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

 

Let x = how long it took her to get back.

 

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.

 

x+2=how long it took her to get to the airport.

 

your author suggests we make a d = rt table to organize our

efforts

 

d=rt table

 

 

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

 

r=d/t

 

to fill in the remaining column.

 

 

d=rt table

 

 

Now, the sentence, "She was able to average 40 mi/hr faster

getting back." gets translated as

 

 

60/x=40+60/(x+2)

 

 

 

Now we have an equation that we can solve. Clear

denominators. This is accomplished by multiplying both sides by

x(x+2)

x(x+2)(60/x)=x(x+2)(40+60/(x+2))

 

 

After we remove parentheses

 

x(x+2)60/x=40x(x+2)+x(x+2)60/(x+2)

and cancel, we are left with

 

60(x+2)=40x(x+2)+60x

 

Remove parentheses

 

60x+12=40x^2+80x+60x

 

 

We have a quadratic. We can simplify quickly by subtracting

60x from both sides, but then we need to transpose all terms to one

side leaving a 0 on the other,

 

0=40x^2+80x-120

 

and factor

 

0=40(x^2+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.

 

0=(x-1)(x+3)

Set the factors = 0

x-1=0,x+3=0,x=1,x=3

 

Check:

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 as being meaningless.

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