(X=3, Y=2) Let $u = \frac{1}{x+y}$ and $v = \frac{1}{x-y}$.
The equations become:
$10u + 2v = 4$ --- $(1)$
$15u - 5v = -2$ --- $(2)$
Multiply $(1)$ by $5$ and $(2)$ by $2$ to eliminate $v$:
$50u + 10v = 20$
$30u - 10v = -4$
Adding these equations: $80u = 16$,so $u = \frac{16}{80} = \frac{1}{5}$.
Substitute $u = \frac{1}{5}$ into $(1)$:
$10(\frac{1}{5}) + 2v = 4 \implies 2 + 2v = 4 \implies 2v = 2 \implies v = 1$.
Now,$\frac{1}{x+y} = \frac{1}{5} \implies x+y = 5$ --- $(3)$
And $\frac{1}{x-y} = 1 \implies x-y = 1$ --- $(4)$
Adding $(3)$ and $(4)$: $2x = 6 \implies x = 3$.
Subtracting $(4)$ from $(3)$: $2y = 4 \implies y = 2$.
Thus,the solution is $x = 3$ and $y = 2$.