Answer
$x$ is rational and is $\frac{1}{\frac{1}{2b}+\frac{1}{2c}-\frac{1}{2a}}$
Work Step by Step
We can rewrite the equations as $\frac{1}{\frac{1}{x} + \frac{1}{y}} = a$, $\frac{1}{\frac{1}{x} + \frac{1}{z}} = b$, $\frac{1}{\frac{1}{y} + \frac{1}{z}} = c$. Therefore, $\frac{1}{x}+\frac{1}{y} = \frac{1}{a}$ and similar for the others. Adding all the equations, we get $\frac{2}{x}+\frac{2}{y}+\frac{2}{z} = \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$. Dividing by 2 and subtracting the third equation from this, we get $\frac{1}{x} = \frac{1}{2b}+\frac{1}{2c}-\frac{1}{2a}, so$ $ x = \frac{1}{\frac{1}{2b}+\frac{1}{2c}-\frac{1}{2a}}$
