IRL_ABooklet

Let $s=x+y$ and $p=xy$, then $x$ and $y$ are the roots of $T^2-sT+p=0$, $$x^2+y^2=s^2-2p(3)$$ and $$x^2+y^2+5x+5y+xy=s^2+5s-p=12,$$ thus $$p=s^2+5s-12.(4)$$ Using (3) and (4), we get $$\begin{array}{rl}{x}^3+y^3& =(x+y)(x^2+y^2-xy)\\ & =s(s^2-3p)=s(s^2-3(s^2+5s-12))\\ & =-2s^3-15s^2+36s=19,\end{array}$$ yielding $$2s^3+15s^2-36s+19=0.(5)$$ 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.