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)

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 $$x=\frac{23}{10}$$
