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

Hint: To convert the pair to a linear equation, put $u=\dfrac{1}{x-1}$ and \[v=\dfrac{1}{y-2}\]. Find the values of u and v by solving both equations formed. Then substitute back the values of u and v, find the value of x and y.

Complete step-by-step answer:
We have been given a pair of equations for which we need to find the value of x and y. The two pair of equations are
$\dfrac{5}{x-1}+\dfrac{1}{y-2}=2$ and $\dfrac{6}{x-1}-\dfrac{3}{y-2}=1$
Now, let us put $\dfrac{1}{x-1}=u$ and \[\dfrac{1}{y-2}=v\] in the above equation. Hence, the equation changes to a linear equation.
\[5u+v=2\] and \[6u-3v=1\]
Now, we need to find the value of u and v. Let us put the equation as
\[5u+v=2\] ……………………………(1)
\[6u-3v=1\] ………………………(2)
From (1) we can write it as,
\[v=2-5u\]
Put this value of v in the equation (2)
\[6u-3\left( 2-5u \right)=1\]
Let us simplify the above expression and find the value of u.
\[6u-6+15u=1\]
\[\Rightarrow 21u=1+6\]
\[\therefore u=\dfrac{7}{21}=\dfrac{1}{3}\]
Thus, we got the value of u as \[\dfrac{1}{3}\] . Now, put the value of in equation (1).
\[v=2-5u=2-5\times \dfrac{1}{3}\]
\[=2-\dfrac{5}{3}=\dfrac{6-5}{3}=\dfrac{1}{3}\]
\[\therefore u=\dfrac{1}{3}\] and \[v=\dfrac{1}{3}\]
Now let us put back the value of u and v.
$\dfrac{1}{x-1}=u$ and \[\dfrac{1}{y-2}=v\]
$\dfrac{1}{x-1}=\dfrac{1}{3}$ and \[\dfrac{1}{y-2}=\dfrac{1}{3}\] , Now let us cross multiply and simplify it
\[3=x-1\] and \[3=y-2\]
\[\Rightarrow x=3+1=4\]
\[y=3+2=5\]
Thus, we got the value of x and y as 4 and 5.
\[\therefore x=4\] and \[y=5\] is the solution of our equation.

Note: If you directly try to solve it without assuming values for u and v, then you might end with a complex equation which can’t be solved. We are asked to convert the given pair of equations to a linear pair in the question. So, assume the values accordingly.