Example 7. For all real numbers $p$ satisfying $0 \leqslant p \leqslant 4$, the range of $x$ that makes the inequality $x^{2}+p x>4 x+p-3$ true is $\qquad$ . (The 7th Junior High School Mathematics Competition in Jiangsu Province)
Analysis: If we solve by conventionally regarding $x$ as the main variable, we need to classify and discuss. By changing the perspective to regard $p$ as the main variable, the original problem can be reduced to finding the range of $x$ for which the linear function $f(p) = (x-1)p + x^2 - 4x + 3 > 0$ holds in the interval $[0,4]$. This way, we can easily determine the range of $x$ by leveraging the monotonicity of the linear function.
$\because x \neq 1$, otherwise the original inequality does not hold.
$\therefore f(p)$ is a linear function.
To ensure that $f(p)$ is always positive for $0 \leqslant p \leqslant 4$, it is equivalent to
$$
\left\{\begin{array}{l}
f(0) > 0, \\
f(4) > 0,
\end{array} \text { i.e., } \left\{\begin{array}{l}
(x-3)(x-1)>0, \\
x-1>0.
\end{array}\right.\right.
$$

Solving this yields $x > 3$ or $x < -1$.