## train word problem

Simplificatation, evaluation, linear equations, linear graphs, linear inequalities, basic word problems, etc.
aluz
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### train word problem

The text I'm using is Intermediate Algebra for College Students by Allen R. Angel (3rd edition), though I imagine this problem would go under beginning algebra. The problem reads:
Two trains leave Chicago, Illinois, traveling in the same direction along parallel tracks. The first train leaves 3 hours before the second, and its speed is 15 miles per hour faster than the second. Find the speed of each train if they are 270 miles apart 3 hours after the second train leaves Chicago.

The solution in the back of the text is "60 mph, 75 mph".

I figured that the first train's speed would be 270 miles/3 hours=90 mph . Since the second train's speed is 15 mph slower than the first train's speed, I thought the answers would be 90 mph for the first train and 75 mph for the second train. Where did I go wrong?

jg.allinsymbols
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### Re: train word problem

A few time intervals will be helpful. During the first 3 hours, only the early and fast train is moving. During the second 3 hours, both the early fast, and the later slow trains are moving.

A rate and distance chart helped me analyze, so the chart combines time with distance.

Each Train_________Speed__________________Distance______________________________
___________________________________t=0________________t=3________________t=6____
Fast Early_________r_________________...________________r*3__________________r*6______
Slow Late__________r-15_____________...________________(r-15)*0_______________(r-15)*3___

Under the 0 column nothing is indicated of time or distance. Under the 3 column is represented the distance (or rate times time) for the end of 3 hours. Under the 6 column is represented the distance for the end of the second three hour period.

Our given description is that at the END OF 6 HOURS, the two trains are 270 miles apart. We have the expressions for their travel distances under the t=6 column, so we can then create the equation:

6r-3(r-15)=270

From that, we can solve for r, the speed of the faster, early train. Then r-15 calculates the speed of the slower later train. Allow me to skip the algebra steps in this posting, but I get r=75 miles per hour. Means slower train is 60 miles per hour.

post note: Were you able to get an equation equivalent to the one that I found?