<s>[INST] 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.
 [/INST] \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} 
</s>