Equations of fraction

$\begin{cases} {\color{#3D99F6}\dfrac xa}+{\color{#D61F06} \dfrac yb} = {\color{#3D99F6}\dfrac 2a}+{\color{#D61F06} \dfrac 1b} \\ {\color{#3D99F6}\dfrac xa}-{\color{#D61F06} \dfrac yb} = {\color{#3D99F6} \dfrac 2a}-{\color{#D61F06} \dfrac 1b} \end{cases}$

By observation, we note that $\color{#3D99F6} x=2$ and $\color{#D61F06} y=1$ , hence $x+y=\boxed{3}$ .

$\large \frac xa + \frac yb = \frac2a + \frac1b ...... (1)$

$\large \frac xb - \frac ya = \frac2b - \frac1a ....... (2)$

From equation $(1),$

$\dfrac{x}{a} + \dfrac{y}{b} = \dfrac{2}{a} + \dfrac{1}{b}$

$\Rightarrow \dfrac{bx+a y}{ab} = \dfrac{2b+a}{ab}$

$\Rightarrow bx+ay = 2b+a$

$\Rightarrow bx = 2b+a-ay$

$\Rightarrow x = \dfrac{2b+a - ay}{b} ........ (3)$

Substituting $x$ in equation $(2),$

$\dfrac{\frac{2b+a-ay}{b}}{b} - \dfrac ya = \dfrac2b - \dfrac 1a$

$\Rightarrow \dfrac{2b+a-ay}{b} \times \dfrac 1b - \dfrac ya = \dfrac2b - \dfrac 1a$

$\Rightarrow \dfrac{2b+a-ay}{b^2} = \dfrac2b - \dfrac 1a + \dfrac ya$

$\Rightarrow \dfrac{2b+a-ay}{b^2} = \dfrac{2a-b+by}{ab}$

$\Rightarrow \dfrac{2b+a-ay}{b} = \dfrac{2a-b+by}{a}$

$\Rightarrow 2ab+a^2- a^2y = 2ab - b^2 + b^2y$

$\Rightarrow a^2 - a^2y = b^2y - b^2$

$\Rightarrow -a^2y - b^2y = -b^2 -a^2$

$\Rightarrow -y (a^2+b^2) = -(a^2+b^2)$

So $y =1$

Substituting $y$ in equation $(3),$

$x = \dfrac{2b + a- a \cdot 1}{b}$

$\Rightarrow x= \dfrac{2b+a-a}{b}$

$\Rightarrow x = \dfrac{2b}{b}$

So $x = 2$

Hence $(x,y) = (2,1) \implies x+ y = \boxed{3}$