Question

Solve $\frac{7x-2y}{xy}=5;\frac{8x+7y}{xy}=15$

Answer 1

$\frac{7x-2y}{xy}=5;\frac{8x+7y}{xy}=15$
$\frac{7x}{xy}-\frac{2y}{xy}=5\&\frac{8x}{xy}+\frac{xy}{xy}=15$
$\therefore \frac{x}{y}-\frac{2}{x}=5\&\frac{8}{y}+\frac{7}{x}=15$
Let $\frac{1}{y}=y\&\frac{1}{x}=x$
$\therefore$ 7y-2x=5 ...(1)
8y+7x=15 ...(2)
$\left(1\right)\times 7$ and $\left(2\right)\times 2$
we get,
$49y-\text{/}14x=35$
$\underset{–}{+16y+\text{/}14x=30}$
$65y=65$
y=1
From (1), $7y-2x=5\Rightarrow x=\frac{7y-5}{2}$
$\therefore x=\frac{7\left(1\right)-5}{2}=\frac{2}{2}$
x=1
$y=\frac{1}{y}\Rightarrow y=\frac{1}{y}=\frac{1}{1}=1$
$x=\frac{1}{x}\Rightarrow x=\frac{1}{x}=\frac{1}{1}=1$
$\therefore$ x=1, y=1.