"Spoooooooky" Russian rational expressions 11

$\mathscr{E} = 1 - \left(\dfrac{2}{a-2}-\dfrac{2}{a+2}\right)\left(a-\dfrac{3a+2}{4}\right)\\ =1-\left(\dfrac{2(a+2)-2(a-2)}{(a-2)(a+2)}\right)\left(\dfrac{4a-(3a+2)}{4}\right)\\ =1-\left(\dfrac{\color{#D61F06}{\cancel{\color{#333333}{8}}2}}{\color{#3D99F6}{\cancel{\color{#333333}{(a-2)}}}(a+2)}\right)\left(\dfrac{\color{#3D99F6}{\cancel{\color{#333333}{a-2}}}}{\color{#D61F06}{\cancel{\color{#333333}{4}}}}\right)\\ =1-\dfrac{2}{a+2}$

$a=2016\\ \mathscr{E} = 1-\dfrac{2}{2016+2}=1-\dfrac{2}{2018}=1-\dfrac{1}{1009}=\dfrac{1008}{1009}$

$\implies x=1008,\;y=1009,\;x+y=1008+1009=\boxed{2017}$

$\begin{aligned} \mathscr{ E } & =1-\left( \frac { 2 }{ a-2 } -\frac { 2 }{ a+2 } \right) \left( a-\frac { 3a+2 }{ 4 } \right) \\ & =1-\left( \frac { 2(a+2) }{ (a-2)(a+2) } -\frac { 2(a-2) }{ (a+2)(a-2) } \right) \left( \frac { 4a }{ 4 } -\frac { 3a+2 }{ 4 } \right) \\ & =1-\left( \frac { 2(a+2)-2(a-2) }{ (a-2)(a+2) } \right) \left( \frac { 4a-(3a+2) }{ 4 } \right) \\ & =1-\left( \frac { 2a+4-2a+4 }{ (a-2)(a+2) } \right) \left( \frac { 4a-3a-2 }{ 4 } \right) \\ & =1-\left( \frac { 8 }{ (a-2)(a+2) } \right) \left( \frac { (a-2) }{ 4 } \right) \\ & =1-\left( \frac { 8(a-2) }{ 4(a-2)(a+2) } \right) \\ & =1-\left( \frac { 2 }{ (a+2) } \right) \\ & =1-\frac { 2 }{ 2018 } \\ & =\frac { 1009 }{ 1009 } -\frac { 1 }{ 1009 } \\ & =\frac { 1008 }{ 1009 } \\ \therefore \quad \quad \quad \frac { x }{ y } & =\frac { 1008 }{ 1009 } \\ \longrightarrow x+y & =1008+1009=\boxed { 2017 } \end{aligned}\\$