Mean-Deviation Form

We can compute the average, $\bar{x}$, of the $x$-values, and introduce a new variable $x_{\ast }=x-\bar{x}$.

x	2	5	7	8
y	1	1	4	3

We can set $y=c_0-c_1{x}_{\ast }$

$$\left[\begin{array}{cc}1& -3.5\\ 1& -0.5\\ 1& 1.5\\ 1& 2.5\end{array}\right]\left[\begin{array}{c}{c}_0\\ c_1\end{array}\right]=\left[\begin{array}{c}1\\ 1\\ 4\\ 3\end{array}\right]$$

Notice that the columns are Orthogonal. Hence, $A^{T}A$ is diagonal. Now we can solve this normal using the normal equation.