4. Linda drove to the airport. which is 60 miles away. On the way there, she got stuck in traffic, but on the way back the traffic was fine. It took her 2 hours longer to get to the airport than it did to get back. If she was able to average 40 miles per hour faster getting back, how long did it take her to get to the airport , and how long did it take to get back?
The first step in solving a word problem is to define the unknown. In this case they're looking for how long it took her to get to the airport and how long it took her to get back. So we start by writing
If we know how long it took her to get back, then we can just add 2 hours to that to get how long it took her to get to the airport.
your authors suggest we make a d = rt table to organize our efforts
We are given that it is 60 miles both to the airport and back.
We can express the time using our unknown. Once we get two columns of these tables filled in we can fill in the remaining column using the appropriate d = rt equation. In this case, we still need to fill in the rate column so we need to use the equation
to fill in the remaining column.
Now, the sentence, "She was able to average 40 mi/hr faster getting back." gets translated as
Now we have an equation that we can solve. Clear denominators. This is accomplished by multiplying both sides by x(x+2)
After we remove parentheses
and cancel, we are left with
We have a quadratic. We can simplify quickly by subtracting 60x from both sides,
120 = 40x2 + 80x
but then we need to transpose all terms to one side leaving a 0 on the other. Subtract 120 from both sides.
0 = 40(x2 + 2x - 3)
0 = 40(x - 1)(x + 3)
Since 40 can never be 0, we can divide both sides by 40. That gets rid of the factor of 40 0n the right, and on the left 0/40 = 0.
Set the factors = 0
x - 1 = 0 x + 3 = 0
x = 1 x = -3
If x = 1, then it took 1 hour to drive the 60 miles back from the airport, so the average speed was 60 mi/hr, 40 mi/hr slower would be 20 mi/hr, and at that rate it would have taken 3 hours to drive to the airport, which would be 2 hours longer than it took to drive back.
But if x = -3, it is not as straightforward to check.
If it took -3 hours to drive to the airport, that means that you were at the airport 3 hours ago, and if you compute your speed, 60/(-3) would be -20mi/hr. That could be interpreted as meaning that you were driving at 20 mi/hr backwards to have been at the airport 3 hours ago if you have just gotten home. In any event, -60 mi/hr is the speed which is 40 mi/hr slower than - 20 mi/hr, and if you were to drive to the airport at -60 mi/hr you would compute the time to be -1 hours which is 2 hours more than -3 hours. So, if you think about it, the -3 hours does check, but, while some students would think it somewhat unimaginative, most authors would have us reject the solution of -3 hours, but how much fun is that?
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