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Algebra-- PLEASE help!

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Pete's Plymouth travels 200 miles at a certain speed. If the car had gone 10 mph faster, the trip would have taken 1 hour less. Find Pete's speed.

I am very confused about how to solve this problem! Please explain well. Thank you so much!!! :-)

  • Algebra-- PLEASE help! - ,

    let his original speed be x mph
    then his original time is 200/x hours

    let his second speed be x+10 mph
    then his second time is 200/(x+10) hours

    so 200/x - 200/(x+10) = 1
    multiply both sides by x(x+10)

    200x + 2000 - 200x = x^2 + 10x
    x^2 + 10x - 2000 = 0
    (x-40)(x+50) = 0
    x = 40 or x= -50 (negative speed makes no sense)

    so Pete's speed is 40 mph.

    check: at 40 mph, it would take him 200/40 = 5 hours
    at 50 mph it would have taken him 200/50 = 4 hours.

    OK then!

  • Algebra-- PLEASE help! - ,

    Lat V be the slower speed and T be the time it takes to go 200 miles at that speed. You can write these two equations:

    T = 200/V

    T = 200/(V + 10) + 1

    Now get rid of the T and create and solve an equation for v alone.

    200 [1/V - 1/(V+10) = 1

    200(V + 10 - V) = V(V+10)
    V^2 + 10V -2000 = 0
    (V-40)(V+50) = 0
    V = 40 since only positive roots are allowed.

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