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Math Help - finding solutions for a mixed variable equation

  1. #1
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    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
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  2. #2
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    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.
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  3. #3
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    Re: finding solutions for a mixed variable equation

    Hello, kiwi5!

    \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.

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  4. #4
    Senior Member abhishekkgp's Avatar
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    Re: finding solutions for a mixed variable equation

    Quote Originally Posted by kiwi5 View Post
    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.
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