Solve the system of equations: x + 3y = 13 and x2 + 3y2 = 43.

$$(13^2 - 2 \cdot 13 \cdot 3y + (3y)^2) + 3y^2 = 43$$ $$169 - 78y + 9y^2 + 3y^2 = 43$$ $$12y^2 - 78y + 169 - 43 = 0$$ $$12y^2 - 78y + 126 = 0$$

$$\frac{12y^2}{6} - \frac{78y}{6} + \frac{126}{6} = 0$$ $$2y^2 - 13y + 21 = 0$$

We can factor this quadratic equation. We need two numbers that multiply to $2 \times 21 = 42$ and add up to $-13$. These numbers are $-6$ and $-7$.

$$2y^2 - 6y - 7y + 21 = 0$$ $$2y(y - 3) - 7(y - 3) = 0$$ $$(2y - 7)(y - 3) = 0$$

This gives two possible values for $y$:
$2y - 7 = 0 \implies 2y = 7 \implies y = \frac{7}{2}$
$y - 3 = 0 \implies y = 3$

Case 1: When $y = 3$

$$x = 13 - 3(3)$$ $$x = 13 - 9$$ $$x = 4$$

Case 2: When $y = \frac{7}{2}$

$$x = 13 - 3\left(\frac{7}{2}\right)$$ $$x = 13 - \frac{21}{2}$$ $$x = \frac{26}{2} - \frac{21}{2}$$ $$x = \frac{5}{2}$$

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

$$\boxed{\left(4, 3\right) \text{ and } \left(\frac{5}{2}, \frac{7}{2}\right)}$$