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$$\frac{xy}{x+y}=\frac{1}{2} \Rightarrow \frac{x+y}{xy}=2$$
We can split that LHS to get
$$\frac{1}{x} + \frac{1}{y} = 2$$
Similarly
$$\frac{1}{y} + \frac{1}{z} = 3$$
$$\frac{1}{x} + \frac{1}{z} = 7$$
Subtracting the first equation from the second equation gives
$$\frac{1}{z} - \frac{1}{x} = 1$$
Then adding this to the last equation gives
$$\frac{2}{z} = 8 \Rightarrow \frac{1}{z} = 4 \Rightarrow z = \frac{1}{4}$$
Hence
$$\frac{1}{x} + 4 = 7 \Rightarrow \frac{1}{x} = 3 \Rightarrow x = \frac{1}{3}$$
And finally
$$\frac{1}{y} = -1 \Rightarrow y = -1$$