<s>[INST] 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)
 [/INST] 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 $x=23/10$.
</s>