1. Factorización de la primera ecuación:
$15x^2 + xy - 2y^2 = 0$
Buscamos factores: $(5x + 2y)(3x - y) = 0$.
Esto nos da:
1) $y = 3x$
2) $y = -\frac{5}{2}x$

2. Sustitución en la segunda ecuación:

Caso 1: $y = 3x$
$$ 7x^2 - 4x(3x) - 3(3x)^2 = -32 \implies 7x^2 - 12x^2 - 27x^2 = -32 $$
$$ -32x^2 = -32 \implies x^2 = 1 \implies x = \pm 1 $$
Si $x = 1 \implies y = 3(1) = 3$. Si $x = -1 \implies y = 3(-1) = -3$.

Caso 2: $y = -2.5x$
$$ 7x^2 - 4x(-2.5x) - 3(-2.5x)^2 = -32 \implies 7x^2 + 10x^2 - 18.75x^2 = -32 $$
$$ -1.75x^2 = -32 \implies x^2 = \frac{32}{1.75} = \frac{3200}{175} = \frac{128}{7} $$
$$ x = \pm \sqrt{\frac{128}{7}} = \pm \frac{8\sqrt{2}}{\sqrt{7}} = \pm \frac{8\sqrt{14}}{7} $$
Para $y = -\frac{5}{2}x$:
$$ y = -\frac{5}{2} \left( \pm \frac{8\sqrt{14}}{7} \right) = \mp \frac{20\sqrt{14}}{7} $$

$$ \boxed{(x, y) \in \left\{ (1, 3), (-1, -3), \left( \frac{8\sqrt{14}}{7}, -\frac{20\sqrt{14}}{7} \right), \left( -\frac{8\sqrt{14}}{7}, \frac{20\sqrt{14}}{7} \right) \right\}} $$