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Math Help - solve using elimination method.

  1. #1
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    solve using elimination method.

    1)3y^2=x
    y^2+(x+2)^2=4

    2)xy=30
    x^2/6=56-2y^2
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  2. #2
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    Hello, todd85!


    - . . . . . 3y .= .x . [1]

    y + (x+2) .= .4 . [2]

    Multiply [2] by 3: . 3y + 3(x+2) .= .12
    . . . Subtract [1]: .-3y . . . . . . . . = .-x

    And we have: . 3(x + 2) .= .12 - x . . 3x + 13x .= .0

    Hence: . x(3x + 13) .= .0 . . x .= .0, -13/3


    In [1], we see that x cannot be negative.

    Therefore, the only solution is: .x = 0, .y = 0




    . .xy .= .30 . 1 . . .[1]
    x/6 .= .56 - 2y . [2]
    Since they insist on Elimination, the approach is trickier.


    From [1], we have: .x .= .30/y . . x .= .900/y . [3]

    From [2], we have: .x + 12y .= .336 . [4]


    Subtract [3] from [4]: .12y .= .336 - 900/y . . 12y - 336 + 900/y .= .0

    Multiply by y/12: . y^4 - 27y - 75 .= .0

    Factor: . (y - 3)(y - 25) .= .0

    . - -.- . . . . . . . . _
    Hence: . y .= .√3, .5
    . - - - . . . . . . . . . ._
    . .And: . x .= .10√3, .6

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  3. #3
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    Im doing this same unit in school you shouldn't always have to use elimination.

    at least in the class that im in we get to choose between substitution and elimination. unless you don't no how to substitution and if you don't send me a message or something.
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