Solve the following pair of equations by reducing them to a pair of linear equations:
$\frac{1}{2x} + \frac{1}{3y} = 2$
$\frac{1}{3x} + \frac{1}{2y} = \frac{13}{6}$

(X=1/2, Y=1/3) Given equations:
$\frac{1}{2x} + \frac{1}{3y} = 2$ ...$(1)$
$\frac{1}{3x} + \frac{1}{2y} = \frac{13}{6}$ ...$(2)$
Let $\frac{1}{x} = p$ and $\frac{1}{y} = q$. Substituting these into the equations:
$\frac{p}{2} + \frac{q}{3} = 2 \Rightarrow 3p + 2q = 12$ ...$(3)$
$\frac{p}{3} + \frac{q}{2} = \frac{13}{6} \Rightarrow 2p + 3q = 13$ ...$(4)$
Multiply equation $(3)$ by $3$ and equation $(4)$ by $2$:
$9p + 6q = 36$ ...$(5)$
$4p + 6q = 26$ ...$(6)$
Subtracting $(6)$ from $(5)$:
$(9p - 4p) + (6q - 6q) = 36 - 26$
$5p = 10 \Rightarrow p = 2$
Substitute $p = 2$ in equation $(3)$:
$3(2) + 2q = 12$
$6 + 2q = 12 \Rightarrow 2q = 6 \Rightarrow q = 3$
Since $p = \frac{1}{x} = 2$,we get $x = \frac{1}{2}$.
Since $q = \frac{1}{y} = 3$,we get $y = \frac{1}{3}$.