It can be helpful to look at 7.28 combined with 7.8.

For 7.8, the end goal was to find $x$ such that $$x\equiv 1\pmod{4}, x\equiv 2\pmod{5}, x\equiv 3\pmod{6}.$$ Although the CRT did NOT apply here, we noted that $x+3$ was a multiple of $4$, $5$, and $6$, so any $x$ that was $3$ less than multiple of $\text{lcm}(4,5,6) = 60$ worked. Thus, $x\equiv 57 \pmod{60}$.

Let's compare this to 7.28. Note 7.8 is 7.28 with $M=3$, as our mods are $1+3=4$, $2+3=5$, and $3+3=6$. 7.28 just wants you to note that the general idea from 7.8 works for additional numbers. For example, if we also required that $x\equiv 4\pmod{7}$ (so $M=4$) we would need $x\equiv 417\pmod{420}$ as $\text{lcm}(4,5,6,7) = 420$.

Hope this helps!