Question 244059
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Let *[tex \Large d] represent the distance traveled by the first car.


Let *[tex \Large r] represent the speed of the first car


Let *[tex \Large t] represent the travel time of the first car


Then *[tex \Large d] represents the distance traveled by the second car also.


And *[tex \Large r\ +\ 15] represents the speed of the second car


And *[tex \Large t\ -\ 1] represents the travel time for the second car.


Since we know that distance is equal to rate times time, we can say with respect to the trip taken by the first car:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ rt\ =\ 180]


And with respect to the trip taken by the second car:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (r\ +\ 15)(t\ -\ 1)\ =\ 180]


Multiply the second equation by *[tex \LARGE \frac{1}{r\ +\ 15}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ -\ 1\ =\ \frac{180}{r\ +\ 15}]


Add 1 to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{180}{r\ +\ 15}\ +\ 1]


Multiply the first equation by *[tex \LARGE \frac{1}{r}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{180}{r}]


Set the two expressions that are equal to Let *[tex \Large t] equal to each other:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{180}{r}\ =\ \frac{180}{r\ +\ 15}\ +\ 1]


Simplify and then solve the resulting quadratic for *[tex \LARGE r].  Exclude the negative root.  The positive root will be the speed of the first car.  Add 15 to find the speed of the second car.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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