Solve a system of equations for positive ($x$, $y$, $z$) I arrived at the system of equations
$$(3+2z)x=3(1-xy)$$
$$(3+7z)y=4(1-xy)$$
$$3y-x+z(y-x)=2xy$$
while working on a geometric problem, with $x$, $y$ and $z$ representing segment lengths of a quadrilateral configuration. The system is inherently quartic. Given its non-standard appearance, I am not sure yet how to rearrange the pieces, which would lead to the solutions smoothly. A math tool I have spills out $(\frac12, \frac13, 1)$, without indicating how. 
Before resorting to brute-force substitution, I would like to have a shot at anyone who may have expertise, or just happens to be conversant in such systems.
 A: \begin{align} 
\begin{cases}
(3+2z)x&=3(1-xy)
,\\
(3+7z)y&=4(1-xy)
,\\
3y-x+z(y-x)&=2xy
.
\end{cases}
\tag{1}\label{1}
\end{align} 
Rearranging the system \eqref{1} as
\begin{align} 
\begin{cases}
\phantom{-}3xy+2zx &= 3-3x
,\\
\phantom{-}4xy+7yz &= 4-3y
,\\
-2xy+yz-zx &=x-3y,
\end{cases}
\tag{2}\label{2}
\end{align} 
we can express $xy,yz,zx$ in terms of $x,y$:
\begin{align} 
15xy &= \phantom{-}7x+36y-13
\tag{3}\label{3}
,\\
15yz &= -4x-27y+16
\tag{4}\label{4}
,\\
5zx &= -11x-18y+14
\tag{5}\label{5}
.
\end{align} 
Equation \eqref{3} gives $y$ in terms of $x$:
\begin{align} 
y &= \frac{7x-13}{3(5x-12)}
\tag{6}\label{6}
.
\end{align}
$z$ in terms of $x$
can be found from
equations \eqref{4} and \eqref{6}:
\begin{align}
z &= -\frac{4x^2-13x+15}{7x-13}
\tag{7}\label{7}
.
\end{align}
Equations \eqref{5}, \eqref{6} and \eqref{7} combined 
result in 
\begin{align}
-\frac{2(2x-1)(5x^3-45x^2+127x-117)}{(7x-13)(5x-12)}
&=0
\tag{8}\label{8}
,
\end{align}
which gives two real solutions for $x$:
\begin{align}
x_1&=\tfrac12
,\\
x_2&=
3+\tfrac1{15}\sqrt[3]{2025+15\sqrt{10545}}
+\frac 8{\sqrt[3]{2025+15\sqrt{10545}}}
\approx 4.54212431
.
\end{align}
Corresponding pairs of $y$ and $z$ are
\begin{align}
y_1&=\tfrac13
,\\
y_2&\approx 0.58492933
,\\
z_1&=1
,\\
z_2&\approx -2.047152
.
\end{align} 
A: This answer is based on Moo's comments. Therefore, all credit due for it should be given to Moo. However, I am taking responsibility for the text that appears below. Those who are not in the know can read about the $\bf{Gr\ddot{o} bner\space Basis}$ here and here.
All three equations of the equivalence system provided by Moo are correct and have the same collection of roots as the system posted by Quanto. As mentioned by Moo, it has 2 real roots and 2 complex roots. The two real roots are given below. 
$$
    \begin{matrix}
    x_1=\frac{1}{2} & \space & x_2=4.542124309455065516 \\
    y_1=\frac{1}{3} & \space & y_2=0.584929333410679641 \\
    z_1=1 & \space & z_2=-2.04715204771556650 \\
    \end{matrix}
$$
Quanto has mentioned in his post that $x, y$ and $z$ represent segment lengths of a quadrilateral configuration, which cropped up in a problem in geometry. Therefore, I think it is safe to assume that $x, y, z\ge 0$. If I am right, as par as Quanto's requirement is concerned, there is only one valid set of solutions available.
A: Solving for $z$ all three equations, after elimination we get
$$
\cases{
8 x+\frac{21}{x}-21 y-\frac{8}{y}-15=0\\
\frac{7 (x-1) x}{x-y}-9 x+\frac{2}{y}+9=0
}
$$
A plot shows the real solutions

