<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] The procedure you followed was correct, including getting the right values of $y$. However, you made mistakes for one value of $x$ (i.e., $\frac{17}{3}$ is not correct) and for both your values of $z$.
However, here's a slightly easier way to deal with the equations instead. You can use the symmetry to multiply your $3$ equations together to get
$$\begin{equation}\begin{aligned}
\left((x+1)(y+1)(z+1)\right)^2 & = 3600 \\
(x+1)(y+1)(z+1) & = \pm 60
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
You can then divide this by each of your $3$ equations to get
$$z + 1 = \pm 3 \implies z = -4, 2 \tag{2}\label{eq2A}$$
$$x + 1 = \pm 5 \implies x = -6, 4 \tag{3}\label{eq3A}$$
$$y + 1 = \pm 4 \implies y = -5, 3 \tag{4}\label{eq4A}$$
This leads to the solutions $(-6,-5,-4)$ and $(4,3,2)$.
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