Recover two unknown observations from a fitted line

The line passes through $(\bar x,\bar y)$. Since $\bar x=(3+2+3+6+1)/5=3$, substitute in the fitted line: $\bar y=2(3)-2=4$. This completes the blank Student working; individual observations need not lie on the fitted line.

The printed preliminary totals are $\sum y=5\bar y=20$, $\sum x=15$, and $\sum x^2=59$. Use the gradient $b=2$:

$\begin{aligned}&2\\ & =\frac{\sum xy-(15)(20)/5}{59-15^2/5}\\ & =\frac{\sum xy-60}{14}.\end{aligned}$

Hence $\sum xy=88$. The completed substitution and result are authored for the blank working space; the source prints the three preliminary totals.

From $\sum y=20$, $p+2+6+q+0=20$, so $p+q=12$. From $\sum xy=88$,

$3p+2(2)+3(6)+6q+1(0)=88\quad\Rightarrow\quad p+2q=22.$

Subtract the first equation to obtain $q=10$, then $p=2$. This completes the Student blank. Do not set every observation equal to $2x-2$: regression minimises squared residuals and does not require every point to lie on the fitted line.