finding solutions for a mixed variable equation

• Jun 17th 2011, 02:10 AM
kiwi5
finding solutions for a mixed variable equation
Hi there, this is a for a calculus question but it's more the algebra I'm having trouble with at the moment.
I've reduced an equation down to:

2y^2 + x^2 - 4xy + 4y - 4x = 0

and need to find all (x,y) pairs that make it true. Could someone show each step in solving this?
Thanks a lot
• Jun 17th 2011, 04:14 AM
TKHunny
Re: finding solutions for a mixed variable equation
1) Did you encounter "Rotation of Axes" in your pre-calculus studies?

2) Without reference to the geometry, you can make a substitution that will solve the problem. $x = 2(u+v)$ and $y = \sqrt{2}(u-v)$ is one such substitution. You can then Complete the Square in u and in v and you'll ne much closer to a solution.
• Jun 17th 2011, 04:19 AM
Soroban
Re: finding solutions for a mixed variable equation
Hello, kiwi5!

Quote:

$\text{I've reduced an equation down to: }\:x^2 -4xy+ 2y^2 -4x + 4y \:=\: 0$
. . Are you sure this is correct?
$\text{and need to find }all\text{ pairs }(x,y)\text{ that make it true. }$ . ALL of them??

I'm quite certain that your equation is incorrect.

It represents a hyperbola . . . and hence has brizillions of solutions.

• Jun 17th 2011, 04:53 AM
abhishekkgp
Re: finding solutions for a mixed variable equation
Quote:

Originally Posted by kiwi5
Hi there, this is a for a calculus question but it's more the algebra I'm having trouble with at the moment.
I've reduced an equation down to:

2y^2 + x^2 - 4xy + 4y - 4x = 0

and need to find all (x,y) pairs that make it true. Could someone show each step in solving this?
Thanks a lot

you can look at it as a quadratic in $x$. the equation becomes $x^2-(4y+4)x+(2y^2+4y)=0$. now apply the formula for solving a quadratic and you will have found $x$ in terms of $y$. that will do.