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

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Question 208282This question is from textbook Introductory Algebra
: rachel allows herself 1hr to reach a sales appointment 50mi away. After she has driven 30mi, she realizes that she must increase her speed by 15mph in order to get there on time. What was her speed for the first 30mi? This question is from textbook Introductory Algebra

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
rachel allows herself 1hr to reach a sales appointment 50mi away. After she has driven 30mi, she realizes that she must increase her speed by 15mph in order to get there on time. What was her speed for the first 30mi?
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1st part of the drive DATA:
distance = 30 mi ; rate = x mph ; time = d/r = 30/x hrs
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2nd part of the drive DATA:
distance 20 mi ; rate = x+15 mph ; time = d/r = 20/(x+15) hrs.
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Equation:
time + time = 1 hr
30/x + 20/(x+15) = 1
30(x+15) + 20x = x(x+15)
50x + 450 = x^2 + 15x
x^2 - 35x - 450 = 0
(x-45)(x+10)= 0
Positive Solution:
x = 45 mph (rate she drove on the 1st 30 miles)
Cheers,
Stan H.