Question

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

Answer 1

The given equations are:
$\frac{5}{x}-\frac{3}{y}=1$ ............(i)
$\frac{3}{2x}+\frac{2}{3y}=5$ .............(ii)

Putting $\frac{1}{x}=u$ and $\frac{1}{y}=v$, we get:
5u − 3v = 1 .............(iii)
⇒ $\frac{3}{2}u+\frac{2}{3}v=5$
$\Rightarrow \frac{9u+4v}{6}=5$
$\Rightarrow 9u+4v=30$ ...............(iv)

On multiplying (iii) by 4 and (iv) by 3, we get:
20u − 12v = 4 .............(v)
27u + 12v = 90 ...........(vi)

On adding (iv) and (v), we get:
47u = 94 ⇒ u = 2
$\Rightarrow \frac{1}{x}=2\Rightarrow x=\frac{1}{2}$

On substituting $x=\frac{1}{2}$ in (i), we get:
$\frac{5}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}-\frac{3}{y}=1$
$\Rightarrow 10-\frac{3}{y}=1\Rightarrow \frac{3}{y}=\left(10-1\right)=9$
$\Rightarrow y=\frac{3}{9}=\frac{1}{3}$
Hence, the required solution is $x=\frac{1}{2}$ and $y=\frac{1}{3}$.