x}\left(n\bar{y}\right)+n\bar{x}\bar{y} \text{ since } \bar{x}=\dfrac{\sum\limits_{i=1}^{n}x_i}{n} \text{ and } \bar{y}=\dfrac{\sum\limits_{i=1}^{n}y_i}{n} \\ &= \sum\limits_{i=1}^{n}x_iy_i-n\bar{x}\bar{y}\text{.}\end{align}

The tricks for problems 2 and three are very similar.