x+y+z

We have $\begin{cases}\dfrac{1}{x}-\dfrac{1}{y}=4 \\ \dfrac{1}{x}-\dfrac{1}{z}=5 \\ \dfrac{1}{y}+\dfrac{1}{z}=3\end{cases}$

Add all three equations: $\left(\dfrac{1}{x}-\dfrac{1}{y}\right)+\left(\dfrac{1}{x}-\dfrac{1}{z}\right)+\left(\dfrac{1}{y}+\dfrac{1}{z}\right)=12$

$\Rightarrow \dfrac{2}{x}=12\Rightarrow \boxed{x=\dfrac{1}{6}}$

Subtract all three equations: $\left(\dfrac{1}{x}-\dfrac{1}{y}\right)-\left(\dfrac{1}{x}-\dfrac{1}{z}\right)-\left(\dfrac{1}{y}+\dfrac{1}{z}\right)=-4$

$\Rightarrow -\dfrac{2}{y}=-4\Rightarrow \boxed{y=\dfrac{1}{2}}$

Subtract the first two equations and add the last: $\left(\dfrac{1}{x}-\dfrac{1}{y}\right)-\left(\dfrac{1}{x}-\dfrac{1}{z}\right)+\left(\dfrac{1}{y}+\dfrac{1}{z}\right)=2$

$\Rightarrow \dfrac{2}{z}=2\Rightarrow \boxed{z=1}$

$1+\dfrac{1}{2}+\dfrac{1}{6}=\boxed{\dfrac{5}{3}}$

Let $a=\dfrac{1}{x}, b=\dfrac{1}{y}$ and $c=\dfrac{1}{z}$

Then the three equations become,

$a-b=4$ $\color{#D61F06}(1)$

$a-c=5$ $\color{#D61F06}(2)$

$b+c=3$ $\color{#D61F06}(3)$

From $\color{#D61F06}(1)$ , $a=4+b$ . From $\color{#D61F06}(2)$ , $a=5+c$ .

$a=a$

$4+b=5+c$

$b=1+c$

Substitute the above equation in $\color{#D61F06}(3)$ . We have

$1+c+c=3$

$2c=2$

$c=1$

Solving for $b$ , we have

$b=1+1=2$

Solving for $a$ , we have

$a=4+2=6$

Now, the values of $x,y$ and $z$ are,

$\boxed{x=\dfrac{1}{6}}$ $\boxed{y=\dfrac{1}{2}}$ $\boxed{z=1}$

Finally,

$x+y+z=\dfrac{1}{6}+\dfrac{1}{2}=1=\dfrac{1}{6}+\dfrac{3}{6}+\dfrac{6}{6}=$ $\boxed{\color{#D61F06}\large\dfrac{5}{3}}$