6.2.1. (12 points) Solve the system of equations $\left\{\begin{array}{l}9 y^{2}-4 x^{2}=144-48 x, \\ 9 y^{2}+4 x^{2}=144+18 x y .\end{array}\right.$

Given the solutions $\left(x_{1} ; y_{1}\right),\left(x_{2} ; y_{2}\right), \ldots,\left(x_{n} ; y_{n}\right)$, write the sum of the squares

$$
x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}+y_{1}^{2}+y_{2}^{2}+\ldots+y_{n}^{2}
$$
Answer: 68.

Solution. From the first equation, we get $(3 y)^{2}=(2 x-12)^{2}$ and sequentially substitute into the second equation $y=\frac{2 x-12}{3}$ and $y=\frac{12-2 x}{3}$. The solutions obtained are: $(x ; y)=(0 ; 4),(0 ;-4),(6 ; 0)$. Answer: $6^{2}+4^{2}+(-4)^{2}=68$.