4. Linda drove to Eureka. She drove 240 miles. On the way back, she was having car troubles and she had to drive 20 mi/hr slower. If it took her 2 hours longer to get back, how fast was she driving on the trip there, and on the trip back?


The first step in solving word problems is to define the unknown. In this problem, there are two unknowns, Linda's speed driving to Eureka and her speed driving back. When it says that she was driving 20 mi/hr slower on the way back, that gives us the speed on the way back as a function of her speed on the way there. So we

let   x = her speed there.


x - 20 = her speed back

Your authors suggest that we make use of   d = rt   tables. We know that the distance both there and back is 240 miles. We also have expressions for the rate. That gives us this much of the table.

Once we have two columns for the table filled in we can use the appropriate form of the   d = rt   equation to fill ion the third column. Since the unfilled column is the one for time, we need to use the equation

t =d/r

to fill in the time. That enables us to fill in the table as follows.

Now the sentence   "It took her 2 hours longer to get back."   can be translated as

Since we now have an equation, we can take the steps to solve it. First, clear denominators. The smallest common denominator is   x(x - 20),   so multiply both sides by that.

The   x(x - 20)   needs to be distributed through the two terms on the right.

This will clear all of our denominators. We assemble the surviving factors.

240x = 2x(x - 20) + 240(x - 20)

Remove parentheses.

240x = 2x2- 40x + 240x - 4800

and first notice that we can subtract   240x   from both sides of the equation. That leaves us with the quadratic

0 = 2x2- 40x - 4800

We thus have a 0 on one side, so we are ready to see if it factors. First notice that there is a common factor of 2 in all the terms which can be factored out.

0 = 2(x2- 20x - 2400)

That makes it a little easier to factor the expression inside. It will factor because there are factors of 2400 which have a difference of 20. That would 40 and 60. So it factors as 

0 = 2(x - 60)(x + 40)

This product will be equal to 0, if and only if one of these factors is equal to 0. The factor of 2 is never equal to 0, so one of the other factors must be equal to zero. Set them equal   0

x - 60 = 0     x + 40 = 0

x = 60       x = -40


If she was driving at 60 mi/hr going to Eureka, then it would have taken her 4 hours to get there. If she drove 20 mi/hr slower on the way back, then she was driving at only 40 mi/hr, and it would have taken her 6 hours, which is 2 hours longer.

But if she was driving at -40 mi/hr, then the computation for the time it would take would be

t = 240/(-40)

t = -6

This could be interpreted as meaning that if she was driving at -40mi/hr, then she was driving 40 mi/hr backward and at that rate she would have been in Eureka 6 hours ago. 20 mi/hr slower than -40 mi/hr would be - 60 mi/hr and at that rate she would have been in Eureka 4 hours ago which is 2 hours later than 6 hours ago. However, many authors will simply reject negative velocities as being meaningless. but how much fun is that?

If one were to reject the answer of -6, it would be because one might think it didn't make sense in the story. The algebra doesn't know what the story is, so it will give you all answers that would make sense in any story.

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