Question

Find $\frac{x}{y}$; when x² + 6y² = 5xy

Answer 1

x^2-5xy+6y^2=0.....x^2-3xy-2xy+6y^2=0....x(x-3y)-2y(x-3y)=0.....(x-2y)(x-3y)=0..... Therefore, x=3y. x/y=3. Or. x=2y...x/y=2..( both values are correct as this is a quadratic equation)

Answer 2

$x^2+6y^2=5xy\\x^2-5xy+6y^2=0\\ \\divide\ all\ by\ y^2\\ \\ \frac{x^2}{y^2} - \frac{5xy}{y^2} + \frac{6y^2}{y^2} =0\\ \\ (\frac{x}{y})^2 -5( \frac{x}{y} )+ 6 =0\\ \\it\ is\ a\ quadratic\ equation\ in\ \frac{x}{y}.\ Solving\ it\ we\ get;\\ (\frac{x}{y})^2 -3( \frac{x}{y} )-2( \frac{x}{y} )+ 6 =0\\ \\( \frac{x}{y} -2)( \frac{x}{y} -3)=0\\ \\ \frac{x}{y} =\boxed{2}\ or\ \frac{x}{y} =\boxed{3}$