Question: A student participates in military training and engages in target shooting, which must be done 10 times. In the 6th, 7th, 8th, and 9th shots, he scored 9.0 points, 8.4 points, 8.1 points, and 9.3 points, respectively. The average score of his first 9 shots is higher than the average score of his first 5 shots. If he wants the average score of 10 shots to exceed 8.8 points, how many points does he need to score at least in the 10th shot? (The points scored in each shot are accurate to 0.1 points)
Solution: From the given, the average score of the first 5 shots is less than
$$
\frac{9.0+8.4+8.1+9.3}{4}=8.7 \text{. }
$$

The total score of the first 9 shots is at most
$$
8.7 \times 9-0.1=78.2 \text{. }
$$

Therefore, the score of the tenth shot must be at least
$$
8.8 \times 10+0.1-78.2=9.9 \text{ (points). }
$$

The above solution, if we set the total score of the first 5 shots as $x$, and the score of the tenth shot as $y$, can be represented by the system of inequalities as
$$
\left.\begin{array}{l}
\left\{\begin{array}{l}
\frac{x}{5}<8.7 \\
x<43.5
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
x<53.2
\end{array} \text{ and the scores are accurate to } 0.1 \text{ points }
\end{array}\right\}
$$

Starting point: Assuming this student scores 9.9 points on the tenth shot, can his average score over 10 shots definitely exceed 8.8 points? The answer is not certain. For example, if the total score of the first 5 shots is 40 points (satisfying the given condition), then the total score of 10 shots is
$$
40+(9.0+8.4+8.1+9.3)+9.9=84.7<88
$$

The system of inequalities
$$
\left\{\begin{array}{l}
x<A \\
x+y<B
\end{array}(B>A)\right.
$$

is not solved by $\left\{\begin{array}{l}x<A \\ y<B-A\end{array}\right.$. Its solution should be, when $x$ takes real values $(x<A)$, $y$ takes values $(y<B-x$. The solution set is
$$
G=\{(x, y) \mid x<A, y<B-x\} .
$$

The graphical solution is as follows:
As shown in Figure 1, draw the line
$$
L_{1}: x+y=B . x+y
$$
$>B$ is the set of points above the line $L_{1}$; draw the line $L_{2}$: $x=A, x<A$, and $y<B-A$ is only the range of $y$ values.

From the above analysis, to find the minimum value of $y$, we must first determine the value of $x$ within its range, for example, $x=C(C<A)$, then $y<B-C$.

For this problem, after solving $\left\{\begin{array}{l}x<53.2\end{array}\right.$, how do we determine the minimum value of $y$? This requires first determining the value of $x$. Since the maximum score for each shot is 10 points, we have
$$
x+(9.0+8.4+8.1+9.3)+10>88 \text{. }
$$

Therefore, $x>43.2$.
From $x<53.2 \Rightarrow y>9.9 \Rightarrow y \geqslant 10 \text{; }$
$$
(iii) \text{ When } x=43.4 \text{, }
$$
$$
43.4+y>53.2 \Rightarrow y>9.8 \Rightarrow y \geqslant 9.9 \text{. }
$$

In summary, to make the average score of 10 shots exceed 8.8 points, we must analyze the total score of the first 5 shots $(x)$. If $x \leqslant 43.2$ points, it is impossible for him to make the average score of 10 shots exceed 8.8 points; only when $x=43.3$ points, 43.4 points, the tenth shot must score at least 10 points and 9.9 points, respectively.