How does $\frac{x + \frac{1}{2}}{\frac{1}{x} + 2} = \frac{x}2$? How does simplifying $\dfrac{x+\frac12}{\frac1x+2}=\dfrac{x}2$
Plugging and chugging seems to prove this, but I don’t understand the algebra behind it.
How would you simplify $\dfrac{x+\frac12}{\frac1x+2}$ to get $\dfrac{x}2$?
I tried multiplying the expression by $x/x$, but that left me with $\frac{x^2+\frac{x}2}{1+2x}$
Is the fact that $\frac1x$ and $2$ are the reciprocals of $x$ and $\frac12$ respectively of any significance?
Thanks!
 A: 
Is the fact that $\frac1x$ and $2$ are the reciprocals of $x$ and $\frac12$ respectively of any significance?

Yes, and in general (assuming that $\,a \ne 0, b \ne 0 ,a + b \ne 0\,$):
$$\require{cancel}
\frac{a+b}{\;\;\dfrac{1}{a}+\dfrac{1}{b}\;\;} = \dfrac{a+b}{\;\;\dfrac{a+b}{ab}\;\;} = \dfrac{\cancel{(a+b)}ab}{\cancel{a+b}} = ab
$$
The posted question is the particular case $\,a=x, b=\dfrac{1}{2}\,$.
A: I'd start by multiplying by $2/2$, to make the expression easier to read.
$$\frac{x + 1/2}{1/x + 2} = \frac{2x + 1}{2/x + 4} = \frac{2x^2 + x}{4x + 2}$$
Factoring the top and bottom:
$$\frac{2x^2 + x}{4x + 2} = \frac{x(2x + 1)}{2(2x + 1)}$$
And cancel:
$$\frac{x(2x + 1)}{2(2x + 1)} = \frac{x}{2}$$
A: When $x\ne -\dfrac12$
$$\dfrac{x+\dfrac12}{\dfrac1x+2}=\dfrac{\dfrac{2x+1}{2}}{\dfrac{2x+1}{x}}=\dfrac x2$$
A: Let $x \not = -1/2$.
Factor the numeratior : 
$x+1/2 = (x/2)(2+1/x)$.
Then : 
$\dfrac{x+1/2}{1/x+2} = $
$\dfrac{(x/2)(2+1/x)}{1/x+2}= x/2.$
A: If $x \neq \frac{-1}{2}$, then $\frac{(x+1/2)}{(1/x+2)} = \frac{x^2+x/2}{1+2x} = \frac{2x^2+x}{2+4x}$. This is multiplying by 2 on top and on the bottom. Note that we can write $x = \frac{2x+4x^2}{2+4x}$. Therefore $\frac{x}{2} = \frac{x+2x^2}{2+4x}$, and the expressions are equivalent.
A: We have that for $x\neq -\frac12$ multiplying both side by $(1/x + 2)$
$$\frac{x + 1/2}{\frac1x + 2}=\frac x 2$$
$$\frac{x + 1/2}{\frac1x + 2}\left(\frac1x +2\right)=\frac x 2 \left(\frac1x +2\right)$$
$$x + \frac12=\frac12+x$$
For $x=-\frac12$ the equality doesn’t hold.
