SOLUTION: Rachel allows herself 1 hr to reach a sales appointment 50 mi away. After she has driven 30 mi, she realizes that she must increase her speed by 15 mph in order to get there on ti

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Rachel allows herself 1 hr to reach a sales appointment 50 mi away. After she has driven 30 mi, she realizes that she must increase her speed by 15 mph in order to get there on ti      Log On


   



Question 142679This question is from textbook
: Rachel allows herself 1 hr to reach a sales appointment 50 mi away. After she has driven 30 mi, she realizes that she must increase her speed by 15 mph in order to get there on time. What was her speed for the first 30mi? This question is from textbook

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Rachel allows herself 1 hr to reach a sales appointment 50 mi away. After she has driven 30 mi, she realizes that she must increase her speed by 15 mph in order to get there on time. What was her speed for the first 30mi?
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1st leg DATA:
distance = 30 mi ; time = x hrs ; rate = 30/x mph
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2nd leg DATA:
distance = 20 mi : time = (1-x) hrs ; rate = 20/(1-x) mph
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EQUATION:
2nd leg rate - 1st leg rate = 15 mph
20/(1-x) - 30/x = 15
4(1-x) - 6/x = 3
4x -6(1-x) = 3x(1-x)
4x -6 + 6x = 3x - 3x^2
3x^2 +7x -6 = 0
3x^2 + 9x - 2x - 6 = 0
3x(x+3) - 2(x+3) =0
(x+3)(3x-2) = 0
Positive solution:
x = 2/3 hrs.
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speed for the 1st leg = 30/(2/3) = 45 mph
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Cheers,
Stan H.