Having a hard time proving this one. I can prove this with a contradiction in the (→) direction but I'm stuck on how to prove this in the (←) direction where x ≠ 1 is the given and ∀x ∈ R[∃y ∈ R(x + y = yx)] is the goal. Any help is MUCH appreciated!
∀x ∈ R[∃y ∈ R(x + y = yx) ↔ x ≠ 1]
You can write out the solution explicitly, $y=x(x-1)^{-1}$ which of course is only defined if $x \neq 1$.
In the reverse direction we are given $x$ such that $x\neq 1$ and asked to find y so that the equation holds. Note that $$x+y=yx\iff x=y(x-1).$$ Since $x\neq 1$ we can take $y=\dfrac{x}{x-1}$.
Here is a proof where we prove both directions at the same time. Let's treat this as a simplification problem: we are asked to simplify $\;\exists y \left ( x + y = yx \right )\;$. The easiest way to do that, is to rewrite $\;x + y = yx\;$ to something of the form $\;y = \ldots\;$.
(Note that I'm leaving out $\;{}\in R\;$ throughout, to reduce visual clutter.)$ \newcommand{\calc}{\begin{align} \quad &} \newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}} \newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} } \newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & } \newcommand{\endcalc}{\end{align}} \newcommand{\ref}[1]{\text{(#1)}} \newcommand{\equiv}{\leftrightarrow} \newcommand{\then}{\rightarrow} \newcommand{\when}{\leftarrow} \newcommand{\true}{\text{true}} \newcommand{\false}{\text{false}} $
So we calculate as follows, for any $\;x\;$: $$\calc \exists y \left ( x + y = yx \right ) \op\equiv\hint{arithmetic -- to bring all $\;y\;$'s to the same side} \exists y \left ( yx - y = x \right ) \op\equiv\hint{arithmetic -- to have a single $\;y\;$} \exists y \left ( y(x - 1) = x \right ) \op\equiv\hint{divide by $\;x-1\;$, splitting off special case $\;x-1=0\;$} \exists y \left ( \left ( x = 1 \;\land\; y \times 0 = 1 \right )\;\lor\; \left ( x \not= 1 \;\land\; y = \frac x {x-1} \right ) \right ) \op\equiv\hint{LHS of $\;\lor\;$ is false} \exists y \left ( x \not= 1 \;\land\; y = \frac x {x-1} \right ) \op\equiv\hint{one-point rule for $\;\exists\;$} x \not= 1 \endcalc$$
The one-point rule for $\;\exists\;$ is the law of logic which says that $$ \exists y \left (y = E \land P(y) \right ) \;\equiv\; P(E) $$ where $\;E\;$ does not contain $\;y\;$: if $\;y\;$ and $\;E\;$ are equal, then we can just substitute $\;E\;$ for $\;y\;$. There are many such laws; see, e.g., another answer of mine.
