Asked by Willow
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! :)
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! :)
Answers
Answered by
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(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
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
Answered by
Damon
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)
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)
Answered by
Willow
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! :)
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! :)
Answered by
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.
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.
Answered by
Willow
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?
Thanks so much though! :)
P.S: Is the correct answer 375 miles away from the station, meaning x=5 and y=3?
Answered by
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
+
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
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.