4・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 $x(x+1)+(3 x+5 y)=24$. Treating $x(x+1)$ and $(3 x+5 y)$ as the roots of the equation $a^{2}-24 a+144=0$, we obtain $x(x+1)=12$ and $3 x+5 y=12$.
Solving $x(x+1)=12$ yields $x_{1}=-4, x_{2}=3$.
Substituting into $3 x+5 y=12$, we get $\quad y_{1}=\frac{24}{5}, y_{2}=\frac{3}{5}$.
$\therefore \quad$ The original system of equations has two solutions $\left(-4, \frac{24}{5}\right),\left(3, \frac{3}{5}\right)$.