SOLUTION: Rachel allows herself 1 hour to reach a sales appointment 50 miles away. After she has driven 30 miles, she realizes that she must increase her speed by 15 mph in order to get ther
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-> SOLUTION: Rachel allows herself 1 hour to reach a sales appointment 50 miles away. After she has driven 30 miles, she realizes that she must increase her speed by 15 mph in order to get ther
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Question 168279This question is from textbook Introductory Algebra
: Rachel allows herself 1 hour to reach a sales appointment 50 miles away. After she has driven 30 miles, 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 miles? This question is from textbook Introductory Algebra
You can put this solution on YOUR website! Rachel allows herself 1 hour to reach a sales appointment 50 miles away. After she has driven 30 miles, 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 miles?
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Let x = speed driven for first 30 miles
x+15 = speed driven for last 20 miles
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And,
Let y = time driving first 30 miles
1-y = time driving for last 20 miles
.
First 30 miles (equation 1):
xy = 30
Last 20 miles (equation 2):
(x+15)(1-y) = 20
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Solve equation 1 for y:
xy = 30
y = 30/x
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Substitute the above into equation 2 and solve for x:
(x+15)(1-y) = 20
(x+15)(1-(30/x)) = 20
Multiplying both sides by x:
(x+15)(x-30) = 20x
FOIL the left side:
x^2-30x+15x-450 = 20x
x^2-15x-450 = 20x
x^2-35x-450 = 0
(x+10)(x-45) = 0
x = {-10, 45}
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We can throw out the negative solution leaving us with:
x = 45 mph
