$x+y=a, xy=b$ then $a(a^2-3b)=19, a^2-b+5a=12$ $b=a^2+5a-12, a(-2a^2-15a+36)=19$ $a=1$ or $a=-\frac{19}{2}$ $a=1: b=-6 \to (x,y)=(3,-2)$ $a=-\frac{19}{2},b=\frac{123}{4}> \frac{a^2}{4} \to (x+y)^2<4xy$ which is impossible

$x+y=a\hspace{5mm}$ $xy=b$ $$a(a^2-3b)=19\Rightarrow a^3-3ab=19$$$$a^2+5a-b=12\Rightarrow b=a^2+5a-12$$$$a^3-3ab=a^3-3a(a^2+5a-12)=-2a^3-15a^2+36a=19\Rightarrow -(a-1)^2(2a+19)=0$$ (1) $a-1=0\Rightarrow y=1-x\hspace{5mm}$ $x^3+y^3=x^3+(1-x)^3=x^3+1-3x+3x^2-x^3=19\Rightarrow x^2-x=6\hspace{5mm}$ ${(x,y)}={(3;-2)}$ and ${(x,y)}={(-2;3)}$ (2) $2a+19=0\Rightarrow x+y=-9,5\hspace{5mm}x^3+y^3=(x+y)(x^2-xy+y^2)=-9,5(x^2-xy+y^2)=19\Rightarrow x^2-xy+y^2=-2$ $$x^2-xy+y^2=(x+y)^2-3xy=(-9,5)^2-3xy=90,25-3xy=-2\Rightarrow 3xy=92,25\Rightarrow xy=30,75$$$$xy=x(-9,5-x)=-x^2-9,5x=30,75$$$$x^2+9,5x+30,75=0\hspace{10mm} \Delta<0$$