Posted by Willow on .
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 
Reiny,
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(t2) 125  t2
didn't they travel the same distance?
so 125(t2) = 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 
Damon,
Train 1 travels for time t at 75 km/hr
Train 2 travels for time (t2) at 125 km/hr
they both go the same distance
distance = rate * time
therefore
75 t = 125 (t2) 
Question 
Willow,
You see, here's what I did.
I changed the "t" in the equation that both of you gave me (75t=125(t2)) 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 
Reiny,
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 
Willow,
Oh, I kind of skimmed what you said and didn't notice the xy=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 
Damon,
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