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Math Help - Simultaneous Equations

  1. #1
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    Simultaneous Equations

    Please solve this:

    x^2+y^2=2
    3x-2y=5

    Please show your working!
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by jerrymalta View Post
    Please solve this:
    x^2+y^2=2
    3x-2y=5

    From the second equation:

    x=\dfrac{5+2y}{3}

    Now, substitute in the first one.

    Please show your working!

    Please, read rule 11 here .

    Fernando Revilla
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  3. #3
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    I thought this would be relatively straightforward but I can't seem to solve myself.

    x = (5 + 2y)/3

    substitute into x^2 + y^2 = 2

    ((5+2y)/3)^2 + y^2 = 2

    25/9 + 20/9y + 4/9y^2 + y^2 = 2

    13/9y^2 + 20/9y + 7/9 = 0

    x 9: 13y^2 + 20y + 7 = 0

    using -b+-rt b^2 - 4ac
    ----------------
    2a

    a=13, b=20, c=7
    y= -13/20 or y = -7/20

    substituting into 3x-2y=5
    when y=-13/20, x = 37/50
    when y = -7/20, x = 43/30

    But this is incorrect because putting these x and y values into circle equation don't produce 2. Can anyone see where I have gone wrong?
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  4. #4
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    Quote Originally Posted by angypangy View Post
    I thought this would be relatively straightforward but I can't seem to solve myself.

    x = (5 + 2y)/3

    substitute into x^2 + y^2 = 2

    ((5+2y)/3)^2 + y^2 = 2

    25/9 + 20/9y + 4/9y^2 + y^2 = 2

    13/9y^2 + 20/9y + 7/9 = 0

    x 9: 13y^2 + 20y + 7 = 0

    using -b+-rt b^2 - 4ac
    ----------------
    2a

    a=13, b=20, c=7

    it's fine to here

    y= -13/20 or y = -7/20

    substituting into 3x-2y=5
    when y=-13/20, x = 37/50
    when y = -7/20, x = 43/30

    But this is incorrect because putting these x and y values into circle equation don't produce 2. Can anyone see where I have gone wrong?
    \displaystyle\ 13y^2+20y+7=0\Rightarrow\ y=\frac{-20\pm\sqrt{400-364}}{26}

    \displaystyle\Rightarrow\ y=\frac{-20\pm6}{26}
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  5. #5
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    Thanks. I realised I was being careless. I will have to learn to be more accurate.
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