Question 237997
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Let *[tex \Large r] represent the rate during the 60 mile leg.  Then *[tex \Large r\ -\ 5] is the rate for the 75 mile leg.


Let *[tex \Large t] represent the time to complete the 60 mile leg.  Then the time to complete the 75 mile leg must be *[tex \Large 8\ -\ t]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{60}{r}]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8\ -\ t\ =\ \frac{75}{r\ -\ 5}]


Solve the second equation for *[tex \Large t] (left as an exercise for the student).  Now you will have two expressions equal to *[tex \Large t].  Set them equal and simplify.  The result will be the following quadratic equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8r^2\ -\ 175r\ -\ 300\ =\ 0]


Factor and solve for *[tex \Large r].  You will obtain two positive roots, but one of them will be < 5, and can be excluded because that would make *[tex \LARGE r - 5 < 0] meaning she went backwards during the 75 mile leg.

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