What's the value of $x$ if $\frac{1}{x}+\frac{1}{y+z}=\frac{1}{2}$ and... Given $x, y, z\in \mathbb{R}$, such that $\frac{1}{x}+\frac{1}{y+z}=\frac{1}{2}$ and $ \frac{1}{y}+\frac{1}{z+x}=\frac{1}{3}$ and $\frac{1}{z}+\frac{1}{x+y}=\frac{1}{4}$. Find the value of $x$.
 A: I suggest to go for $X=\frac 1x$, $Y=\frac 1y$ and $Z=\frac 1z$ because it has the advantage to simplify the denominators.
The equations become 
$\dfrac{XY+YZ+ZX}{Y+Z}=\dfrac 12\quad;\quad\dfrac{XY+YZ+ZX}{X+Z}=\dfrac 13\quad;\quad\dfrac{XY+YZ+ZX}{X+Y}=\dfrac 14$
And it gives a simple system to solve
$\begin{cases}
Y+Z=2k\\
X+Z=3k\\
X+Y=4k\\
XY+YZ+ZX=k\end{cases}$
For instance $(2)-(1)$ gives $X-Y=k$ and reporting in $(3)$ gives $2Y=3k$.
This is all similar easy calculation and we end up with $X=\frac 52 k,Y=\frac 32k,Z=\frac 12k$ and $\dfrac{23k^2}4=k$
Finally $x=\frac 1X=\frac 25\times\frac{23}{4}=\frac {23}{10}$.
A: Let $xy=c$, $xz=b$ and $yz=a$.
Thus, $$\frac{x+y+z}{c+b}=\frac{1}{2},$$
$$\frac{x+y+z}{c+a}=\frac{1}{3}$$ and
$$\frac{x+y+z}{a+b}=\frac{1}{4},$$
which gives
$$\frac{c+a}{c+b}=\frac{3}{2}$$ and
$$\frac{a+b}{c+b}=2.$$
From here we obtain $$b=\frac{3}{5}a$$ and
$$c=\frac{1}{5}a$$ or
$$xz=\frac{3}{5}yz$$ and
$$xy=\frac{1}{5}yz,$$ which gives
$$y=\frac{5}{3}x$$ and
$$z=5x.$$
Id est, $$\frac{1}{x}+\frac{1}{\frac{5}{3}x+5x}=\frac{1}{2},$$ which gives
$$x=2.3.$$
