3. Given that $x$, $y$, and $z$ are positive numbers, and satisfy:
$$
\left\{\begin{array}{l}
x+y+x y=8, \\
y+z+y z=15, \\
z+x+z x=35 .
\end{array}\right.
$$

Then $x+y+z+x y=$ $\qquad$
3. 15 .

From $x+y+x y=8 \Rightarrow(1+x)(1-y)=9$.
Similarly,
$$
\begin{array}{l}
(1+y)(1+z)=16, \\
(1+z)(1+x)=36 .
\end{array}
$$

Then $(1+x)(1+y)(1+z)=3 \times 4 \times 6$.
Solving, we get $z=7, x=\frac{7}{2}, y=1$.
Therefore, $x+y+z+x y=15$.