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Math Help - Finding the Number of ordered pairs that satisfy an equation. Please help.

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    Finding the Number of ordered pairs that satisfy an equation. Please help.

    Finding the Number of ordered pairs that satisfy an equation. Please help.-question-2.jpg

    Find the total number of ordered pairs (x,y) which satisfy the equation x^2-4xy+5y^2+2y-4 = 0



    .... I just know how to complete the square. but I'm not sure how to count the number of ordered pairs. Thanks.
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    Re: Finding the Number of ordered pairs that satisfy an equation. Please help.

    Quote Originally Posted by kokoro123 View Post
    Click image for larger version. 

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    Find the total number of ordered pairs (x,y) which satisfy the equation x^2-4xy+5y^2+2y-4 = 0



    .... I just know how to complete the square. but I'm not sure how to count the number of ordered pairs. Thanks.
    \displaystyle \begin{align*} x^2 - 4x\,y + 5y^2 + 2y - 4 &= 0 \\ x^2 - 4x\,y + \left(-2y\right)^2 - \left(-2y\right)^2 + 5y^2 + 2y - 4 &= 0 \\ \left(x - 2y\right)^2 - 4y^2 + 5y^2 + 2y - 4 &= 0 \\ \left(x - 2y\right)^2 + y^2 + 2y - 4 &= 0 \\ \left(x - 2y\right)^2 + y^2 + 2y + 1^2 - 1^2 - 4 &= 0 \\ \left(x - 2y\right)^2 + \left(y + 1\right)^2 - 5 &= 0 \end{align*}

    This is an ellipse. It's clear there are an infinite number of points that satisfy this equation, since the relation is continuous.

    (x - 2y)^2 + (y + 1)^2 - 5=0 - Wolfram|Alpha
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    Re: Finding the Number of ordered pairs that satisfy an equation. Please help.

    thanks to Prove It for all the grunt work ...

    Quote Originally Posted by Prove It View Post
    \left(x - 2y\right)^2 + \left(y + 1\right)^2 - 5 = 0

    This is an ellipse. It's clear there are an infinite number of points that satisfy this equation, since the relation is continuous.
    the original question (in the attachment) asks for total number of integer ordered pairs (x,y)

    so, two possible cases to consider ...

    (x-2y)^2 = 4 \, , \, (y+1)^2 = 1

    or

    (x-2y)^2 = 1 \, , \, (y+1)^2 = 4

    ... leave it for you to finish.
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