The solution set of the system of equations ${\log }_{12}x(\frac{1}{{\log }_{x}2}+{\log }_{2}y)={\log }_{2}x$ and ${\log }_{2}x.\left({\log }_3(x+y)\right)=3{\log }_{3}x$ is

Given , ${\log }_{12}x(\frac{1}{{\log }_{x}2}+{\log }_{2}y)={\log }_{2}x$

${\log }_{2}x\left({\log }_3(x+y)\right)=3{\log }_{3}x$

$\Rightarrow {\log }_{12}x(\frac{1}{{\log }_{x}2}+{\log }_{2}y)={\log }_{2}x$

$\Rightarrow \frac{\log x}{\log 12}[{\log }_{2}x+{\log }_{2}y]={\log }_{2}x$

$\Rightarrow \frac{\log x}{\log 12}[{\log }_{2}x+{\log }_{2}y]=\frac{\log x}{\log 2}$

$\Rightarrow \frac{1}{\log 12}\left[{\log }_{2}xy\right]=\frac{1}{\log 2}$

$\Rightarrow \frac{1}{\log 12}\left[\frac{\log xy}{\log 2}\right]=\frac{1}{\log 2}$

$$\begin{gathered} \Rightarrow \log xy=\log 12 \\ \Rightarrow xy=12\to \left(1\right) \\ \Rightarrow {\log }_{2}x\left({\log }_3(x+y)\right)=3{\log }_{3}x \\ \Rightarrow \frac{\log x}{\log 2}\left(\frac{\log (x+y)}{\log 3}\right)=\frac{3\log x}{\log 3} \\ \Rightarrow \log (x+y)=3\log 2 \end{gathered}$$

$\Rightarrow \log (x+y)=\log 8$

$$\begin{gathered} \Rightarrow x+y=8\to \left(2\right) \\ x-y=\sqrt{{(x+y)}^2-4xy} \\ x-y=\sqrt{64-48} \\ x-y=\sqrt{16} \\ x-y=\pm 4 \end{gathered}$$