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Math

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x^2 + 9y^2 = 37

x - 2y = -3

  • Math - ,

    x = 2y-3, so we have

    (2y-3)^2 + 9y^2 = 37
    4y^2 - 12y + 9 + 9y^2 = 37
    13y^2 - 12y - 28 = 0
    (y-2)(13y+14) = 0

    so, solutions are (1,2) and (-67/3,-14/13)

  • Math - ,

    for 2nd equation, we get x = 2y-3

    thus plugging into first equation gets

    (2y-3)^2+9y^2 = 37.

    4y^2-12y+9+9y^2 = 13y^2-12y+9=37 = 13y^2-12y-28 = 0.

    Solving with the quadratic formula gives us: 2,-14/13

  • Math - ,

    I assume you are solving.

    from the 2nd: x = 2y-3
    sub into the 1st

    (2y-3)^2 + y^2 - 37 = 0
    4y^2 - 12y + 9 + 9y^2 - 37 = 0
    13y^2 - 12y - 28 = 0
    (y-2)(13y + 14) = 0
    y = 2 or y = -14/13

    if y = 2, then x = 2-3 = -1
    if y = -14/13, x = -28/13 - 3 = -67/13

    points of intersection:
    (-1,2) and (-67/13 , -14/13)

  • Math - ,

    arrgghhh!

    for first point:
    y = 2, then x = 4-3 = 1 , just like Steve had above

    point is (1,2)

  • Math - ,

    so, is this a typo fest, or what?
    Good job, Knights.

    Hope you can dig out the correct solutions, Princess!

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