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Math Help - Motorboat and Current

  1. #1
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    Motorboat and Current

    A motorboat traveling with the current can go 30 miles in 2 hrs. Against the current , it takes 3 hrs. to go the same distance. Find the rate of the motorboat in calm water and the rate of the current!

    My set up:

    Let x = speed of motorboat in still water

    Let y = speed of current

    See attachment for table I created.

    From the table, I created the following system of equations but got the wrong answers.

    2x + 2y = 30
    3x - 3y = 30

    What did I do wrong?
    Attached Thumbnails Attached Thumbnails Motorboat and Current-table.jpg  
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  2. #2
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    Hello, sologuitar!


    Your table is correct and your equations are correct.
    It must be your algebra.


    We have: . \begin{array}{ccccccc}2x + 2y \;=\; 30 & \Rightarrow & x + y \;=\; 15 & [1] \\ 3x - 3y \;=\;30 & \Rightarrow & x - y \;=\; 10 & [2] \end{array}


    Add [1] and [2]: . 2x \:=\:25 \quad\Rightarrow\quad x \:=\:\frac{25}{2}


    Substitute into [1]: . \frac{25}{2} + y \:=\:15 \quad\Rightarrow\quad y \:=\:\frac{5}{2}


    Therefore: . \begin{array}{cc}\text{Motorboat} & 12\frac{1}{2}\text{ mph} \\ \\[-4mm]\text{Current} & 2\frac{1}{2}\text{ mph} \end{array}

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  3. #3
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    soroban..

    Quote Originally Posted by Soroban View Post
    Hello, sologuitar!


    Your table is correct and your equations are correct.
    It must be your algebra.


    We have: . \begin{array}{ccccccc}2x + 2y \;=\; 30 & \Rightarrow & x + y \;=\; 15 & [1] \\ 3x - 3y \;=\;30 & \Rightarrow & x - y \;=\; 10 & [2] \end{array}


    Add [1] and [2]: . 2x \:=\:25 \quad\Rightarrow\quad x \:=\:\frac{25}{2}


    Substitute into [1]: . \frac{25}{2} + y \:=\:15 \quad\Rightarrow\quad y \:=\:\frac{5}{2}


    Therefore: . \begin{array}{cc}\text{Motorboat} & 12\frac{1}{2}\text{ mph} \\ \\[-4mm]\text{Current} & 2\frac{1}{2}\text{ mph} \end{array}
    Believe it or not, I got the same answers but decided to reject the fractions. How about that?
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