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Area triangle, showing it can be represented by given equation Watch

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    The diagram of the triangle can be seen on the past paper in this link, it's question 10i:

    http://www.mei.org.uk/files/papers/c107ju_6gfd3.pdf

    The triangle's base length is 2x - 3 and its perpendicular height is x + 1, it's area is 9cm^2, show that 2x^2 - x - 21 = 0

    I got this so far:
    1/2(base x height)
    1/2(2x-3)(x+1) = 9
    1/2(2x^2 + 2x -3x -3) = 9
    1/2(2x^2 - x -3) = 9
    I can already see, I'm not going to get the right answer, the markscheme says you multiply the 9 by 2, but why?
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    Everything you've done is right, what the markscheme is doing is multiplying both sides by 2 so as to get rid of the half on the left hand side. With a simple rearrange you then have the right answer.
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    (Original post by Magenta96)
    The diagram of the triangle can be seen on the past paper in this link, it's question 10i:

    http://www.mei.org.uk/files/papers/c107ju_6gfd3.pdf

    The triangle's base length is 2x - 3 and its perpendicular height is x + 1, it's area is 9cm^2, show that 2x^2 - x - 21 = 0

    I got this so far:
    1/2(base x height)
    1/2(2x-3)(x+1) = 9
    1/2(2x^2 + 2x -3x -3) = 9
    1/2(2x^2 - x -3) = 9
    I can already see, I'm not going to get the right answer, the markscheme you muliply the 9 by 2, but why?
    9/0.5 = 9x2
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    (Original post by Magenta96)
    The diagram of the triangle can be seen on the past paper in this link, it's question 10i:

    http://www.mei.org.uk/files/papers/c107ju_6gfd3.pdf

    The triangle's base length is 2x - 3 and its perpendicular height is x + 1, it's area is 9cm^2, show that 2x^2 - x - 21 = 0

    I got this so far:
    1/2(base x height)
    1/2(2x-3)(x+1) = 9
    1/2(2x^2 + 2x -3x -3) = 9
    1/2(2x^2 - x -3) = 9
    I can already see, I'm not going to get the right answer, the markscheme you muliply the 9 by 2, but why?
    How else would you suggest getting rid of the 1/2
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    (Original post by Magenta96)
    The diagram of the triangle can be seen on the past paper in this link, it's question 10i:

    http://www.mei.org.uk/files/papers/c107ju_6gfd3.pdf

    The triangle's base length is 2x - 3 and its perpendicular height is x + 1, it's area is 9cm^2, show that 2x^2 - x - 21 = 0

    I got this so far:
    1/2(base x height)
    1/2(2x-3)(x+1) = 9
    1/2(2x^2 + 2x -3x -3) = 9
    1/2(2x^2 - x -3) = 9
    I can already see, I'm not going to get the right answer, the markscheme you muliply the 9 by 2, but why?
    I think it is a mistake to think your answer is wrong when you have used a different method to the examiner.

    Some marks scheme actually show several possible solutions to a problem. If I did a question today by a particular method I might well do it a different way next week. It's rather like a car journey - often several acceptable routes and you don't necessarily know in advance which is the best route/solution.
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    I understand now, thank you everyone
 
 
 
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