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
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
75 t = 125 (t-2)
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?
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.
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?
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
0 = 125 y - 75 y -150
150 = 50 y
y = 3
then go back for x
x = y+2 = 5
Math - A train leaves a station and travels north at a speed of 45mph. Four ...
Algembra - a train leaves a station and travels north at 85 km/hr. Four hours ...
Algebra - A train leaves a city heading west and travels at 50 miles per hour. ...
pre algebra - train A leaves a station at 1:15 PM,traveling 60 mph.Train B ...
Algebra 1 - a train leaves a city heading west and travels at 50 miles per hour...
Math - Train A and train B leave station going in opposite directions. Train B ...
algebra - Train A leaves a station traveling at 80km/h. Two hours later, train B...
physics - A train moving with constant velocity travels 190 m north in 12 and an...
Physics - A train moving with constant velocity travels 190 north in 14s and an ...
Math Word Problem - A metra commuter train leaves Union Station in Chicago at 12...