<s>[INST] Solve in $\mathbb{R^{3}}$ : $\begin{cases}x+y+xy=19\\y+z+yz=11\\x+z+xz=14\end{cases}$ Solve in $\mathbb{R^{3}}$ the following system : 

$$\begin{cases}x+y+xy=19\\y+z+yz=11\\x+z+xz=14\end{cases}$$

My try but I think not complete : 
The system equivalent : 
$$\begin{cases}(x+1)(y+1)=20\\(y+1)(z+1)=12\\(x+1)(z+1)=15\end{cases}$$
Then from first equation we have :
$$x+1=\frac{20}{y+1}$$
So third equation give : 
$$z+1=3\frac{y+1}{4}$$
Second equation $\implies $ 
$(y+1)^{2}=16$ then $y=3,-5$ 
This mean $x=\frac{17}{3}$ or $x=-6$ also $z=3$ or $z=-3$ 
I'm correct or no ??
 [/INST] Alternatively, take logs in
$$\begin{cases}(x+1)(y+1)=20\\(y+1)(z+1)=12\\(x+1)(z+1)=15\end{cases}. $$
Let $l_x=\ln(|x+1|)$, $l_y=\ln(|y+1|)$, and $l_z=\ln(|z+1|)$ and happily solve
$$\begin{array}{ccc|c}
1 & 1 & 0 & \ln(20)\\
0 & 1 & 1 & \ln(12)\\
1 & 0 & 1 & \ln(15)
\end{array}
$$
in just two steps for $l_x$, $l_y$, and $l_z$.
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