$4 \cdot 102$ Solve the system of equations
$$
\left\{\begin{array}{l}
x(x+1)(3 x+5 y)=144 \\
x^{2}+4 x+5 y=24
\end{array}\right.
$$
[Solution] From (2) we get $x(x+1)+(3 x+5 y)=24$
Considering $x(x+1),(3 x+5 y)$ as the roots of the equation $a^{2}-24 a+144=0$, we get $x(x+1)=12$ and $3 x+5 y=12$.

Solving $x(x+1)=12$ yields $\quad x_{1}=-4, x_{2}=3$.
Substituting into $3 x+5 y=12$ gives $, \quad y_{1}=\frac{24}{5}, y_{2}=\frac{3}{5}$.
$\therefore$ The original system of equations has two solutions $\left(-4, \frac{24}{5}\right),\left(3, \frac{3}{5}\right)$.