Question

$$\begin{gathered} \frac{9}{x}-\frac{4}{y}=8, \\ \frac{13}{x}+\frac{7}{y}=101\left(x\ne 0,y\ne 0\right). \end{gathered}$$

Answer 1

The given equations are:
$\frac{9}{x}-\frac{4}{y}=8$ ............(i)
$\frac{13}{x}+\frac{7}{y}=101$ .............(ii)

Putting $\frac{1}{x}=u$ and $\frac{1}{y}=v$, we get:
9u − 4v = 8 .............(iii)
13u + 7v = 101 ...........(iv)

On multiplying (iii) by 7 and (iv) by 4, we get:
63u − 28v = 56 .............(v)
52u + 28v = 404............(vi)

On adding (v) and (vi), we get:
115u = 460 ⇒ u = 4
$\Rightarrow \frac{1}{x}=4\Rightarrow x=\frac{1}{4}$
On substituting $x=\frac{1}{4}$ in (i), we get:
$\frac{9}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}-\frac{4}{y}=8$
$\Rightarrow 36-\frac{4}{y}=8\Rightarrow \frac{4}{y}=\left(36-8\right)=28$
$\Rightarrow y=\frac{4}{28}=\frac{1}{7}$
Hence, the required solution is $x=\frac{1}{4}$ and $y=\frac{1}{7}$.