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Maths simultaneous equations help!!! Watch

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    My text book has the following equation

    x^2+y^2=10
    x+y=4

    I tried both substitution and elimination but didn't get the right answer that the answer book states, being x=3 y=1, or x=1 y=3.

    Please can anyone help me!!!
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    Using substitution:

    1. x^2 + y^2 = 10
    2. x + y = 4
    x = 4 - y

    substituting x into (1)
    (4-y)^2 + y^2 = 10
    (4 - y)^2 = 16 - 8y + y^2
    16 - 8y + y^2 + y^2 = 10
    2y^2 - 8y + 6 = 0
    y^2 - 4y + 3 = o
    (y - 3)(y - 1) = 0

    therefore y = 3, and y = 1

    substituting these values into the second equation:

    y = 3
    x = 4 - 3
    therefore x = 1

    y = 1
    x = 4 - 1
    therefore x = 3



    Hope this helps.
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    (Original post by Lauren99)
    Using substitution:

    1. x^2 + y^2 = 10
    2. x + y = 6 x = 4 - y

    substituting x into (1)
    (4-y)^2 + y^2 = 10 (4 - y)^2 = 16 - 8y + y^2
    16 - 8y + y^2 + y^2 = 10
    2y^2 - 8y + 6 = 0
    y^2 - 4y + 3 = o
    (y - 3)(y - 1) = 0

    therefore y = 3, and y = 1

    substituting these values into the second equation:

    y = 3 x = 4 - 3 therefore x = 1

    y = 1 x = 4 - 1 therefore x = 3



    Hope this helps.
    Where did x+y=6 come from???
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    (Original post by Want_To_Achieve)
    Where did x+y=6 come from???
    I have figured it out now by reading the rest

    Thanks this really helped
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    (Original post by Want_To_Achieve)
    My text book has the following equation

    x^2+y^2=10
    x+y=4

    I tried both substitution and elimination but didn't get the right answer that the answer book states, being x=3 y=1, or x=1 y=3.

    Please can anyone help me!!!
    you know y = 4 - x
    meaning you know that y^2 is equals to (x-4)^2
    and when you put that back into the equation to just get x in the equation and simplify it down, you get 2x^2 - 8x + 6 = 0 which you can factorise to give you
    (2x-6) and (x-1)

    that means that x = 3 or 1, and then sub that back into x + y = 4 to get the y values.
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    nevermind lol
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    (Original post by Lauren99)
    Using substitution:

    1. x^2 + y^2 = 10
    2. x + y = 6 x = 4 - y

    substituting x into (1)
    (4-y)^2 + y^2 = 10 (4 - y)^2 = 16 - 8y + y^2
    16 - 8y + y^2 + y^2 = 10
    2y^2 - 8y + 6 = 0
    y^2 - 4y + 3 = o
    (y - 3)(y - 1) = 0

    therefore y = 3, and y = 1

    substituting these values into the second equation:

    y = 3 x = 4 - 3 therefore x = 1

    y = 1 x = 4 - 1 therefore x = 3



    Hope this helps.
    Of course it helps, you literally DID the bloody question for him/her step by step as if you were the one doing it in an exam.
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    (Original post by Want_To_Achieve)
    Where did x+y=6 come from???
    Sorry, typo.
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    (Original post by RDKGames)
    Of course it helps, you literally DID the bloody question for him/her step by step as if you were the one doing it in an exam.
    Well I needed someone who could go through a step by step method. So saying that she DONE the question is not what you should say.
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    (Original post by Want_To_Achieve)
    Well I needed someone who could go through a step by step method. So saying that she DONE the question is not what you should say.
    So what do you want me to say?

    Yes she did the question for you, that's what going step-by-step implies. You already knew what you needed to do so it was your algebraic manipulation was was wrong at some point. You could've easily shown your working out and we would've pointed out the mistake without even the effort to go through the question.
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    (Original post by RDKGames)
    Of course it helps, you literally DID the bloody question for him/her step by step as if you were the one doing it in an exam.
    CALM DOWN :lol:
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    (Original post by fatima1998)
    CALM DOWN :lol:
    NO! I WILL NOT STAND FOR THIS!! :fuhrer:
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    (Original post by RDKGames)
    NO! I WILL NOT STAND FOR THIS!! :fuhrer:
    :unimpressed: you have to
    SO CALM It's done so no point
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    (Original post by Lauren99)
    Using substitution:

    1. x^2 + y^2 = 10
    2. x + y = 4
    x = 4 - y

    substituting x into (1)
    (4-y)^2 + y^2 = 10
    (4 - y)^2 = 16 - 8y + y^2
    16 - 8y + y^2 + y^2 = 10
    2y^2 - 8y + 6 = 0
    y^2 - 4y + 3 = o
    (y - 3)(y - 1) = 0

    therefore y = 3, and y = 1

    substituting these values into the second equation:

    y = 3
    x = 4 - 3
    therefore x = 1

    y = 1
    x = 4 - 1
    therefore x = 3



    Hope this helps.
    Thanks for this step by step solution. I also had no clue how to solve this question, now I can. I'm also confident I can apply the techniques you've shown here to many different questions. On top of that, it didn't take long at all!

    This is a much better way of helping someone rather than giving people little pushes in the right direction.

    It's like being teased by a girl you're about to have sex with and nobody wants that do they?
 
 
 
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