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    The question is, find all real numbers x, y and z which satisfy the simulataneous equations

     x^2 - 4y + 7 = 0, y^2 - 6z + 14 = 0 \ \text{and} \ z^2 - 2x - 7 = 0

    I got x = 1, y = 2 and z = 3, are those the only solutions?

    Thanks.
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    (Original post by 0x2a)
    The question is, find all real numbers x, y and z which satisfy the simulataneous equations

     x^2 - 4y + 7 = 0, y^2 - 6z + 14 = 0 \ \text{and} \ z^2 - 2x - 7 = 0

    I got x = 1, y = 2 and z = 3, are those the only solutions?

    Thanks.
    http://www.wolframalpha.com/input/?i...over+the+reals
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    I forgot the ultimate cheating machine.

    Thank you.
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    Yep.

    add all of these equations to each other, and factorise (completing the square) and you get:

    (x-1)^{2}+(y-2)^{2}+(z-3)^{2}=0 (a single point)
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    (Original post by Hasufel)
    Yep.

    add all of these equations to each other, and factorise (completing the square) and you get:

    (x-1)^{2}+(y-2)^{2}+(z-3)^{2}=0 (a single point)
    Is that a sphere then?
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    (Original post by keromedic)
    Is that a sphere then?
    The equation is in the form which gives a sphere, but it's radius is 0.
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    (Original post by 0x2a)
    The equation is in the form which gives a sphere, but it's radius is 0.
    Does that make it a sphere in the complex plane?
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    (Original post by keromedic)
    Does that make it a sphere in the complex plane?
    The z is only used here as another "axis" along with the x and the y. A shape of the form  (x - x_1)^2 + (y - y_1)^2 + (z - z_1)^2 = r^2 would appear in  \mathbb{R}^3 , unless you're explicitly dealing with complex numbers.
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    (Original post by keromedic)
    Does that make it a sphere in the complex plane?
    It makes it a point.
 
 
 
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