Solve simultaneously for x and y: 5x + y = 3 and 3x2 − 2xy = y2 − 105

We can factor the quadratic expression:

$$(x+6)(x-4) = 0$$

The roots are $x = -6$ and $x = 4$.

Since the parabola $y = x^2 + 2x - 24$ opens upwards (the coefficient of $x^2$ is positive), the expression is greater than or equal to zero outside the roots.
Therefore, the solution is when $x$ is less than or equal to the smaller root or greater than or equal to the larger root.

$$x \le -6 \quad \text{or} \quad x \ge 4$$

The solution is $\boxed{x \le -6 \text{ or } x \ge 4}$.

$$3x^2 - 6x + 10x^2 = (9 - 30x + 25x^2) - 105$$ $$13x^2 - 6x = 25x^2 - 30x - 96$$

Move all terms to one side to form a standard quadratic equation:

$$0 = 25x^2 - 13x^2 - 30x + 6x - 96$$ $$0 = 12x^2 - 24x - 96$$

Divide the entire equation by 12 to simplify:

$$x^2 - 2x - 8 = 0$$

Factor the quadratic expression:

$$(x-4)(x+2) = 0$$

This gives two possible values for $x$:

$$x = 4 \quad \text{or} \quad x = -2$$

For $x = 4$:

$$y = 3 - 5(4) = 3 - 20 = -17$$

For $x = -2$:

$$y = 3 - 5(-2) = 3 + 10 = 13$$

The solutions are the pairs $(x, y)$:

$$\boxed{(4, -17) \text{ and } (-2, 13)}$$