Let first consider the equations one by one and will be solving one by one ;
[tex]{:\implies \quad \sf x+\dfrac{1}{y}=4}[/tex]
Multiplying both sides by y will lead ;
[tex]{:\implies \quad \sf xy+1=4y}[/tex]
[tex]{:\implies \quad \boxed{\sf xy=4y-1\quad \cdots \cdots(i)}}[/tex]
Now, consider the second equation which is ;
[tex]{:\implies \quad \sf y+\dfrac{1}{x}=\dfrac14}[/tex]
Multiplying both sides by x will yield
[tex]{:\implies \quad \sf xy+1=\dfrac{x}{4}}[/tex]
[tex]{:\implies \quad \sf xy=\dfrac{x}{4}-1}[/tex]
[tex]{:\implies \quad \boxed{\sf xy=\dfrac{x-4}{4}\quad \cdots \cdots(ii)}}[/tex]
As LHS of both equations (i) and (ii) are same, so equating both will yield;
[tex]{:\implies \quad \sf 4y-1=\dfrac{x-4}{4}}[/tex]
Multiplying both sides by 4 will yield
[tex]{:\implies \quad \sf 16y-4=x-4}[/tex]
[tex]{:\implies \quad \sf 16y=x}[/tex]
Dividing both sides by y will yield :
[tex]{:\implies \quad \boxed{\bf{\dfrac{x}{y}=16}}}[/tex]
Hence, the required answer is 16
