$xyz\ne0$; $\dfrac{1}{x}+\dfrac{1}{y+z}=\dfrac{x+y+z}{x(y+z)}=\dfrac{1}{2}\Longrightarrow x+y+z\ne0$ and $xy+xz=2(x+y+z)$. Similarly: $yz+xy=3(x+y+z); xz+yz=4(x+y+z)$. Results: $2xy=(xy+xz)+(yz+xy)-(xz+yz)=x+y+z\Longrightarrow xy=\dfrac{x+y+z}{2}$; $xz=(xy+xz)-xy=\dfrac{3(x+y+z)}{2}$; $yz=(yz+xy)-xy=\dfrac{5(x+y+z)}{2}$. $\dfrac{y}{x}=\dfrac{yz}{xz}=\dfrac{5}{3}\Longrightarrow y=\dfrac{5x}{3}$; $\dfrac{z}{x}=\dfrac{yz}{xy}=5\Longrightarrow z=5x$. Replacing in the first equation: $\dfrac{1}{x}\cdot\left(1+\dfrac{1}{\dfrac{5}{3}+5}\right)=\dfrac{1}{2}\Longrightarrow$ $\Longrightarrow x=\dfrac{23}{10}\Longrightarrow y=\dfrac{23}{6}; z=\dfrac{23}{2}$. Hence, the solution of the system is: $(x,y,z)=\left(\dfrac{23}{10},\dfrac{23}{6},\dfrac{23}{2}\right)$.

we set $x+y+z=l$ so transforming our equations we get: $2l=x(y+z)$ $3l=y(z+x)$ $4l=z(x+y)$ so we get $xy=\frac{l}{2} , xz=\frac{3l}{2}, zy=\frac{5l}{2}$ which gives us $5x=3y=z=k$ now solving for $k$ we get: $k=\frac{23}{2}$ which gives: $\boxed{(x,y,z)=\left(\frac{23}{10},\frac{23}{6},\frac{23}{2}\right)}$

$\color{blue} \boxed{\textbf{SOLUTION}}$ $\dfrac{1}{x}+\dfrac{1}{y+z}=\dfrac{\sum x}{x(y+z)}=\dfrac{1}{2}$ $\implies xy+xz=2(\sum x)$ Similarly, $yz+xy=3(\sum x)$ $xz+yz=4(\sum x)$ Now, $xy=2(\sum x)-xz=2(\sum x)-[4(\sum x)-yz]=-2(\sum x)+yz=-2(\sum x)+3(\sum x)-xy=\sum x - xy$ $\implies xy=\frac{\sum x}{2}$ Similarly, $xz=\dfrac{3(\sum x)}{2}$ $yz\dfrac{5(\sum x)}{2}$ Then we get, $y=\dfrac{5x}{3}$ and $z=5x$ Plugging these is $(1),$ $\dfrac{1}{x} + \left(\dfrac{1}{\dfrac{5x}{3}+5x}\right)=\dfrac{1}{x}\cdot\left(1+\dfrac{1}{\dfrac{5}{3}+5}\right)=\dfrac{1}{2}$ $\implies x=\dfrac{23}{10}\Longrightarrow y=\dfrac{23}{6}, z=\dfrac{23}{2}$ So, $\boxed{(x,y,z)=\left(\frac{23}{10},\frac{23}{6},\frac{23}{2}\right)}$