Solve the following pair of equations by reducing them to a pair of linear equations:
$\frac{4}{x} + 3y = 14$
$\frac{3}{x} - 4y = 23$

(A) Given equations are:
$\frac{4}{x} + 3y = 14$ $...(1)$
$\frac{3}{x} - 4y = 23$ $...(2)$
Let $\frac{1}{x} = p$. Substituting this in the equations,we get:
$4p + 3y = 14$ $...(3)$
$3p - 4y = 23$ $...(4)$
Using the elimination method,multiply equation $(3)$ by $4$ and equation $(4)$ by $3$:
$16p + 12y = 56$ $...(5)$
$9p - 12y = 69$ $...(6)$
Adding equations $(5)$ and $(6)$:
$25p = 125$
$p = 5$
Since $p = \frac{1}{x}$,we have $\frac{1}{x} = 5$,which implies $x = \frac{1}{5}$.
Substitute $p = 5$ in equation $(3)$:
$4(5) + 3y = 14$
$20 + 3y = 14$
$3y = 14 - 20$
$3y = -6$
$y = -2$
Thus,the solution is $x = \frac{1}{5}$ and $y = -2$.