MHB Find Half a Number's Reciprocal Increased by Half its Reciprocal

[karush]

$\tiny{3.1.2}$
The reciprocal of half a number increased by half the recipical of the number is $\dfrac{1}{2}$
$\begin{array}{rl}
n= & \textit{the number} \\ \\
\dfrac{n}{2}= &\textit{half the number}\\ \\
\dfrac{2}{n} = &\textit{the reciprocal of half the number}\\ \\
\dfrac{1}{2n}= & \textit{half the reciprocal of the number}\\ \\
\dfrac{2}{n}+\dfrac{1}{2n} &=\dfrac{1}{2}\\ \\
&\textit{Multiply every term by 6n to cancel denominators}\\ \\
12+3=15 &=3n\quad\therefore n=5
\end{array}$
hopefully :unsure:

 

[karush]

ahhh victory...

 

[HOI]

$\tiny{3.1.2}$
The reciprocal of half a number increased by half the recipical of the number is $\dfrac{1}{2}$
$\begin{array}{rl}
n= & \textit{the number} \\ \\
\dfrac{n}{2}= &\textit{half the number}\\ \\
\dfrac{2}{n} = &\textit{the reciprocal of half the number}\\ \\
\dfrac{1}{2n}= & \textit{half the reciprocal of the number}\\ \\
\dfrac{2}{n}+\dfrac{1}{2n} &=\dfrac{1}{2}\\ \\
&\textit{Multiply every term by 6n to cancel denominators}\\ \\
12+3=15 &=3n\quad\therefore n=5
\end{array}$
hopefully :unsure:

Very good. But why "Multiply every term by 6n"? There is no "3" in any of the denominators. Multiplying by 2n is sufficient:
$2n\left(\frac{2}{n}+ \frac{1}{2n}\right)= 2n\left(\frac{1}{2}\right)$
$4+ 1= n$ so $n= 5$.