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Algebra

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A train leaves Danville Junction and travels north at a speed of 75 km/h. Two hours later, a second train leaves on a parallel track and travels north at 125 km/h. How far from the station will they meet?

I have no idea what the system of equations would be for this problem, or how to find the system of equations.

Once I have the system of equations, I should be able to solve it myself.

Thanks! :)

  • Algebra - ,

    I used to encourage my students to make a chart for these kind of problems
    the rows of the chart are the different situations and the columns would be titled
    D(distance) R(rate) and T(time)

    ............D ........R ......T
    1st train: 75t ------ 75 ---- t
    2nd train: 125(t-2)-- 125 --- t-2

    didn't they travel the same distance?
    so 125(t-2) = 75t

    Notice I only used one variable, if your teacher insists that you use two of them,
    define first time as x
    the second time as y
    then the difference in their times is 2
    ---> x - y = 2
    and the second equation would be 75x = 125y

  • Algebra - ,

    Train 1 travels for time t at 75 km/hr
    Train 2 travels for time (t-2) at 125 km/hr
    they both go the same distance
    distance = rate * time
    therefore
    75 t = 125 (t-2)

  • Question - ,

    You see, here's what I did.

    I changed the "t" in the equation that both of you gave me (75t=125(t-2)) to an x so I could use elimination to solve the problem.

    That made that equation be
    125x - 250 = 75x
    +
    75x = 125 y

    That's where my problem comes along... it gets all weird and doesn't work. Is there another different equation I could have for the first equation in x and y form? Because I know that the second equation is basically the same as the first just w/ x's and y's, so is there another totally different equation?

    Thanks! :)

  • Algebra - ,

    You can't just toss around x's and y's indiscriminately.

    I gave you the corresponding equations if you have to use x and y

    your equation of 75x = 125y contradicts your equation 125x - 250 = 75x
    Of course things got weird.
    you don't have an equation that states that the difference in their times is 2

    Read my alternate solution.

  • Algebra - ,

    Oh, I kind of skimmed what you said and didn't notice the x-y=2... I'm sorry!

    Thanks so much though! :)

    P.S: Is the correct answer 375 miles away from the station, meaning x=5 and y=3?

  • Algebra - ,

    125x - 250 = 75x
    +
    75x = 125 y
    ------------------------
    but what is y in your system?
    You have just written the same fact twice.
    To make a system of two equations you need to state two facts
    first they go the same distance, at 75 for x hours and the other at 125 for y hours
    that is your
    75 x = 125 y
    now the other fact you are given is
    x = y + 2
    multiply that by 75 to use elimination
    75 x = 75 y + 150
    so
    0 = 125 y - 75 y -150
    150 = 50 y
    y = 3
    then go back for x
    x = y+2 = 5

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