Question

Solve the following pair of equations by reducing them to a pair of linear equations :

(5/x-1) + (1//y-2) = 2(6/x-1) - (3/y-2) = 1

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{given}\mathrm{equations}\mathrm{are}: \\ \left(\frac{5}{x-1}\right)+\left(\frac{1}{y-2}\right)=2...........\left(1\right) \\ \left(\frac{6}{x-1}\right)-\left(\frac{3}{y-2}\right)=1............\left(2\right) \\ \mathrm{put}\frac{1}{x-1}=p\mathrm{and}\frac{1}{y-2}=q\mathrm{in}\mathrm{given}\mathrm{equations}\mathrm{we}\mathrm{get}, \\ 5p+q=2...........\left(3\right) \\ \mathrm{and}6p-3q=1...........\left(4\right) \\ \mathrm{Multiplying}\left(3\right)\mathrm{by}6\mathrm{and}\left(4\right)\mathrm{by}5\mathrm{we}\mathrm{get}, \\ 30p+6q=12............\left(5\right) \\ 30p-15q=5............\left(6\right) \\ \mathrm{Now}\mathrm{subtracting}\left(6\right)\mathrm{from}\left(5\right)\mathrm{we}\mathrm{get}, \\ 21q=7 \\ \Rightarrow q=\frac{7}{21}=\frac{1}{3} \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}q\mathrm{in}\left(3\right)\mathrm{we}\mathrm{get}, \\ 5p+\frac{1}{3}=2 \\ \Rightarrow 5p=2-\frac{1}{3} \\ \Rightarrow 5p=\frac{6-1}{3} \\ \Rightarrow 5p=\frac{5}{3} \\ \Rightarrow p=\frac{5}{3\times 5}=\frac{1}{3} \end{gathered}$$

$$\begin{gathered} \mathrm{Hence},\frac{1}{x-1}=\frac{1}{3}\mathrm{and}\frac{1}{y-2}=\frac{1}{3} \\ \Rightarrow x-1=3\mathrm{and}y-2=3 \\ \Rightarrow x=3+1\mathrm{and}y=3+2 \\ \Rightarrow x=4\mathrm{and}y=5 \\ \mathrm{Hence},x=4 \\ \mathrm{and}y=5 \end{gathered}$$