Solution 1: By adding the first two equations, we can get $$x y+z+x z+y=(x+1)(y+z)=91$$ From the third equation we have $$(x+1)+(y+z)=19+1=20$$ so $x+1$ and $y+z$ are the two roots of $t^{2}-20 t+91=0$ by Vieta's theorem. As the quadratic equation can be decomposed into $$(t-7)(t-13)=0$$ we know that either $x=6, y+z=13$ or $x=12, y+z=7$. - If $x=12$, by the first equation we have $12 y+z=40$, and substituting $y+z=7$ we have $11 y=33$, $y=3$ and $z=4$. - If $x=6$, by the first equation we have $6 y+z=40$, and substituting $y+z=13$ we have $5 y=27$, $y=5.4$ and $z=7.6$. Hence, the two solutions are $(12,3,4)$ and $(6,5.4,7.6)$. Solution 2: Viewing $x$ as a constant, the equations become three linear equations in two variables $y$ and $z$. This system can only have a solution if $$\operatorname{det}\left[\begin{array}{ccc} x & 1 & 40 \\ 1 & x & 51 \\ 1 & 1 & 19-x \end{array}\right]=0$$ Expanding out the determinant, we have $$\begin{aligned} & x^{2}(19-x)+51+40-51 x-40 x-(19-x)=0 \\ & \quad \Longrightarrow x^{3}-19 x^{2}+90 x-72=0 \\ & \quad \Longrightarrow(x-1)\left(x^{2}-18 x+72\right)=0 \\ & \Longrightarrow(x-1)(x-6)(x-12)=0 \end{aligned}$$ so $x=1,6$, or 12. If $x=1$, the system has no solutions, and if $x=6$ or 12, we can find $y$ and $z$ as in the first solution.