SOLUTION: On a 135-mile bycycle excursion, Maria averaged 5 mph faster for the first 60 miles than she did for the last 75 miles. The entire trip took 8 hours. Find the rate for the first 60

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Question 173690: On a 135-mile bycycle excursion, Maria averaged 5 mph faster for the first 60 miles than she did for the last 75 miles. The entire trip took 8 hours. Find the rate for the first 60 miles.
Answer by Mathtut(3670)   (Show Source): You can put this solution on YOUR website!
you have to set this up as 2 different parts of a lonnnnnnnnnggggggg trip.
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lets call the 2nd part of the trips rate r and time t and
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so then we will call the first part of the trips rate r+5 and the time 8-t(total time minus 2nd part of the trips time(t))
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so
75=rt......................eq 1
60=(r+5)(8-t)..........eq 2
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lets rewrite eq 1 to t=75/r and plug t's value into eq 2
:
60=(r+5)(8-(75/r))
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60=(r+5)((8r-75)/r).........now multiply both sides by r
:
60r=(r+5)(8r-75)
:


:
throw out negative answer
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therefore rate on 75 mile part of trip is 15mph
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rate on 60 mile part of trip is mph
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Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=21025 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: 15, -3.125. Here's your graph:
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