imc

The first two equations in the problem are $xy=24$ and $xz=48$. Since $xyz\ne 0$, we have $\frac{xy}{xz}=\frac{24}{48}⟹2y=z$. We can substitute $z=2y$ into the third equation $yz=72$ to obtain $2y^2=72⟹y=6$ and $2y=z=12$. We replace $y$ into the first equation to obtain $x=4$. Since we know every variable's value, we can substitute them in to find $x+y+z=4+6+12=\boxed{22}$.