One, (20 points) A certain audio-visual company has calculated that a blank disc can record 2 episodes of a TV drama plus 1 skit, or 2 skits plus 3 pop songs. A student plans to record 7 episodes of a TV drama, 11 skits, and 20 pop songs. How many blank discs does the student need to prepare at minimum to record all the programs? Please explain your reasoning.
(Note: The time for each episode of the TV drama, each skit, and each song is the same, and the time for each skit is greater than the time for each song.)
Let the recordable time of each blank disc be $s$, the time occupied by each episode of TV drama, each sketch, and each pop song be $x, y \backslash z (y>z)$, respectively. The total time for the student to record 7 episodes of TV drama, 11 sketches, and 20 pop songs is $w$. According to the problem, we have:
$$
s=2 x+y, s=2 y+3 z, w=7 x+11 y+20 z .
$$

Thus, $x=\frac{1}{2} y+\frac{3}{2} z$.
Therefore, $w=7 x+11 y+20 z$
$=14 x+7 y+4 y+20 z-7 x$
$=7(2 x+y)+4 y+20 z-7 x$
$=7(2 x+y)+4 y+20 z-7\left(\frac{1}{2} y+\frac{3}{2} z\right)$
$=7(2 x+y)+4 y+20 z-\frac{7}{2} y-\frac{21}{2} z$
$=7(2 x+y)+\frac{1}{2} y+\frac{19}{2} z$
$=7(2 x+y)+\frac{1}{2} y+\frac{13}{2} z+3 z$
$=8 s+\left(\frac{13}{2} z-\frac{3}{2} y\right)$.
When $\frac{13}{2} z \leqslant \frac{3}{2} y$, $w=8 s$;
When $\frac{13}{2} z>\frac{3}{2} y$, $w>8 s$.
Also, $\frac{1}{2} y+\frac{19}{2} z<2 s$, thus,
When $y \geqslant \frac{13}{3} z$, at least 8 blank discs need to be prepared;
When $y<\frac{13}{3} z$, at least 9 blank discs need to be prepared.