system of equations with 3 variables in denominator I'm having problems with this system of equations:
$$\frac{3}{x+y}+\frac{2}{x-z}=\frac{3}{2}$$
$$\frac{1}{x+y}-\frac{10}{y-z}=\frac{7}{3}$$
$$\frac{3}{x-z}+\frac{5}{y-z}=-\frac{1}{4}$$

I've tried substitution method but that took me nowhere, I'm not sure I can solve it and would appreciate if anyone showed me the proper method to solve this problem.
 A: 
$$\frac{3}{x+y}+\frac{2}{x-z}=\frac{3}{2}$$
  $$\frac{1}{x+y}-\frac{10}{y-z}=\frac{7}{3}$$
  $$\frac{3}{x-z}+\frac{5}{\color{red}{x-z}}=-\frac{1}{4}$$

I assume that the part in red should be $y-z$; if not, please clarify.
Hint. Letting:
$$a=\frac{1}{x+y} \;,\; b=\frac{1}{x-z} \;,\;c=\frac{1}{y-z}$$
turn the system into:
$$\left\{\begin{array}{rcr}
3a&+&2b&&&=&\tfrac{3}{2}\\
a&&&-&10c&=&\tfrac{7}{3}\\
&&3b&+&5c&=&-\tfrac{1}{4}
\end{array}\right.$$
Perhaps this looks easier?
A: I hope you mean the following.
$$\frac{3}{x+y}+\frac{2}{x-z}=\frac{3}{2},$$
$$\frac{1}{x+y}-\frac{10}{y-z}=\frac{7}{3},$$
$$\frac{3}{x-z}+\frac{5}{y-z}=-\frac{1}{4}.$$
If so, let $\frac{1}{x+y}=c$, $\frac{1}{x-z}=b$ and $\frac{1}{y-z}=a$.
Thus,
$$3c+2b=\frac{3}{2},$$ $$c-10a=\frac{7}{3}$$ and
$$3b+5a=-\frac{1}{4}.$$
The last two equations give $$6b+c=\frac{11}{6},$$ which with first gives
$$c=\frac{1}{3},$$ $$b=\frac{1}{4}$$ and from here we'll get $$a=-\frac{1}{5}.$$ 
Thus, $$x+y=3,$$ $$x-z=4$$ and $$y-z=-5,$$ which gives
$$(x,y,z)=(6,-3,2).$$
