10th Class Mathematics Pair of Linear Equations in Two Variables Question Bank MCQs - Pair of Linear Equations in Two Variables

The value of x and y in \[\frac{5}{x+1}-\frac{2}{y-1}=\frac{1}{2};\,\frac{10}{x+1}+\frac{2}{y-1}=\frac{5}{2},\]where \[x\ne -1\] and \[y\ne 1\], is

A) \[x=4,y=5\]

B) \[x=5,y=4\]

C) \[x=\frac{1}{4},\,y=\frac{1}{5}\]

D) None of these

Correct Answer: A

Solution :

Let \[\frac{1}{x+1}=u\]and \[\frac{1}{y-1}=v\], then the given

pair of linear equations becomes

\[5u-2v=\frac{1}{2}\] ...(i)

and \[10u+2v=\frac{5}{2}\] ...(ii)

On adding Eqs. (i) and (ii), we get

\[15u=\frac{1}{2}+\frac{5}{2}\Rightarrow 15u=\frac{6}{2}\Rightarrow u=\frac{1}{5}\]

On putting \[u=\frac{1}{5}\]in Eq. (i), we get

\[5\times \frac{1}{5}-2v=\frac{1}{2}\Rightarrow v=\frac{1}{4}\]

Now, \[u=\frac{1}{5}\Rightarrow \frac{1}{x+1}=\frac{1}{5}\]

\[\Rightarrow \,x+1=5\,\,\Rightarrow x=4\]

And \[v=\frac{1}{4}\Rightarrow \,\frac{1}{y-1}=\frac{1}{4}\Rightarrow y-1=4\Rightarrow y=5\]

Hence, x = 4 and y = 5, which is the required unique solution.