Question 181158
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Let <i>r</i> be the riding rate, then the walking rate must be <i>r</i> - 10.  Let *[tex \Large d_r] be the riding distance and the walking distance  be *[tex \Large d_w].  And since she walked for 2 hours of a 5.5 hour trip, the riding time must have been 3.5 hours.


Using the basic equation for distance, rate, and time, i.e. *[tex \Large d = rt], we can describe the walking part of the trip as:


*[tex \Large \text{          }\math d_w=2(r - 10)]


and the riding part of the trip:


*[tex \Large \text{          }\math d_r=3.5r]


But we know that the total distance, which can be expressed as *[tex \Large d_w + d_r] is 57 miles, so:


*[tex \Large \text {          }\math d_w + d_r = 57 \ \ \rightarrow \ \ 2(r - 10) + 3.5r = 57 \ \ \rightarrow \ \ 2r - 20 + 3.5r = 57 \ \ \rightarrow \ \ 5.5r = 77 \ \ \rightarrow \ \ r = 14] 





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