Question

If $x$ and $y$ both are positive and $(2x^2-5y^2):xy=1:3$, find $x:y$.

Answer 1

Given,

$(2x^2-5y^2):xy=1:3$

or, $\frac{2x^2-5y^2}{xy}=13$

or, $2x^2-3xy-5y^2=0$

or, $2x^2-5xy+2xy-5y^2=0$

or, $x(2x-5y)+y(2x-5y)=0$

or, $(x+y)(2x-5y)=0$

Since $x,y$ are given to be positive then $2x=5y$ or, $x:y=5:2$.