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)

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. 
