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Math Help - System of Linear Equations - Two Variable - Distance Time Speed Question

  1. #1
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    System of Linear Equations - Two Variable - Distance Time Speed Question

    hey im stumped at developing a formula for this question:
    Chris walks at 8 km/h and runs at 12 km/h. One day he walks and runs on the way from his house to the library. It takes him 20 minutes. On his way back from the library he runs twice as far and the journey home takes 17.5 minutes. How far is his house from the library?
    Could you please show me how you did the question?
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  2. #2
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    Let D be the total distance to the library.
    Let d be the distance covered running *to* the library.
    This means D - d is the distance covered walking *to* the library.

    From the problem you know d/(12 km/h) + (D-d)/(8 km/h) = 20 min

    The distance covered running *from* the library is 2d.
    This means D - 2d is the distance covered walking *from* the library.

    From the problem you know 2d/(12 km/h) + (D - 2d)/(8 km/h) = 17.5 min

    Solve for D
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  3. #3
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    Quote Originally Posted by snowtea View Post
    Let D be the total distance to the library.
    Let d be the distance covered running *to* the library.
    This means D - d is the distance covered walking *to* the library.

    From the problem you know d/(12 km/h) + (D-d)/(8 km/h) = 20 min

    The distance covered running *from* the library is 2d.
    This means D - 2d is the distance covered walking *from* the library.

    From the problem you know 2d/(12 km/h) + (D - 2d)/(8 km/h) = 17.5 min

    Solve for D
    thanks so much
    when I did it as you told me to i got D=180 and the answer in my math book said 3 km
    so then i did the same thing but i changed 20 min into 1/3 hour and 17.5 min into 7/24 hour and I got 3 km for D
    most of the time these units mess me up
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  4. #4
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    Yes, you always have to make sure your units match, which is why I included them.
    Good job working out the correct answer.
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  5. #5
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    Hello, who192!

    Another approach . . .


    \text}Chris walks at 8 km/hr and runs at 12 km/hr. \;One day he walks}
    \text{and runs from his house to the library. \;It takes him 20 minutes.}

    \text{On his way back, he runs twice as far and the journey home}
    \text{takes 17.5 minutes. \;How far is his house from the library?}

    Going to the library, let \,R = distance he ran, \,W = distance he walked.

    He ran \,R km at 12 km/hr. .This took \frac{R}{12} hours.

    He walked \,W km at 8 km/hr. .This took \frac{W}{8} hours.

    His total time is: . 20\text{ minutes }=\:\frac{1}{3}\text{ hour.}

    . . Then: . \displaystyle \frac{R}{12} + \frac{W}{8} \:=\:\frac{1}{3} \quad\Rightarrow\quad 2R + 3W \:=\:8 .[1]



    Going home, he ran 2R km and walked W-R km.

    He ran 2R km at 12 km/hr. .This took \frac{2R}{12} hours.

    He walked W-R km at 8 km/hr. .This took \frac{W-R}{8} hours.

    His total time is: . 17.5\text{ minutes }=\:\frac{7}{24}\text{ hour.}

    . . Then: . \dfrac{2R}{12} + \dfrac{W-R}{8} \:=\:\dfrac{7}{24} \quad\Rightarrow\qud R + 3W \:=\:7 .[2]



    Subtract [1] - [2]: . R \:=\:1 \quad\Rightarrow\quad W \:=\:2


    The total distance is: . R + W \:=\:1 + 2 \:=\:3\text{ km.}

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