348. Solve the systems of equations:
a) $\left\{\begin{array}{l}x^{2}-3 x y-4 y^{2}=0, \\ x^{3}+y^{3}=65 ;\end{array}\right.$
b) $\left\{\begin{array}{l}x^{2}+2 y^{2}=17, \\ 2 x y-x^{2}=3\end{array}\right.$
$\triangle$ a) The first equation of the system is homogeneous. We divide it term by term by $y^{2}$. In this process, no solutions are lost: although the pair $(0 ; 0)$ satisfies the first equation, it does not satisfy the second. We get:

$$
\left(\frac{x}{y}\right)^{2}-3 \frac{x}{y}-4=0
$$

Let $\frac{x}{y}=t$. Then

$$
t^{2}-3 t-4=0 ; \quad t_{1}=4, \quad t_{2}=-1
$$

$4^{*}$

99

Let $t=4$, i.e., $x=4 y$. From the second equation, we have:

$$
64 y^{3}+y^{3}=65, \quad y^{3}=1 ; \quad y=1, \quad x=4
$$

Let $t=-1$, i.e., $x=-y$. In this case, the second equation has no solutions.

b) Neither of the equations in the system is homogeneous. However, a homogeneous equation can be obtained if we equalize the free terms of the equations in the system by multiplying the first by 3, the second by 17, and subtracting the resulting equations, so that the free term disappears:

$$
\left\{\begin{array}{l}
3 x^{2}+6 y^{2}=51 \\
34 x y-17 x^{2}=51
\end{array}\right.
$$

from which

$$
20 x^{2}-34 x y+6 y^{2}=0, \quad 10 x^{2}-17 x y+3 y^{2}=0 .
$$

A homogeneous equation has been formed. The further solution is the same as in the previous problem.

Answer: a) $(4 ; 1) ;$ b) $(3 ; 2),(-3 ;-2),(\sqrt{3} / 3 ; 5 \sqrt{3} / 3),(-\sqrt{3} / 3 ;-5 \sqrt{3} / 3)$.