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could someone check these simultaneous equation answers? Watch

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    the equations were

    x+2y=13
    x^2+y^2=9

    I got the answers y=40/3 x=-43/3 And y=4 x=5

    whenever I plug these in they're wrong though whats the answer
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    (Original post by jonjoshelvey21)
    the equations were

    x+2y=13
    x^2+y^2=9

    I got the answers y=40/3 x=-43/3 And y=4 x=5

    whenever I plug these in they're wrong though whats the answer
    Well that means you've went wrong somewhere in your working. Try again by substituting x=13-2y into x^2+y^2=9
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    (Original post by jonjoshelvey21)
    the equations were

    x+2y=13
    x^2+y^2=9

    I got the answers y=40/3 x=-43/3 And y=4 x=5

    whenever I plug these in they're wrong though whats the answer
    First you rearange equation 1 by subtracting 2y from both sides to get the equation x=13-2y

    Then you substitute this value into equation 2 to get (13-2y)^2+y^2=9 which you the expand to 169-52y+4y^2+y^2=9

    This is simplified to 5y^2-52y+160=0

    From my experiance I can see from this equation that in can't be factorised unless you use imaganary numbers which is too high a level for this. Are you sure you typed the eqautions in properly?
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    (Original post by PandahatDude)
    First you rearange equation 1 by subtracting 2y from both sides to get the equation x=13-2y

    Then you substitute this value into equation 2 to get (13-2y)^2+y^2=9 which you the expand to 169-52y+4y^2+y^2=9

    This is simplified to 5y^2-52y+160=0

    From my experiance I can see from this equation that in can't be factorised unless you use imaganary numbers which is too high a level for this. Are you sure you typed the eqautions in properly?
    yeah sorry should be for the seconds equation x squared minus y squared equals nine
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    (Original post by jonjoshelvey21)
    yeah sorry should be for the seconds equation x squared minus y squared equals nine
    Okay, so you would instead end up with 3y^2-52y+160=0

    This can then be put into the quadratic formula to find y=\frac{40}{3} or y=4

    You then substitute these values into the equation x=13-2y to find that when y=\frac{40}{3},x=-\frac{41}{3} and when y=4,x=5

    I hope this helped
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    (Original post by PandahatDude)
    Okay, so you would instead end up with 3y^2-52y+160=0

    This can then be put into the quadratic formula to find y=\frac{40}{3} or y=4

    You then substitute these values into the equation x=13-2y to find that when y=\frac{40}{3},x=-\frac{41}{3} and when y=4,x=5

    I hope this helped
    thanks it did a lot. btw what are imaginary numbers ?
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    (Original post by jonjoshelvey21)
    thanks it did a lot. btw what are imaginary numbers ?
    If you ever take further maths in A Level then you get what is known as imaginary numbers. It is used when you are trying to square root a negative number which is impossible as any number times by itself is alway positive (e.g.-1^2=1) and therefore this is a problem when trying to solve complex eqautions which need you to find the sqaure root of a negative.

    We say that the \sqrt{-1}=i. This then leads to lots of other crazy things where you can plot imaginary numbers alongside "real" numbers on a graph etc. I won't go into to much detail but if you want try putting the \sqrt{-1} in your calculator and it will come up with a math error as the number does not exist.
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    (Original post by PandahatDude)
    If you ever take further maths in A Level then you get what is known as imaginary numbers. It is used when you are trying to square root a negative number which is impossible as any number times by itself is alway positive (e.g.-1^2=1) and therefore this is a problem when trying to solve complex eqautions which need you to find the sqaure root of a negative.

    We say that the \sqrt{-1}=i. This then leads to lots of other crazy things where you can plot imaginary numbers alongside "real" numbers on a graph etc. I won't go into to much detail but if you want try putting the \sqrt{-1} in your calculator and it will come up with a math error as the number does not exist.
    oh right very interesting thank you. I will research into it a bit further
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    (Original post by jonjoshelvey21)
    oh right very interesting thank you. I will research into it a bit further
    Imaginary numbers are A level further maths. I don't think you need to know this yet.
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    (Original post by joyoustele)
    Imaginary numbers are A level further maths. I don't think you need to know this yet.
    I'm kind of wishing id also taken further maths now. it sounds really interesting lol
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