Question

Solve the following systems of equations:
$\frac{xy}{x+y}=\frac{6}{5}$

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

Answer 1

Rewriting the equations by inversing them dividing them into partial fractions ;

$\frac{xy}{x+y}=\frac{6}{5}$

$\Rightarrow \frac{x+y}{xy}=\frac{5}{6}$

$\Rightarrow \frac{x}{xy}+\frac{y}{xy}=\frac{5}{6}$

$\Rightarrow$ $\frac{1}{x}+\frac{1}{y}=\frac{5}{6}$ ---(1)

$\frac{xy}{(y-x)}=6$

$\Rightarrow \frac{y-x}{xy}=\frac{1}{6}$

$\Rightarrow \frac{y}{xy}-\frac{x}{xy}=\frac{1}{6}$

$\Rightarrow \frac{1}{x}-\frac{1}{y}=\frac{1}{6}$ ---(2)

Adding two equations ;

$\frac{1}{x}+\frac{1}{y}+\frac{1}{x}-\frac{1}{y}=\frac{5}{6}+\frac{1}{6}$

$\Rightarrow \frac{2}{x}=\frac{5+1}{6}$

$\Rightarrow \frac{2}{x}=1$

$\Rightarrow x=2$

Substituting $x$ in $\frac{1}{x}+\frac{1}{y}=\frac{5}{6}$

$\frac{1}{2}+\frac{1}{y}=\frac{5}{6}$

$\Rightarrow \frac{1}{y}=\frac{5}{6}-\frac{1}{2}$

$\Rightarrow \frac{1}{y}=\frac{5-3}{6}$

$\Rightarrow \frac{1}{y}=\frac{2}{6}$

$\Rightarrow \frac{1}{y}=\frac{1}{3}$

$\Rightarrow y=3$

$\therefore x=2\text{and}y=3$