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Thread: distance time problem

  1. #1
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    distance time problem

    Q: After going a certain distance in an automoble, a driver found that if he had gone 3 miles an hour faster, he would have traveled the distance in 1 hour less time. and that if he had gone 5 miles an hour faster, he would have gone the distance in 1 hour and 30 minutes less.
    What was the distance?

    my attempt:
    let x = distance and y = time

    $\frac{x}{60}+\frac{3}{60} = \frac{x}{y -60}$

    $\frac{x}{60}+\frac{5}{60} = \frac{x}{y -90}$

    $(x+3)(y-60)=60x$

    $(x+5)(y-90)=60x$

    $-150x+5y=450-xy$

    $-120x+3y=180-xy$

    Im stuck at this point i dont think i got the right sim equations
    can someone please help me
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  2. #2
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    Re: distance time problem

    Q: After going a certain distance in an automoble, a driver found that if he had gone 3 miles an hour faster, he would have traveled the distance in 1 hour less time. and that if he had gone 5 miles an hour faster, he would have gone the distance in 1 hour and 30 minutes less.
    What was the distance?
    $v = \text{ original speed in mph}$

    $t = \text{ original time in hrs}$


    $vt = d$

    $(v+3)(t-1) = d$

    $(v+5)(t-1.5) = d$

    3 equations, 3 unknowns ... see what you can do.
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  3. #3
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    Re: distance time problem

    Quote Originally Posted by skeeter View Post
    $v = \text{ original speed in mph}$

    $t = \text{ original time in hrs}$


    $vt = d$

    $(v+3)(t-1) = d$

    $(v+5)(t-1.5) = d$

    3 equations, 3 unknowns ... see what you can do.
    thank you very much
    im sorry im not allowed to add to ur reputation
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