Solve the following pair of equations by reducing them to a pair of linear equations:
$\frac{5}{x-1} + \frac{1}{y-2} = 2$
$\frac{6}{x-1} - \frac{3}{y-2} = 1$

(A) Let us substitute $\frac{1}{x-1} = p$ and $\frac{1}{y-2} = q$. Then the given equations can be written as:
$5p + q = 2$ $...(1)$
$6p - 3q = 1$ $...(2)$
To solve these,multiply equation $(1)$ by $3$:
$15p + 3q = 6$ $...(3)$
Adding equation $(2)$ and $(3)$:
$(6p - 3q) + (15p + 3q) = 1 + 6$
$21p = 7$
$p = \frac{7}{21} = \frac{1}{3}$
Substitute $p = \frac{1}{3}$ into equation $(1)$:
$5(\frac{1}{3}) + q = 2$
$\frac{5}{3} + q = 2$
$q = 2 - \frac{5}{3} = \frac{6-5}{3} = \frac{1}{3}$
Now,substitute back the values of $p$ and $q$:
$\frac{1}{x-1} = \frac{1}{3} \implies x-1 = 3 \implies x = 4$
$\frac{1}{y-2} = \frac{1}{3} \implies y-2 = 3 \implies y = 5$
Thus,the solution is $x = 4, y = 5$.