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Math Help - Confused about how to solve simple problem...?

  1. #1
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    Confused about how to solve simple problem...?

    The distance between two cities is 10km more than one third of the distance between them?

    I understand the answer is 15 km - but how do i put the answer in an algebraic equation?
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  2. #2
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    Quote Originally Posted by EyesThatSparkle02 View Post
    The distance between two cities is 10km more than one third of the distance between them?

    I understand the answer is 15 km - but how do i put the answer in an algebraic equation?
    If d is the distance between the two cities then

    d = 10 + \frac{1}{3}d

    \frac{2}{3}d = 10

    d = \frac{3}{2}\cdot 10

    d = 15.
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  3. #3
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    Let the distance between the two cities be X

    X is 10km more than a third the distance between them
    Hint: When they say 1/3 distance between them, they're referring to X

    Algebraically
    X=10+\frac{1}{3}X
    \frac{2}{3}X=10
    X=15

    Edit: Beaten by Prove It, I'm slow tonight
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    Thanks guys,

    New Question: (Similar)
    If a 5 cm piece of wire is cut into two parts such that a square is formed by bending one part will have four times the area of a square formed by bending the other part, what is the length of the longer part?
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  5. #5
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    Quote Originally Posted by EyesThatSparkle02 View Post
    Thanks guys,

    New Question: (Similar)
    If a 5 cm piece of wire is cut into two parts such that a square is formed by bending one part will have four times the area of a square formed by bending the other part, what is the length of the longer part?
    Let P_1 = x be the perimeter of one of the squares (length of one of the pieces of wire).

    Its side length is therefore L_1 = \frac{1}{4}x.

    Thus its area is A_1 = L_1^2

     = \left(\frac{1}{4}x\right)^2

    = \frac{1}{16}x^2.



    Let P_2 = 5 - x be the perimeter of the other piece.

    Clearly P_1 + P_2 = 5.

    Its side length is therefore L_2 = \frac{1}{4}(5 - x).

    Thus its area is A_2 = \left[\frac{1}{4}(5 - x)\right]^2

     = \frac{1}{16}(5 - x)^2

     = \frac{1}{16}(25 - 10x - x^2)

     = -\frac{1}{16}x^2 - \frac{5}{8}x + \frac{25}{16}.


    We also know that A_2 = 4A_1

    Therefore

    -\frac{1}{16}x^2 - \frac{5}{8}x + \frac{25}{16} = \frac{1}{4}x^2

    0 = \frac{5}{16}x^2 + \frac{5}{8}x - \frac{25}{16}

    0 = x^2 + 2x - 5

     x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-5)}}{2(1)}

     = \frac{-2 \pm \sqrt{24}}{2}

     = \frac{-2 \pm 2\sqrt{6}}{2}

     = -1 \pm \sqrt{6}.


    Since you can not have a negative length

    x = -1 + \sqrt{6}.


    Remembering that the other length  = 5 - x

     = 5 - (-1 + \sqrt{6})

     = 5 + 1 - \sqrt{6}

     = 6 - \sqrt{6}.


    Put this into the calculator and you will know how many cm this length is.
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