Solve the pair of equations:
$\frac{2}{x} + \frac{3}{y} = 13$
$\frac{5}{x} - \frac{4}{y} = -2$

(N/A) Let us write the given pair of equations as:
$2\left(\frac{1}{x}\right) + 3\left(\frac{1}{y}\right) = 13$ $...(1)$
$5\left(\frac{1}{x}\right) - 4\left(\frac{1}{y}\right) = -2$ $...(2)$
These equations are not in the standard form $ax + by + c = 0$. However,if we substitute $\frac{1}{x} = p$ and $\frac{1}{y} = q$ in equations $(1)$ and $(2)$,we get:
$2p + 3q = 13$ $...(3)$
$5p - 4q = -2$ $...(4)$
Now,we solve these linear equations using the elimination method. Multiply equation $(3)$ by $4$ and equation $(4)$ by $3$:
$8p + 12q = 52$ $...(5)$
$15p - 12q = -6$ $...(6)$
Adding equations $(5)$ and $(6)$:
$23p = 46 \implies p = 2$
Substituting $p = 2$ in equation $(3)$:
$2(2) + 3q = 13 \implies 4 + 3q = 13 \implies 3q = 9 \implies q = 3$
Since $p = \frac{1}{x}$ and $q = \frac{1}{y}$,we have:
$\frac{1}{x} = 2 \implies x = \frac{1}{2}$
$\frac{1}{y} = 3 \implies y = \frac{1}{3}$
Thus,the solution is $x = \frac{1}{2}$ and $y = \frac{1}{3}$.