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Math Help - Point of intersection

  1. #1
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    Point of intersection

    x^2 + 4y^2=20 and x+2y=6


    I solved for y and set them equal to each other but the x's cancel out. What else could I do?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Hmm

    Quote Originally Posted by MathNeedy18 View Post
    x^2 + 4y^2=20 and x+2y=6


    I solved for y and set them equal to each other but the x's cancel out. What else could I do?
    I think you made a mistake...solving and setting them equal I got \frac{-x}{2}+3=\sqrt{5-\frac{x^2}{4}}
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  3. #3
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    Quote Originally Posted by Mathstud28 View Post
    I think you made a mistake...solving and setting them equal I got \frac{-x}{2}+3=\sqrt{5-\frac{x^2}{4}}

    I think I am doing the algebra wrong because when I solve that, the 1/4 cancel out.
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  4. #4
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    Quote Originally Posted by MathNeedy18 View Post
    x^2 + 4y^2=20 and x+2y=6


    I solved for y and set them equal to each other but the x's cancel out. What else could I do?
     x^2 +(2y)^2=20 and 2y=6-x

    subbing into the first equation we get

    x^2+(6-x)^2=20

    x^2+36-12x+x^2=20 \iff 2x^2-12x+16=0

    2(x^2-6x+8)=0 \iff 2(x-4)(x-2)=0

    so x=2 or x=4 sub these values back into the linear equation

    2y=6-2 \iff y=2

    2y=6-4 \iff y=1

    so we get the two points (2,2) and (4,1)
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