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Math Help - System of Equations

  1. #1
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    System of Equations

    Please help with the following problem. Thank You!

    Two planes leave Pittsburgh and Philadelphia at the same time. Each going to the other city. One plane flies 25mph faster than the other plane. Find the air speed of each plane if the cities are 275 miles apart and the planes pass each other after 40 minutes of flying time.
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    Quote Originally Posted by ss103 View Post
    Please help with the following problem. Thank You!

    Two planes leave Pittsburgh and Philadelphia at the same time. Each going to the other city. One plane flies 25mph faster than the other plane. Find the air speed of each plane if the cities are 275 miles apart and the planes pass each other after 40 minutes of flying time.
    let v = speed of slower plane in mph

    v+25 = speed of the faster plane in mph

    note that 40 min = \frac{2}{3} hr

    (speed)(time) = distance

    v\left(\frac{2}{3}\right) + (v+25)\left(\frac{2}{3}\right) = 275

    solve for v
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  3. #3
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    thanks! but we have to find the speed of each plane, separately. v is just the speed of one plane
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    Quote Originally Posted by ss103 View Post
    Please help with the following problem. Thank You!

    Two planes leave Pittsburgh and Philadelphia at the same time. Each going to the other city. One plane flies 25mph faster than the other plane. Find the air speed of each plane if the cities are 275 miles apart and the planes pass each other after 40 minutes of flying time.
    Let the speed of first plane (which leaves Pittsburgh) = x mph
    Let the speed of second plane = (x + 25) mph

    Let the two planes pass each other at distance "y" from Pittsburgh, after 40 min.

    So, distance travelled by first plane in 40 min = y
    Speed = x
    Time = 40 min = 40/60 h = 2/3 h

    Distance = speed . Time

    so, y = x \times \frac{2}{3}

    y = \frac{2}{3}\;x

    3y = 2x .....................................(1)

    Now, distance travelled by second plane in 40 min = (275 - y)
    Speed = (x + 25)
    Time = 40 min = 40/60 h = 2/3 h

    Distance = speed . Time

    so, 275 - y = (x+25) \times \frac{2}{3}

    275 - y = \frac{2}{3}\;(x+25)

    3(275 - y) = 2(x+25)

    825 - 3y = 2x + 50

    2x + 3y = 775 .................................(2)

    Now, solve these two eqns (1) and (2). FINISH it.
    The speeds of two planes will be 193.75 mph and 218.75 mph.
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