Question

Daca
$\frac{x + y \sqrt{3} }{x - y \sqrt{3} } = 2$
atunci determinati
$\frac{x}{y}$
​

Answer 1

$\it \frac{x + y\sqrt{3} }{x - y\sqrt{3} } = 2 <=> \it x + y\sqrt{3} = 2(x - y\sqrt{3}) <=> x + y\sqrt{3} = 2x - 2y\sqrt{3} <=> \\ x - 2x = -2y\sqrt{3} - y\sqrt{3} <=> -x = -3y\sqrt{3} ~~/ \cdot (-1) <=> x = 3y\sqrt{3}$

$\it \frac{x}{y} = \frac{3y\sqrt{3} }{y} = \frac{3 \cdot y \cdot \sqrt{3} }{y} = 3\sqrt{3}$