Let $s = x + y$ and $p = xy$, then $x$ and $y$ are the roots of $T^2 - sT + p = 0$,
$$x^2 + y^2 = s^2 - 2p \quad (3)$$
and
$$x^2 + y^2 + 5x + 5y + xy = s^2 + 5s - p = 12,$$
thus
$$p = s^2 + 5s - 12. \quad (4)$$
Using (3) and (4), we get
$$\begin{align*} x^3 + y^3 &= (x + y)(x^2 + y^2 - xy) \\ &= s(s^2 - 3p) = s(s^2 - 3(s^2 + 5s - 12)) \\ &= -2s^3 - 15s^2 + 36s = 19, \end{align*}$$
yielding
$$2s^3 + 15s^2 - 36s + 19 = 0. \qquad (5)$$
It is easy to detect that $s = 1$ is a root of this cubic, and by factoring out $s-1$, we find the resulting quadratic $2s^2+17s-19$ which has roots $s = 1$ and $s = -19/2$. Corresponding to $s = 1$ we find $p = -6$ from (4). To find $x$ and $y$, we solve $T^2-T-6 = 0$ which has roots $-2$ and $3$. We get $(x, y) = (-2, 3)$ or $(x, y) = (3, -2)$, both satisfying the original equations.

Corresponding to $s = -19/2$ we find $p = 123/4$. The equation $T^2 + \frac{19}{2}T + \frac{123}{4} = 0$ has negative discriminant $s^2 - 4p = -\frac{131}{4}$ hence no real solutions.