if x is a real number $x \not =\ 1 $, then there exists y which is also a real number $ ((y+1) \div ( y-2) ) = x .$ Prove it's converse also.
Logical Argument:
given: $x \not = 1$
Goal: $ \exists y ((y+1) \div ( y-2) ) = x .$
Proof by contradiction
$ \forall y ((y+1) \div ( y-2) ) \not = x --------(1) $
Take $y =-1$
which make $(1) $ to $0$ but $x$ can be zero bcoz the given is $x \not = 1 $ leads to $0 \not = 0$. Hence a contradiction.
Converse: Take $ x=1 $ then $ (y+1) \div (y-2) = 1 $ For this to be true, $ y+1 = y-2$ which means $+1=-2$ but that's not possible. Hence it's contradiction.
Suggestion of better way is recommended.
Your second argument is fine. Your first one is not. I'll leave it for you to contemplate your solution a little bit. Meanwhile, here's the solution:
In order to prove the existence of $y$ here, you'll have to construct such one. Assuming the contrary isn't constructive in such situations. To construct $y$, scratch on the draft:
$$\frac{y + 1}{y - 2} = x \implies y + 1 = xy - 2x \implies y - xy = -(1 + 2x) \implies y(1-x) = -(1 + 2x)$$
And since we are allowed to divide by $1 - x$:
$$y = \frac{2x + 1}{x - 1}$$
Now because we illegally found $y$, it remains is to show that this $y$ verifies the requirement, and verify that $y = 2$ can't happen. Then we are done.