If 2x^2 - 3xy + y^2 = 0, find the ratio of x:y.
The first step would be to divide through by y:
$\displaystyle \frac{2x^2}{y} - 3x + y = 0$
Then divide through by x:
$\displaystyle \frac{2x}{y} - 3 + \frac{y}{x} = 0$
Now, we move the last two terms to the other side of the equation:
$\displaystyle \frac{2x}{y} = 3 - \frac{y}{x}$
Then we divide through by 2:
$\displaystyle \frac{x}{y} = \frac{3}{2} - \frac{y}{2x}$
And there you go.
Hi snighda,
the previous two methods given by HallsofIvy and Diagonal
are the main variations.
$\displaystyle 2x^2-3xy+y^2=0$ may be converted to a quadratic in $\displaystyle \frac{x}{y}$
$\displaystyle \frac{2x^2-3xy+y^2}{y^2}=\frac{0}{y^2}$
$\displaystyle 2\left(\frac{x}{y}\right)^2-3\left(\frac{x}{y}\right)+1=0$
This is now of the form $\displaystyle aq^2+bq+c=0$
for which the solutions are $\displaystyle q=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
Hence $\displaystyle 2\left(\frac{x}{y}\right)^2-3\left(\frac{x}{y}\right)+1=0$ for $\displaystyle \frac{x}{y}=\frac{3\pm\sqrt{9-4(2)}}{2(2)}=\frac{3\pm1}{4}$
You achieve the exact same result by writing the expression as a multiplication of factors
$\displaystyle 2x^2-3xy+y^2=(2x+u)(x+v)$
$\displaystyle uv=y^2,\ 2xv+ux=x(2u+v)=x(-3y)$
therefore $\displaystyle u=v=-y$
$\displaystyle 2x^2-3xy+y^2=0\ \Rightarrow\ (2x-y)(x-y)=0$
Two numbers multiplied give zero, hence either one of them can be zero.
which yields two solutions, x=y and 2x=y
from which the exact same ratios obtained from the quadratic may be written.