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

Algebra ->  Polynomials-and-rational-expressions -> 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      Log On


   

Question 208274This 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 ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
1 hr to reach a sales appointment 50mi 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 30 mi?
:
Let s = speed for the 1st 30 mi
:
The total travel time has to be 1 hr
Write a time equation: Time = dist%2Fspeed
:
30%2Fs + 20%2F%28%28s%2B15%29%29 = 1
Multiply equation by s(s+15)
30(s+15) + 20s = s(s+15)
:
30s + 450 + 20s = s^2 + 15s
:
50s + 450 = s^2 + 15s
:
Arrange as a quadratic equation on the right
0 = s^2 + 15s - 50s - 450
:
s^2 - 35s - 450 = 0
Factor
(s-45)(s+10) = 0
Positive solution:
s = 45 mph for the 1st 30 miles
;
:
Check solution by finding the times
30/45 = .67
20/60 = .33
------------
total = 1.0 hr