By multiplying both sides with $4$ we get, $$ \frac{-15}{2x-1}+\frac{-15}{2y-1}+2x+1+2y+1=4x+4y $$Let $(m,n)=(2x-1,2y-1)$ and note that $mn$ is not $0$. $$ m+n+\frac{15}{m}+\frac{15}{n}=0\implies (m+n)\cdot (1+\frac{15}{mn})=0 $$Thus we have two cases; $$ m+n=0\implies x=-y $$$$ mn=-15\implies (x,y)=(1,-7),(2,-2),(3,-1),(8,0) $$(and symmetrics of those solutions) It's easy to check that all of these are the solutions of the problem.

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multiply by 4 Motivation denominator will cut Guves two eqn after simplification $x+y=1$ or $x+y=2xy+8$ Thus 8 solutions

on re-writing the equation we get : $\frac{x^2-4}{2x-1}-x+\frac{y^2-4}{2y-1}-y=0$ which gives us that $\frac{-4x^2+4x-16}{2x-1}=\frac{4y^2-4y+16}{2y-1}$ which yields $2-2(x+y)=-15\left(\frac{2-2(x+y)}{(2x-1)(2y-1)}\right)$ Case1:-$x+y=1$ we set $y=1-x$ and plug back to observe that it satisfies the original equation for every $x \in \mathbb{Z}$ hence we get $(x,y) =(z,1-z)$ for $z\in \mathbb{Z}$ Case2:-$x+y \neq 1$ $(2x-1)(2y-1)=-15$ on solving for every possible pair we get that $(x,y) \in \{(-7,1),(-2,2),(-1,3),(0,8),(1,-7),(2,-2),(3,-1),(8,0)\}$ so clubbing up we get $\boxed{(x,y)=(z,1-z) \hspace{0.01cm} \text{for each} \hspace{0.01cm} z \in \mathbb{Z} \text{and} \hspace{0.01cm} (x,y) \in \{(-7,1),(-2,2),(-1,3),(0,8),(1,-7),(2,-2),(3,-1),(8,0)\}}$ $\blacksquare$