How do I formalize this proof? "Prove that if $x\ne 1$ then there is a real number $y$ such that $\frac{y+1}{y-2} = x$"

Question

Consider trivial problem:

Prove that if $x ≠ 1$ then there is a real number $y$ such that $\dfrac{y+1}{y-2} = x$

Let's solve for y

$$\frac{y+1}{y-2} = x$$ $$y+1 = (y-2)x$$ $$y+1 = yx - 2x$$ $$y - yx = - 2x - 1 $$ $$y(1-x) = - 2x - 1 $$ $$y = \frac{- 2x - 1}{1-x}$$

Now, although it is obvious that if $x ≠ 1$ then $y$ is defined, but how do I express it formally? I believe that something like "It can be seen that $y$ is defined for any $x$, provided that $x ≠1$ $\Box$" Won't be a rigorous conclusion.

I want to write formal proof of the conjecture above (including all the necessary steps), how should I write it?

Answer 1

In my opinion, it is one step before the last one who needs the reminder, meaning: after you get $\;(1-x)y=-2x-1\;$ , then you bring up that principle (or axiom): "we can now divide both sides by $\;1-x\;$ if $\;x\neq1\;$" (as then we divide by a non-zero number!), and thus we get

$$\require{cancel}\frac{\cancel{(1-x)}y}{\cancel{1-x}}=\frac{-2x-1}{1-x}\implies y=\frac{-2x-1}{1-x}=\frac{2x+1}{x-1}$$

Answer 2

When solving an equation you are actually assuming that such a y exists (which is what you wanted to prove!). In order to prove that a solution y to $\frac{y+1}{y-2}=x$ exists for all $x\neq 1$ you can simply start by setting $y=\frac{-2x-1}{1-x}$, and then show that this y actually solves the equation, which is essentially doing your calculations in reverse.