Prove that $ x $ is rational where $ x $ is mixed with other variables Hi I am trying to complete the following question in my practice:
Suppose a, b, c are integers and x, y and z are non-zero real numbers that satisfy the following equations:
\begin{equation} \cfrac{xy}{x+y} = a \quad and \quad \cfrac{xz}{x+z} = b \quad and \quad \cfrac{yz}{y+z} = c \end{equation}
Prove that $ x $ is rational
Source: Discrete Mathematics with Applications 4th edition
I tried combining the equations and logically conclude that $ x $ is rational but to no avail as I do not know how to separate $ x $ from the rest of the variables. The only starting point that I have is the theorem for rational numbers:
\begin{equation} if \; x \in \mathbb{Q}, \; x = \cfrac{a}{b} \; where \; a,b \in \mathbb{Z} \; and \; b \neq 0 \end{equation}
I have no other idea how to carry on from here. Could someone please advise me?
 A: By rewriting we get
$$\frac{xy}{x+y} = \frac{1}{\frac{1}{x}+\frac{1}{y}}$$
so we can rewrite the equations as $$\frac{1}{x}+\frac{1}{y} = \frac{1}{a}, ~~\frac{1}{x}+\frac{1}{z} = \frac{1}{b}, ~~\frac{1}{y}+\frac{1}{z} = \frac{1}{c}.$$
This is a regular system of linear equations in $\frac{1}{x}, \frac{1}{y}$ and $\frac{1}{z}$, and since all the coefficients are in $\mathbb{Q}$, the solutions must be in $\mathbb{Q}$ as well. Feel free to solve that last system if you like.
A: Since $x,y,z$ are nonzero one can define $u=1/x,v=1/y,w=1/z,$ and it also follows from the given equations that none of $a,b,c$ are zero. So we can also put $d=1/a,e=1/b,f=1/c.$ Now the equations, after taking reciprocals, are $u+v=d,u+w=e,v+w=f$ which can be solved rationally for $u,v,w.$
A: Suppose $x$ is irrational. Then
$$ \frac{\frac{xy}{x+y}}{\frac{xz}{x+z}} = \frac ab $$, since $b \ne 0$. Now
$$ \frac{xy(x+z)}{xz(x+y)} = \frac ab \implies \frac{x+z}{x+y} = \frac{az}{by} = q \in \Bbb Q$$, since $y \ne 0$
So $x+z = qx+qy \implies (1-q)x = qy - z \in \Bbb Q$, but then $1-q \notin \Bbb Q$ ! Thus, leading to a contradiction.
Note that this same argument could be applied to $y, z$ as well, so it shows at once that all of them have to be rational.
A: Inverting the equations shows $\,x^{-1}\!+y^{-1},\,\ z^{-1}\!+x^{-1},\,\ \color{#c00}{y^{-1}\!+z^{-1}}\in \Bbb Q,\  $ i.e. are all rational. 
Adding them $\ 2(x^{-1}\!+y^{-1}\!+z^{-1}) =:q\in \Bbb Q\ $ so $\ x^{-1} =  q/2-(\color{#c00}{y^{-1}\!+z^{-1}}) \in\Bbb Q.\ \ $   QED
