\(\Rightarrow\dfrac{5}{x}=\dfrac{1}{6}+\dfrac{y}{3}\)

\(\Rightarrow\dfrac{5}{x}=\dfrac{1}{6}+\dfrac{2y}{6}\)

\(\Rightarrow\dfrac{5}{x}=\dfrac{2y+1}{6}\)

\(\Rightarrow x\left(2y+1\right)=30\)

But 2y + 1 is a odd => x is a even number

=> \(x\in\left\{-30;-10;-6;-2;2;6;10;30\right\}\)

We have:

So \(\left(x,y\right)=\left(-30;-1\right);\left(-10;-2\right);\left(-6;-3\right);\left(-2;-8\right);\left(2;7\right);\left(6;2\right);\left(10;1\right);\left(30;0\right)\)

⇒5x=16+y3⇒5x=16+y3

⇒5x=16+2y6⇒5x=16+2y6

⇒5x=2y+16⇒5x=2y+16

⇒x(2y+1)=30⇒x(2y+1)=30

But 2y + 1 is a odd => x is a even number

=> x∈{−30;−10;−6;−2;2;6;10;30}x∈{−30;−10;−6;−2;2;6;10;30}

We have:

x -30 -10 -6 -2 2 6 10 30
2y+1 -1 -3 -5 -15 15 5 3 1
y -1 -2 -3 -8 7 2 1 0

So (x,y)=(−30;−1);(−10;−2);(−6;−3);(−2;−8);(2;7);(6;2);(10;1);(30;0)