How do you solve in terms of y 4x^2-2xy+3y^2=2

[JKCB]

Homework Statement

Homework Equations

The Attempt at a Solution

4x^2-2xy=2-3y^2
x(4x-2y)=2-3y^2 then I'm stuck[/B]

 

[Delta2]

View the equation as a quadratic $ay^2+by+c=0$ with $a=3, b=-2x, c=4x^2$, that is if you want to find $y=f(x)$.
if you want to find $x=f(y)$ then view it as a quadratic $ax^2+bx+c=0$ with $a=4,b=-2y, c=3y^2$.

 

[JKCB]

What do I do with the 2 at the end of the original equation?

 

[SammyS]

Homework Statement

Homework Equations

The Attempt at a Solution

4x^2-2xy=2-3y^2
x(4x-2y)=2-3y^2 then I'm stuck[/B]

Please include a complete statement of your problem in the text of the post itself, no matter what you state in the title.

 

[Delta2]

What do I do with the 2 at the end of the original equation?

Ah yes sorry, I thought it was =0, just incorporate 2 inside c, so $c=4x^2-2$ (or $c=3y^2-2$)

 

[JKCB]

Ah! Well that's going to be fun! Thank you so much!