question_answer

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

\[\frac{1}{x}-\frac{4}{y}=2\]

\[\frac{1}{x}+\frac{3}{y}=9\]

Answer:

Given, \[\frac{1}{x}-\frac{4}{y}=2\]

\[\frac{1}{x}+\frac{3}{y}=9\]

Let                    \[\frac{1}{x}=u,\frac{1}{y}=v\]

So,                   \[u-4v=2\]                                             ?(i)

\[u+3v=9\]                                          ?(ii)

On solving eq. (i) and eq. (ii)

\[_{\begin{smallmatrix}  u\,\,+\,\,3v\,\,=\,\,9 \\  -\,\,\,\,\,\,\,\,\,-  \\  \overline{\,\,\,\,\,-7v\,\,=\,\,-7} \end{smallmatrix}}^{u\,\,-\,\,4v\,\,=\,\,2}\]

\[v=1\]

Putting the value of v in eq. (i)

\[u-4v=2\]

\[u-4\times 1=2\]

\[u-4=2\]

\[u=2+4\]

\[u=6\]

So                    \[v=1\Rightarrow \frac{1}{y}=1,y=1\]

\[u=6\Rightarrow \frac{1}{x}=6,x=\frac{1}{6}\]

Hence, \[x=\frac{1}{6}\]  and \[y=1\]