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October 1, 2014

October 1, 2014

Posted by **Willow** on Monday, February 23, 2009 at 4:29pm.

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**, Monday, February 23, 2009 at 4:42pmI 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 -
**Damon**, Monday, February 23, 2009 at 4:42pmTrain 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 -
**Willow**, Monday, February 23, 2009 at 4:52pmYou 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 -
**Reiny**, Monday, February 23, 2009 at 5:01pmYou 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**, Monday, February 23, 2009 at 5:06pmOh, 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 -
**Damon**, Monday, February 23, 2009 at 5:06pm125x - 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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