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.
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.
Uniqueness follows straight by solving the equation for $y$, since $x \neq 6$ then $$ xy+x=6y \Longleftrightarrow xy-6y=-x\Longleftrightarrow (x-6)y=-x \Longleftrightarrow y=\frac{x}{6-x} $$
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.
