In general, you end up with two equations, of the form
$$
C = na + rb \\
D = ra + sb
$$
Here $n$ is the number of points, $r$ is the sum of the $x$-values, and $s$ is the sum of the squared $x$-values. In the example, $n = 5$, $r = 15$, and $s = 55$. I'm using letters rather than numbers so that you can see the general pattern.
I'm going to show how to perform algebraic operations on these two equations until we have formulas that express $a$ and $b$ in terms of $n$, $r$, and $s$; these formulas are the only ones you then need.
What I'm going to do is to multiply each equation by a constant, like changing $1 = 3 - 2$ into $4 = 12 - 8$ by multiplying everything by 4. Doing so doesn't change the solutions of the equations, as long as the constant I'm using isn't zero, but it can help make them easier to work with.
First, I'll multiply the top one by $s$:
$$
sC = sna + rs b \\
D = ra + sb
$$
That's allowed, because the number $s$, which is the sum of the squares of a bunch of different numbers, will always be positive, so it's not zero.
Now I'm going to multiply the bottom equation by $r$, which might be zero. If it is, the new equation that I'll get will be just $0 = 0$, in disguise. So I'll keep both the new equation and the old one, to make sure I haven't lost any information:
\begin{align}
sC &= sna + rs b \\
D &= ra + sb\\
rD &= r^2a + rsb
\end{align}
Now I can subtract the bottom equation from the top one, to get this
\begin{align}
sC - rD &= (sna - r^2a) + rs b - rsb\\
D &= ra + sb.
\end{align}
Notice how the two $rsa$ terms in the first equation cancel -- that's why I chose $s$ and $r$ as multipliers in the earlier steps. So now we have
\begin{align}
sC - rD &= (sn - r^2)a\\
D &= ra + sb.
\end{align}
From the first equation, we can solve for $a$ by dividing both sides by $r^2 - ns$. That's only valid if $r^2 - ns$ is nonzero. It's a slightly tricky fact about the least-squares set-up that this number will always be nonzero, as long as you have at least two distinct $x$-values. (If you had only one, it'd be tough to choose a single line to "best" fit through it, wouldn't it?) Anyhow, here's the result:
\begin{align}
\frac{sC - rD}{sn - r^2} &= a\\
D &= ra + sb.
\end{align}
Now that we have a formula for $a$, in the second equation, the values $D, r, a,$ and $s$ are all known, so we can find $b$. To do that in algebra, I'll bring the $ra$ to the other side:
\begin{align}
\frac{sC - rD}{sn - r^2} &= a\\
D -ra &= sb.
\end{align}
and then divide through by $s$ (which is nonzero, so this is OK):
\begin{align}
\frac{sC - rD}{sn - r^2} &= a\\
\frac{D -ra}{s} &= b.
\end{align}
Now we can substitute the answer we got for $a$ into the second equation to get a final result:
\begin{align}
\frac{sC - rD}{sn - r^2} &= a\\
\frac{D -r\frac{sC - rD}{sn - r^2}}{s} &= b
\end{align}
which can be simplified a little bit (with some algebraic steps I'm not going to go through) to get
\begin{align}
\frac{sC - rD}{sn - r^2} &= a\\
\frac{-rC +nD}{sn - r^2} &= b
\end{align}
And that is the general solution for any least squares problem. Written out in terms of the things in the original article, you get
\begin{align}
Q = n(\sum x_i^2) - (\sum x_i)^2 \\
\frac{(\sum x_i^2)(\sum y_i) - (\sum x_i)(\sum y_i^2)}{Q} &= a\\
\frac{-(\sum x_i)(\sum y_i) + n(\sum x_i y_i)}{Q} &= b
\end{align}
where I've defined the number $Q$ to make the second and third lines a little simpler.
Just to finish this off, let's look at the original problem, where
$$
r = 15 \\
s = 55 \\
n = 5 \\
C = 25 \\
D = 88
$$
Then
\begin{align}
\frac{sC - rD}{sn - r^2} &= a\\
\frac{-rC +nD}{sn - r^2} &= b
\end{align}
becomes
\begin{align}
\frac{55 \cdot 25 - 15 \cdot 88}{55 \cdot 5 - 15^2} &= a\\
\frac{-15 \cdot 25 +5 \cdot 88}{55 \cdot 5 - 15^2} &= b
\end{align}
which simplifies to
\begin{align}
\frac{55}{50} &= a\\
\frac{65}{50} &= b
\end{align}
\begin{align}
1.1 &= a\\
1.3 &= b
\end{align}
