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    Name:  1486678318676-2132571362.jpg
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Size:  417.9 KB hi , for this question I am stuck on the y values I obtained. From my working , there are two y values worked out for each solution of x. But the mark scheme only has on y value for each x value.Attachment 618352618356

    ThanksAttachment 618352618356
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    (Original post by coconut64)
    Name:  1486678318676-2132571362.jpg
Views: 19
Size:  417.9 KB hi , for this question I am stuck on the y values I obtained. From my working , there are two y values worked out for each solution of x. But the mark scheme only has on y value for each x value.Attachment 618352618356

    ThanksAttachment 618352618356
    Mark scheme attachment doesn't seem to be working, but you're supposed to substitute x = 2, -2 (assuming this is correct) back into the y = -x equation, not into the original curve equation.
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    (Original post by tajtsracc)
    Mark scheme attachment doesn't seem to be working, but you're supposed to substitute x = 2, -2 (assuming this is correct) back into the y = -x equation, not into the original curve equation.
    why is that?
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    (Original post by coconut64)
    why is that?
    When \frac{dy}{dx}=0, you have found that y=-x so you need to find where y=-x on the curve. To do that you need to solve these simultaneous equations:

    y=-x
    x^2+2xy-3y^2+16=0

    You solved these equations to get x=2 and x=-2. If you remember (GCSE) you can normally plug in these into either of your simultaneous equations to get the y values. So it's best to choose the simpler one y=-x. You can use the more complicated equation but that will give you two y solutions and only one of them will also be valid in y=-x so you'd have to check.

    So it makes sense to always plug the x values into the simpler equation. Does this make sense?
 
 
 
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