Find x+y+z

$x+y$ = $4$ ........................[1]

$x+z$ = $16$ ......................[2]

$z+y$ = $8$ ........................[3]

subtracting [1] from [2] we get,

$x+z-x-y$ = $16-4$

or, $z-y$ = $12$ ............................................[4]

now, adding [3] and [4] we get,

$z+y+z-y$ = $8+12$

or, $2z$ = $20$

or, $z=10$

so, $x=16-10$ ..............[using equation 2]

or, $x=6$

and, $y=4-6$ or, $y=-2$ ........[using equation 1]

so, $x+y+z=10+6-2=14$

we see that $\begin{cases} x+y=4--(1)\\y+z=8--(2)\\z+x=16--(3)\end{cases}$ sum them up $x+y+y+z+z+x=4+8+16$ $2x+2y+2z=28\longrightarrow 2(x+y+z)=28$ $x+y+z=\dfrac{28}{2}=\boxed{14}$

\[\begin{array} & (&x &+ &y & & &)= &4 \\ &+ (& & &y &+ &z &)= &8 \\ &+ (&x & & & + &z &)= &16 \\ \hline \\ &= (&2x &+ &2y & + &2z &)= &28 \\ \\ \hline &= (&x &+ &y & + &z &)= &\boxed{14}

\end{array}\]

2x + 2y + 2z = 28, x + y + z = 14