Question 1187999
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Distance equals rate times time.


Let *[tex \Large d] represent the distance between the two cities, and let *[tex \Large r] represent the speed of the slower traveler.


So the slower traveler's trip can be modeled by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ 3r]


since he took 3 hours to complete his trip.


The faster traveler's speed can be represented by *[tex \Large r\,+\,20] because he is going 20 miles per hour faster than the slower guy.  Therefore his trip over the same distance can be modeled by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ 2(r\,+\,20)]


Since *[tex \Large d\ =\ d] we can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3r\ =\ 2(r\,+\,20)]


Solving for *[tex \Large r] we get:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r\ =\ 40]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ (3\text{ hours})(40\text{ mph})\ =\ 120\text{ miles}]

																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
*[illustration darwinfish.jpg]

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
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