Find $x$ if $\frac {1} {x} + \frac {1} {y+z} = \frac {1} {2}$ I found this question from past year's maths competition in my country. I've tried any possible way to find it, but it is just way too hard.

Find $x$ if \begin{align}\frac {1} {x} + \frac {1} {y+z} &= \frac {1} {2}\\ \frac {1} {y} + \frac {1} {x+z} &= \frac{1}{3}\\ \frac {1} {z} + \frac {1} {x+y} &= \frac {1} {4}\end{align}
$(A)\;\frac 32$
$(B)\;\frac {17}{10}$
$(C)\;\frac {19}{10}$
$(D)\;\frac {21}{10}$
$(E)\;\frac {23}{10}$

EDIT: I'm very sorry guys, it should be $\frac {1} {x+y}$ not $xy$, I'm sorry for the typos (idk what is wrong with me)

\begin{align}
\frac {1} {x} + \frac {1} {y+z} &= \frac {1} {2}\\ \frac {1} {y} + \frac {1} {x+z} &= \frac{1}{3}\\ \frac {1} {z} + \frac {1} {x+y} &= \frac {1} {4}
\end{align}
Is equivalent to the system
\begin{align}
xy+xz&=2(x+y+z)
\\
xy+yz&=3(x+y+z)
\\
xz+yz&=4(x+y+z),
\end{align}
Solving this system as a linear system of
equations in terms of $xy,xz$ and $yz$ and $x+y+z$,
we arrive at
\begin{align}
xy&=\tfrac12 (x+y+z), \quad(1)
\\
yz&=\tfrac52 (x+y+z),
\\
xz&=\tfrac32 (x+y+z).
\end{align}
Dividing these equations pairwise, we get
\begin{align}
z&=5x
\\
y&=\tfrac53 x,
\end{align}
combining with (1) we have
\begin{align}
\tfrac{5}{3}x^2-\tfrac{23}{6}x&=0
\end{align}
and since $x\ne0$, the answer is $x=\tfrac{23}{10}$.
