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)
 A: \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 $x=\tfrac{23}{10}$.
A: Let $x=1/p$, $y=1/q$, and $z=1/r$.  The three equations become
$$\begin{align}
{1\over2}&=p+{qr\over q+r}={s\over q+r}\\
{1\over3}&=q+{rp\over r+p}={s\over r+p}\\
{1\over4}&=r+{pq\over p+q}={s\over p+q}\\
\end{align}$$
where 
$$s=pq+qr+rp$$
Thus
$$\begin{align}
2s&=q+r\\
3s&=r+p\\
4s&=p+q\\
\end{align}$$
From this it follows that 
$$(q+r)+(p+q)=2(r+p)\implies2q=r+p=3s\implies q={3\over2}s$$
and the others follow:
$$r=2s-q={1\over2}s\quad\text{and}\quad p=4s-q={5\over2}s$$
But now we have
$$s=pq+qr+rp={15\over4}s^2+{3\over4}s^2+{5\over4}s^2={23\over4}s^2$$
hence $s=4/23$, so $p=10/23$, and thus $x=23/10$.
A: multiplying (2) by $x$ $(x \ne 0)$ we obtain
$$\frac{1}{z}=\frac{1}{4}-\frac{1}{xy}$$
with (3) we obtain $$y=\frac{12(x^2-1)}{4x^2-3x}$$ (I)
from (1) and (3) we get 
$$\frac{x^2y}{3}-x^2+1=\frac{xy}{4}$$
plugging (I) in this equation  and simplifying we get 
$$-60 x^5+119 x^4+156 x^3-288 x^2-72 x+144=0$$
with five real solutions, no of them from (A) to (E)
the system after the correction has the solution $$x=\frac{23}{10}$$
A: Choosing
$p = \frac{1}{x} $
$q = \frac{1}{y}  $
$r = \frac{1}{z} $
will certainly help writing the equations more neatly, I guess.
Then your equations can be written as:
$$2p + 2qr = 1 \\ 3q + 3pr = 1 \\ 4r + 4pq = 1$$ 
using first two equations you get:
$$ 2p + 2qr = 3q + 3pr \\ \implies r = \frac{3q - 2p}{3p - 2q}$$
Substitute in equations and solve as solved by Alan above. 
