\(\Rightarrow\dfrac{x}{6}-\dfrac{1}{30}=\dfrac{2}{y}\)

\(\Rightarrow\dfrac{5x}{30}-\dfrac{1}{30}=\dfrac{2}{y}\)

\(\Rightarrow\dfrac{5x-1}{30}=\dfrac{2}{y}\)

\(\Rightarrow y\left(5x-1\right)=60\)

But \(5x-1\equiv4\left(mod5\right)\)

\(\Rightarrow5x-1\in\left\{-6;-1;4\right\}\)

If 5x - 1 = -6

=> 5x = -5 => x = -1

And y = 60 : -6 = -10

If 5x-1 = -1 

=> 5x = 0 => x = 0

And y = 60 : -1 = -60

If 5x-1 = 4 

=> 5x = 5 => x = 1

And y = 60 : 4 = 15

So \(\left(x,y\right)=\left(-1;-10\right);\left(0;-60\right);\left(1;15\right)\)

⇒x6−130=2y⇒x6−130=2y

⇒5x30−130=2y⇒5x30−130=2y

⇒5x−130=2y⇒5x−130=2y

⇒y(5x−1)=60⇒y(5x−1)=60

But 5x−1≡4(mod5)5x−1≡4(mod5)

⇒5x−1∈{−6;−1;4}⇒5x−1∈{−6;−1;4}

If 5x - 1 = -6

=> 5x = -5 => x = -1

And y = 60 : -6 = -10

If 5x-1 = -1 

=> 5x = 0 => x = 0

And y = 60 : -1 = -60

If 5x-1 = 4 

=> 5x = 5 => x = 1

And y = 60 : 4 = 15

So (x,y)=(−1;−10);(0;−60);(1;15