problem: If $j$, $k$, and $l$ are positive with $jk=24$, $jl = 48$, and $kl=18$, find $j+k+l$.
solution: Since $$j=\frac{24}{k}=\frac{48}{l}$$we have $l = 2k$. So $18
  = 2k^2$, which means that $9 = k^2$. Since $k$ has to be positive, this implies that $k = 3$. This means that $j = 8$, and $l = 6$. Hence $j+k+l = \boxed{17}$.
  
  OR
  
  Take the product of the equations to get $jk\cdot jl
  \cdot kl = 24 \cdot 48 \cdot 18$. Thus $$(jkl)^2 = 20736.$$So $(jkl)^2 =
  (144)^2$, and we have $jkl = 144$. Therefore, $$j = \frac{jkl}{kl} = \frac{144}{18} = 8.$$From this it follows that $k=3$ and $l=6$, so the sum is $8+3+6=\boxed{17}$.