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(A) Let $u = \frac{1}{x-1}$ and $v = \frac{1}{y-2}$.Substituting these into the given equations,we get:$5u + v = 2$  --- $(1)$$6u - 3v = 1$ --- $(2)$M

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$x = 4, y = 5$

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(A) Let $u = \frac{1}{x-1}$ and $v = \frac{1}{y-2}$.Substituting these into the given equations,we get:$5u + v = 2$  --- $(1)$$6u - 3v = 1$ --- $(2)$Multiply equation $(1)$ by $3$ to eliminate $v$:$15u + 3v = 6$ --- $(3)$Adding equation $(2)$ and $(3)$:$(6u - 3v) + (15u + 3v) = 1 + 6$$21u = 7$$u = \frac{7}{21} = \frac{1}{3}$Substitute $u = \frac{1}{3}$ into equation $(1)$:$5(\frac{1}{3}) + v = 2$$\frac{5}{3} + v = 2$$v = 2 - \frac{5}{3} = \frac{6-5}{3} = \frac{1}{3}$Now,substitute back the values of $u$ and $v$:$\frac{1}{x-1} = \frac{1}{3} \implies x-1 = 3 \implies x = 4$$\frac{1}{y-2} = \frac{1}{3} \implies y-2 = 3 \implies y = 5$Thus,the solution is $x = 4, y = 5$.

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

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Solve the following pairs 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$