Question

$\frac{x+y}{xy}=2,\frac{x-y}{xy}=6(x\ne 0,y\ne 0).$

Answer 1

The given equations are:
$\frac{x+y}{xy}=2$
$\Rightarrow \frac{x}{xy}+\frac{y}{xy}=2$
$\Rightarrow \frac{1}{y}+\frac{1}{x}=2$ ...........(i)

and $\frac{x-y}{xy}=6$
$\Rightarrow \frac{x}{xy}-\frac{y}{xy}=6$
$\Rightarrow \frac{1}{y}-\frac{1}{x}=6$ ...........(ii)
Putting $\frac{1}{y}=v$ and $\frac{1}{x}=u$, we get:
v + u = 2 ..........(iii)
v − u = 6 .............(iv)
On adding (iii) and (iv), we get:
2v = 8 ⇒ v = 4
$\Rightarrow \frac{1}{y}=4\Rightarrow y=\frac{1}{4}$
On substituting $y=\frac{1}{4}$ in (i), we get:
$\frac{1}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}+\frac{1}{x}=2$
$\Rightarrow 4+\frac{1}{x}=2\Rightarrow \frac{1}{x}=\left(2-4\right)=-2$
$\Rightarrow x=\frac{-1}{2}$
Hence, the required solution is $x=\frac{-1}{2}$ and $y=\frac{1}{4}$.