Question

$$x\Delta y=\frac{(x+y)}{y}$$

Solve for x

$x=\frac{y}{\Delta y^{2}-1}$
$\Delta \neq \frac{1}{y^{2}}\text{ and }y\neq 0$

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Solve for y (complex solution)

$\left\{\begin{matrix}y=\frac{\sqrt{4\Delta x^{2}+1}+1}{2x\Delta }\text{; }y=\frac{-\sqrt{4\Delta x^{2}+1}+1}{2x\Delta }\text{, }&\Delta \neq 0\text{ and }x\neq 0\\y=-x\text{, }&x\neq 0\text{ and }\Delta =0\end{matrix}\right.$

Solve for y

$\left\{\begin{matrix}y=\frac{-\sqrt{4\Delta x^{2}+1}+1}{2x\Delta }\text{; }y=\frac{\sqrt{4\Delta x^{2}+1}+1}{2x\Delta }\text{, }&\left(|x|\leq \frac{\sqrt{-\frac{1}{\Delta }}}{2}\text{ or }\Delta >0\right)\text{ and }\Delta \geq -\frac{1}{4x^{2}}\text{ and }\Delta \neq 0\text{ and }x\neq 0\\y=-x\text{, }&x\neq 0\text{ and }\Delta =0\end{matrix}\right.$