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A-Level Maths [Help] - Simultaneous equation Watch

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x+y = 9

x^2 - 3xy+2y^2 = 0

    I know how to do it and what to do etc.
    I may have made a mistake somewhere but was just wondering if someone could tell me what answer they got when they attempted this.

    If possible could you make  x = 9-y a this is what I done and it would be helpful knowing where i went wrong

    I got a negative root (used quadratic formula) but i don't think i am suppose to get one but not sure

    Thanks
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    (Original post by AdeptDz)
    

x+y = 9

x^2 - 3xy+2y^2 = 0

    I know how to do it and what to do etc.
    I may have made a mistake somewhere but was just wondering if someone could tell me what answer they got when they attempted this.

    If possible could you make  x = 9-y a this is what I done and it would be helpful knowing where i went wrong

    I got a negative root (used quadratic formula) but i don't think i am suppose to get one but not sure

    Thanks
    Okay so you got x=9-y

    Therefore subbing it into the other one would give you (9-y)^2-3y(9-y)+2y^2=0

    \Rightarrow (y^2-18y+81)+(3y^2-27y)+2y^2=0

    and have a go from there.
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    Post your working and I'll check it.
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    (Original post by RDKGames)
    Okay so you got x=9-y

    Therefore subbing it into the circle/ellipse would give you (9-y)^2-3y(9-y)+2y^2=0

    \Rightarrow (y^2-18y+81)+(3y^2-27y)+2y^2=0

    and have a go from there.
    I added those stuff to get

     54 - 18y + 6y^2 = 0



then i divided all this by 6 to get 

y^2 - 3y + 9 = 0
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    (Original post by AdeptDz)
    I added those stuff to get

     54 - 18y + 6y^2 = 0



then i divided all this by 6 to get 

y^2 - 3y + 9 = 0
    You added wrong. The 54 doesn't even appear anywhere, where did you pull that from??
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    (Original post by RDKGames)
    You added wrong.
    I can't see where
    (i editted post btw if maybe you saw the old one)
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    (Original post by AdeptDz)
    I can't see where
    (i editted post btw if maybe you saw the old one)
    Check again. Like for one, where do you even pull the 54 from when the 81 is the ONLY constant term??
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    (Original post by RDKGames)
    Check again. Like for one, where do you even pull the 54 from when the 81 is the ONLY constant term??
    oh yeh thanks I see where i went wrong..
    I done 27 instead of 27y so i got 81-27 = 54.
    Thanks ill try it with the correction
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    (Original post by RDKGames)
    Okay so you got x=9-y

    Therefore subbing it into the other one would give you (9-y)^2-3y(9-y)+2y^2=0

    \Rightarrow (y^2-18y+81)+(3y^2-27y)+2y^2=0

    and have a go from there.
    6y^2 + 45y + 81
    ?
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    (Original post by AdeptDz)
    6y^2 + 45y + 81
    ?
    Correct numbers though signs aren't all right. Also =0.
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    Is 3y^2+15y+27

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    Sorry 2y^2

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    (Original post by moshe015)
    Is 3y^2+15y+27

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    1. Close (even with 2y^2 you're wrong)
    2. You forgot the equals sign, so technically you cannot divide by anything.
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    (Original post by RDKGames)
    Correct numbers though signs aren't all right. Also =0.
    6y^2 - 45y + 81
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    No need to be pedantic

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    (Original post by AdeptDz)
    6y^2 - 45y + 81
    = 0

    Yep. You can carry on from there
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    (Original post by RDKGames)
    = 0

    Yep. You can carry on from there
    Thanks
    I make like 100 mistakes on the easiest things. It's so annoying
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    (Original post by moshe015)
    No need to be pedantic

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    Hard not to be when you're technically wrong. Especially with the wrong sign on the second term.
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    True that. I just saw a quadratic and simplified it. Didnt check to see if the sign was correct in the original question.

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