solving equations .

let $u=y+z,v=z+x,w=x+y$ , then

$uv=30,uw=15,vu=18$

Multiplying the three equations and extracting the square root gives

$wuv=\pm 90$

Combining this result (use wuv=+90)with each of the equations gives

$u=3, v=6, w=5$

Therefore,

$3=y+z,6=z+x,5=x+y$

Adding these gives

$2y+2z+2x=14$

Dividing both sides by $2$ gives

$x+y+z=\boxed{7}$

$\begin{cases} (x+y)(z+x) & = 30 &...(1) \\ (y+z)(x+y) & = 15 &...(2) \\ (z+x)(y+z) & = 18 &...(3) \end{cases}$

$\begin{cases} \dfrac {(1)}{(2)}: & \dfrac {z+x}{y+z} = 2 & \implies z + x = 2y + 2z & \implies x = 2y + z \\ \dfrac {(2)}{(3)}: & \dfrac {x+y}{z+x} = \dfrac 56 & \implies 6x+6y = 5z+5x & \implies x = 5z - 6y \end{cases}$

$\implies 2y+z=5z-6y \implies 8y = 4z \implies z = 2y$ and $x = 2y+z = 4y$ . From $(1): (x+y)(z+x) = (5y)(6y) = 30y = 30$ $\implies y = 1$ taking only the positive value. Then $x=4$ and $y=2$ and $x+y+z = 4+1+2 = \boxed{7}$ .