Question

$$\begin{gathered} \frac{3}{x}-\frac{1}{y}+9=0, \\ \frac{2}{x}+\frac{3}{y}=5\left(x\ne 0,y\ne 0.\right) \end{gathered}$$

Answer 1

The given equations are:
$\frac{3}{x}-\frac{1}{y}+9=0$
⇒ $\frac{3}{x}-\frac{1}{y}=-9$ ............(i)
$\frac{2}{x}+\frac{3}{y}=5$ .............(ii)
Putting $\frac{1}{x}=u$ and $\frac{1}{y}=v$, we get:
3u − v = −9 .............(iii)
2u + 3v = 5 ...........(iv)

On multiplying (iii) by 3, we get:
9u − 3v = −27 .............(v)

On adding (iv) and (v), we get:
11u = −22 ⇒ u = −2
$\Rightarrow \frac{1}{x}=-2\Rightarrow x=\frac{-1}{2}$

On substituting $x=\frac{-1}{2}$ in (i), we get:
$\frac{3}{-{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}-\frac{1}{y}=-9$
$\Rightarrow -6-\frac{1}{y}=-9\Rightarrow \frac{1}{y}=\left(-6+9\right)=3$
$\Rightarrow y=\frac{1}{3}$
Hence, the required solution is $x=\frac{-1}{2}$ and $y=\frac{1}{3}$.