Question

$\frac{xy}{x+y}=\frac{6}{5},\frac{xy}{x-y}=6,\mathrm{where}x+y\ne 0\mathrm{and}y-x\ne 0.$

Answer 1

The given equations are:
$\frac{xy}{x+y}=\frac{6}{5}\Rightarrow 6x+6y=5xy$
$\Rightarrow \frac{6x+6y}{xy}=5\Rightarrow \frac{6x}{xy}+\frac{6y}{xy}=5$
$\Rightarrow \frac{6}{y}+\frac{6}{x}=5$
$\Rightarrow \frac{6}{x}+\frac{6}{y}=5$...........(i)
and $\frac{xy}{y-x}=6\Rightarrow 6y-6x=xy$
$\Rightarrow \frac{6y-6x}{xy}=1\Rightarrow \frac{6y}{xy}-\frac{6x}{xy}=1$
$\Rightarrow \frac{6}{x}-\frac{6}{y}=1$ ...........(ii)
Putting $\frac{1}{x}=u$ and $\frac{1}{y}=v$, we get:
6u + 6v = 5 ..........(iii)
6u − 6v = 1 .............(iv)
On adding (iii) and (iv), we get:
12u = 6 ⇒ $u=\frac{1}{2}$
$\Rightarrow \frac{1}{x}=\frac{1}{2}\Rightarrow x=2$
On substituting x = 2 in (i), we get:
$\frac{6}{2}+\frac{6}{y}=5\Rightarrow 3+\frac{6}{y}=5$
$\Rightarrow \frac{6}{y}=\left(5-3\right)=2$
$\Rightarrow 2y=6\Rightarrow y=3$
Hence, the required solution is x = 2 and y = 3.