# 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.

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

Of course things got weird.
you don't have an equation that states that the difference in their times is 2

• 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