If $15 (2 x^{2} - y^{2}) = 7 x y,$ , find $x : y;$ if $x$ and $y$ both are positive.

Open in App

Solution

$15 (2 x^{2} - y^{2}) = 7 x y$
$⟹ 30 x^{2} - 7 x y - 15 y^{2} = 0$
solving for quadratic equation for $x$
We get $x = \frac{7 y \pm 43 y}{60}$
$x = 5 y / 6, - 3 y / 10$
$6 x - 5 y = 0$ $10 x + 3 y = 0$
$⟹ \frac{x}{y} = \frac{5}{6}$ or $\frac{x}{y} = \frac{- 3}{10}$
Since $x$ and $y$ are positive so
$x : y 5 : 6$

Suggest Corrections

Similar questions

15 (2 x^{2} - y^{2}) = 7 x y,

x

y

(2 x^{2} - 5 y^{2}) : x y = 1 : 3

x : y

{(x + y)}^{2} = 144

x^{2} - y^{2} = 24

x

x

y

x^{2} + y^{2}

Join BYJU'S Learning Program