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Math Help - linear equation problem solving

  1. #1
    Newbie white's Avatar
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    Exclamation linear equation problem solving

    Need help on these questions..

    question 1- show that 2/c = 1/a+1/b

    attempt:
    2ab=bc + ac...

    question 2 - Xiu travels from town A to B at u km/h and then returns at v km/h. Town A is d km from town B
    a) if i t takes T hours to travel from A to B, find the time taken:
    i) the entire trip in terms of T, u and v

    attempt:
    time = distance/ speed
    u + v =speed back and forth
    distance is d km
    so d/u+v ??

    question 3- A man walks 70 km. He walks x km at 8 km/h and y km at 10km/h
    a) fins his average speed in terms of x and y.

    attempt:
    average speed is = distance/ time
    70/8x+10y ??



    thanks!
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  2. #2
    Lord of certain Rings
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    Quote Originally Posted by white View Post
    Need help on these questions..

    question 1- show that 2/c = 1/a+1/b

    attempt:
    2ab=bc + ac...
    How can we prove 2/c = 1/a+1/b without knowing anything about c,a and b?! So please post the complete question...


    question 2 - Xiu travels from town A to B at u km/h and then returns at v km/h. Town A is d km from town B
    a) if it takes T hours to travel from A to B, find the time taken:
    i) the entire trip in terms of T, u and v

    attempt:
    time = distance/ speed
    u + v =speed back and forth
    distance is d km
    so d/u+v ??
    Wrong...

    (a) says "it takes T hours to travel from A to B".So there is no "back and forth" trip.

    d km is the distance from A to B.

    time = distance/ speed

    T = \frac{d}{u}

    Let the "back and forth trip" take time 't', then the total time is:

    t = \frac{d}{u} + \frac{d}{v}

    But d  = uT, so:

    t = \frac{uT}{u} + \frac{uT}{v} = T\left(1 + \frac{u}{v}\right)

    Thus t = T\left(1 + \frac{u}{v}\right) is the time taken for the entire trip in terms of T, u and v.

    question 3- A man walks 70 km. He walks x km at 8 km/h and y km at 10km/h
    a) fins his average speed in terms of x and y.

    attempt:
    average speed is = distance/ time
    70/8x+10y ??
    average speed = total distance/total time

    \text{total distance }= x + y = 70

    \text{total time }= \frac{x}{8} + \frac{y}{10}
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  3. #3
    Newbie fresh_'s Avatar
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    question 1
    a man on a bicycle rides one third of the way from town A to town B at a speed a km/h and the remainder of the way at 2blm/h

    show that 2/c = 1/a+1/b
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  4. #4
    Lord of certain Rings
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    Quote Originally Posted by fresh_ View Post
    question 1
    a man on a bicycle rides one third of the way from town A to town B at a speed a km/h and the remainder of the way at 2blm/h

    show that 2/c = 1/a+1/b
    Ok but what is c?
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  5. #5
    Newbie white's Avatar
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    sorry guys , i didnt email the full quessiton to fresh .
    here it is..
    question 1
    a man on a bicycle rides one third of the way from town A to town B at a speed a km/h and the remainder of the way at 2blm/h
    a) if the distance between the two towns is 9km find the time taken to ride form A to B.

    if the man travelled at a uniform rate of 3c km/h, he could jave ridden from A to B and back again in the same time

    b)show that 2/c = 1/a+1/b
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  6. #6
    Lord of certain Rings
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    Quote Originally Posted by white View Post
    sorry guys , i didnt email the full quessiton to fresh .
    here it is..
    question 1
    a man on a bicycle rides one third of the way from town A to town B at a speed a km/h and the remainder of the way at 2blm/h
    a) if the distance between the two towns is 9km find the time taken to ride form A to B.

    if the man travelled at a uniform rate of 3c km/h, he could jave ridden from A to B and back again in the same time

    b)show that 2/c = 1/a+1/b
    Since the total distance is 9 km, one third is 3 km. Thus the total time taken to move from A to B is:

    T = \frac{3}{a}+\frac{6}{2b}

    So to move from A to B and back again, he travels 18 km at a uniform rate of 3c km/hr, the time he takes is \frac{18}{3c}.

    According to the question, this is same as T.

    So

    T =\frac{18}{3c} = \frac{3}{a}+\frac{6}{2b}

    \frac{6}{c} = \frac{3}{a}+\frac{3}{b}

    \frac{2}{c} = \frac{1}{a}+\frac{1}{b}
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