[tex]2x=3y=6z=(x+y+z)\cdot n\\
\text{Daca facem un artificiu de calcul obtinem:}\\
\dfrac{x}{\frac{1}{2}}=\dfrac{y}{\frac{1}{3}}=\dfrac{z}{\frac{1}{6}}=\dfrac{x+y+z}{\frac{1}{n}} (1)\\
\text{Dar din egalitatea proportiilor avem ca:}
\dfrac{x}{\frac{1}{2}}=\dfrac{y}{\frac{1}{3}}=\dfrac{z}{\frac{1}{6}}=\dfrac{x+y+z}{\frac{1}{2}+\frac{1}{3}+\frac{1}{6}} (2)\\
\text{Din (1) si (2) obtinem:}\\
\\
\dfrac{1}{n}=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\\
\\
\dfrac{1}{n}=\dfrac{3+2+1}{6}\\
[/tex]
[tex]\dfrac{1}{n}=\dfrac{6}{6}=1\Rightarrow \boxed{n=1}[/tex]
