Question

2x^2-y^2=7/15xy find value of x and y such that x and y are positive

Answer 1

Answer:

$(x,y)=(5,6),(10,12),...$

Step-by-step explanation:

$2x^2-y^2=7/15xy$

$30x^2-7xy-15y^2=0$

$(6x-5y)(5x+3y)=0$

∴$x=\frac{5}{6} y$ or $x=-\frac{5}{3} y$

When $x=\frac{5}{6} y$, let $y=6n$ ( $n$ is positive )

∴$x=5y$, $y=6n$

When $x=-\frac{5}{3} y$, let $y=3n$ ( $n$ is positive )

∴$x=-5n$, $y=6n$ ( $x$ is negative, not counting )

∴$(x,y)=(5,6),(10,12),...$