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    I know the method but have spent ages wondering why I can't get the right values in the simultaneous equation:
    y=3x-1
    x^2-y^2+12x=0

    Can anyone help/solve please?


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    You know that

    y = 3x - 1

    therefore, substitute that into y^2 to get a quadratic in x (since you have y in terms of x), and solve as you would solve a quadratic normally. Then, once you've ascertained x, substitute that into

    y = 3x - 1

    to find the corresponding y value(s).
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    Do you know what the answers are supposed to be? Also does the answer have to be whole?

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    I know that's the method but it is a non calculator question and I am assuming the answer is relatively basic numbers, I end up with the simplified quadratic -8x^2+6x+1=0, is this correct by this stage?


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    (Original post by Shadowninja107)
    I know that's the method but it is a non calculator question and I am assuming the answer is relatively basic numbers, I end up with the simplified quadratic -8x^2+6x+1=0, is this correct by this stage?


    Posted from TSR Mobile
    



y = 3x-1

    Therefore, y^2 = (3x-1)^2

    Substituting (3x-1)^2 into x^2 - y^2 +12x = 0 as the y^2:

    x^2 - (3x-1)^2 + 12x = 0

    Solve from there.
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    (Original post by Shadowninja107)
    I know that's the method but it is a non calculator question and I am assuming the answer is relatively basic numbers, I end up with the simplified quadratic -8x^2+6x+1=0, is this correct by this stage?


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    yep that's what I get, now just quadratic formula both of them and plug both answers back into the original y=3x-1 .
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    You should get:
    When x = 2.193, y = 5.579
    When x = 0.057, y=-0.829

    It depends on how you round it because you have to use the quadratic formula.
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    (Original post by Shadowninja107)
    I know that's the method but it is a non calculator question and I am assuming the answer is relatively basic numbers, I end up with the simplified quadratic -8x^2+6x+1=0, is this correct by this stage?


    Posted from TSR Mobile
    Bring all the terms on the other side so that it becomes positive rather than negative. It makes things much more easier. So you should get 8x^2-18x+1. It's 18x because 6x +12x =18x. Be careful of your signs.
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    Or do you just want me to show you the whole working out?
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    You should get 8x^2-18x+1=0 and from there, use the x=(-(b)^2± √(b)^2-4(ac))/2(a) formula to get the values of x, then substitue back into any equation of your choice to get the y values

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    (Original post by Brainiac_98)
    You should get 8x^2-18x+1=0 and from there, use the x=(-(b)^2± √(b)^2-4(ac))/2(a) formula to get the values of x, then substitue back into any equation of your choice to get the y values

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    Yeah, exactly!
 
 
 
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