Proving existence of a unique real number I am working on the following question:
For all $x \in \mathbb{R}$, $x \neq 6$, there exists a unique real number $y$ such that $xy+x=6y$.
Now I have the existence part. That there exists a $$y=\frac{x}{6-x}.$$
To show uniqueness I know that I must show that if there is any other number say $z=x/(6-x)$, then $z$ must equal $y$. But I am not exactly sure how to show that part. 
 A: Supposing $x\neq6$, we have
\begin{align}
xy+x=6y&\iff xy-6y=-x\\
&\iff y(x-6)=-x\\
&\iff y=\frac{-x}{x-6}=\frac{x}{6-x}.
\end{align}
In particular, the $[\Longrightarrow]$ implications show that if $y$ is a solution, then it must equal $\frac{x}{6-x}$. Hence you have uniqueness for free.
A: Uniqueness follows straight by solving the equation for $y$, since $x \neq 6$ then 
$$
xy+x=6y \Longleftrightarrow xy-6y=-x\Longleftrightarrow (x-6)y=-x \Longleftrightarrow y=\frac{x}{6-x} 
$$
A: As others have said, you get the uniqueness for free as you show existence in this problem. But if you wanted to show it separately, you would suppose that there exist $y$ and $z$ so that both $xy+x=6y$ and $xz+x=6z$. Then solve for $x$ to find 
$$x=\frac{6y}{y+1}=\frac{6z}{z+1}.$$
(Note that we definitely have not divided by zero here, or else the previous equations reduce to $0=-6$.) From here, we cross multiply to find
\begin{align*}
6y(z+1)&=6(y+1)z\\
6yz+6y&=6yz+6z\\
6y&=6z\\
y&=z
\end{align*}
So if the problem has two solutions, they must be identical. 
That's a bit unnecessary for this problem, but could be good practice for others.
