        \problemdata       % Takes 5 *mandatory* arguments
        {xy-cases1}             % latex-friendly label for the prob.
        {proofs}  % The topic(s) of the problem content.
        {}               % Source (if known)
        {F01 PS1-6, S00} % Usage (list ones you are aware of).
        {}        % Last revision info (author, date).

\begin{problem}
Suppose $x$ is a real number.  Prove \textbf{by cases} 
of whether $x=1$ or $x \neq 1$, that there is a
$y \in \reals$ such that $(y+1)/(y-2)=x$ if and only if $x \ne 1$.

\solution{\begin{description}

\item[Case 1 ($x \not = 1$): ] We need to prove that there exists $y \in
\reals$ such that $(y+1)/(y-2)=x$.  In particular, we can solve for $y$
algebrically and conclude that $(2x+1)/(x-1)$ is the desired $y$.  This
$y$ is a well-defined real number because $x \neq 1$.  Therefore, $\exists
y \in \reals\; ((y+1)/(y-2) = x)$ by existential generalization.

\item[Case 2 ($x = 1$): ]
We need to prove that no $y \in \reals$ exists 
such that $(y+1)/(y-2)=x$. 
Suppose, for contradiction, that such a $y$ did exist.  Then:
\begin{align}
\frac{y+1}{y-2} & =  1  &\tag{because $x = 1$} \\
y+1 & = y-2  &\tag{multiply both sides by $y-2$} \\
1 & =  -2 &\tag{subtract $y$ from both sides},
\end{align}
a contradiction.

\end{description}
}     
\end{problem}