Question

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

Answer 1

Step: Find $x:y$

Given: $15(2x^2-y^2)=7xy,$

Divide both sides by $15xy$, we get

$\Rightarrow \frac{15(2x^2-y^2)}{15xy}=\frac{7xy}{15xy}$

$\Rightarrow \frac{2x^2}{xy}-\frac{y^2}{xy}=\frac{7}{15}$

$\Rightarrow 2\frac{x}{y}-\frac{y}{x}=\frac{7}{15}$

Assume, $\frac{x}{y}=z$

$\therefore 2z-\frac{1}{z}=\frac{7}{15}$

$\Rightarrow \frac{2z^2-1}{z}=\frac{7}{15}$

$\Rightarrow 30z^2-15=7z$

$\Rightarrow 30z^2-7z-15=0$

$\Rightarrow 30z^2-25z+18z-15=0$

$\Rightarrow 5z(6z-5)+3(6z-5)=0$

$\Rightarrow (6z-5)(5z+3)=0$

$\therefore 6z-5=0\text{and}5z+3=0$

$\therefore z=\frac{5}{6}\text{and}z=-\frac{3}{5}$

$\therefore \frac{x}{y}=\frac{5}{6}\text{and}\frac{x}{y}=-\frac{3}{5}$

Since, $x\text{and}y$ both are positive so $\frac{x}{y}$ can't be negative.

Hence the value of$\frac{x}{y}\text{is}\frac{5}{6}\text{or}x:y=5:6.$