<s>[INST] Proving a goal with an existential quantifier and making sure it covers all cases I'm trying to prove the following theorem:

Suppose x is a real number. Prove that if x $\neq$ 1 then there is a
  real number y such that $\frac{y + 1}{y - 2}$ = x.

The logical structure of the sentence is:
x $\neq$ 1 $\implies$ $\exists$y($\frac{y + 1}{y - 2}$ = x)
I first suppose x $\neq$ 1 so my goal becomes $\exists$y($\frac{y + 1}{y - 2}$ = x).
How to Prove it: A Structured Approach says the following regarding goals with existential quantifiers:

To prove a goal of the form $\exists$x P(x): Try to find a value of x
  for which you think P(x) will be true. Then start your proof with “Let
  x = (the value you decided on)” and proceed to prove P(x) for this
  value of x.

Let's say I chose y = 5. I can therefore let y = 5, conclude that $\frac{y + 1}{y - 2}$ = 2, and since my hypotheses say x $\neq$ 1 and 2 $\neq$ 1 finish my proof.
However this looks like I'm proving a theorem about a certain value of x instead or proving the theorem for all values of x except for 1. I feel I'm proving the following instead:
"Prove that if x = 2 then there is a real number y such that $\frac{y + 1}{y - 2}$ = x"
How should I deal with goals with existential quantifiers instead, to make sure I prove all the cases?
 [/INST] You should solve for $y$ in terms of $x$. Once you find it, you can then go back and choose $y$ to be this number and show that it works. Messing around, we see that:
\begin{align*}
\frac{y + 1}{y - 2} &= x \\
y + 1 &= xy - 2x \\
2x + 1 &= xy - y = y(x - 1) \\
y &= \frac{2x + 1}{x - 1}
\end{align*}
We're now ready to prove what we want.

Proof: Suppose that $x \neq 1$. Then consider $y = \frac{2x + 1}{x - 1}$ (which is a well-defined real number, since we're not dividing by zero). Observe that:
\begin{align*}
\frac{y + 1}{y - 2}
&= \frac{\frac{2x + 1}{x - 1} + 1}{\frac{2x + 1}{x - 1} - 2} \\
&= \frac{(2x + 1) + (x - 1)}{(2x + 1) - 2(x - 1)} \\
&= \frac{(2x + 1) + (x - 1)}{(2x + 1) + (-2x + 2)} \\
&= \frac{3x}{3} \\
&= x
\end{align*}
as desired. $~~\blacksquare$
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