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Simultaneous equations-one linear, one quadratic! watch

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    Solve x+y=2 and 3x^2-2c+y^2=2 simultaneously.

    Well, i got up to 3y^2-12y+12-2y+4+y^2=2 so far?! Is that right? What next?
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    is the -2c meant to be -2c or -2x?
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    (Original post by El Stevo)
    is the -2c meant to be -2c or -2x?
    sorry, -2x!
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    next collect all the terms together to get a quadratic in y. solve for y, and sub into one of the equations to get x. then you will have a pair of solutions for x and y.
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    I got y=1, y=3/2 and corres. values of x are x=1 x=1/2. hope that helps
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    Solve x+y=2 and 3x^2-2c+y^2=2 simultaneously.

    [1] x + y = 2 -----> change this to y=2 - x

    [2] 3x² - 2x + y² = 2

    In [2] put (2-x) instead of y

    [2] 3x² - 2x + (2-x)² = 2

    ...3x² - 2x + 4 - 4x + x² = 2
    Rearrange in the form ax² + bx + c = 0
    ...4x² - 6x + 2 = 0
    Take out a factor of 2
    ...2(2x² - 3x +1) = 0
    ...2x² - 3x +1=0
    Solving this using the quadratic formula gives, x = 0.5 and x = 1
    Substituting into either equation to find the corresponding y values gives,
    (0.5, 1.5) and (1, 1)
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    thank you!!!
 
 
 
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Updated: January 22, 2005

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