4. Linda drove to Los Angeles to visit her boyfriend. She drove 450 miles. On the way back, she was having car troubles and she had to drive 10 mi/hr slower. If it took her an hour and a half longer to get back, how fast was she driving on the trip down, and on the trip back?

Since this is a word problem, we first define our unknown. In this problem they are looking for two unknowns: how fast she was driving on the way down and how fast on the way back. Since we know that she was driving 10 mi/hr slower on the way back, if we know how fast she was driving on the way down, we will be able to figure out how fast she was driving on the way back as well. So we

and

Your authors suggest that we set up a *d* = *rt* table.

Once we have two columns in these tables filled out, we can fill
in the third. We use the equation *t* = *d/r*.

We can now express the fact that it took an hour and a half longer to get back as

Since this is a rational equation, we need to first clear
denominators. The smallest common denominator is 2*x*(*x* - 10). Multiply
both sides.

Since there are two terms on the right side, we need to multiply them both.

Assemble the surviving factors.

Remove parentheses.

We see that we have a quadratic. If we just subtract 900x from both sides, we will get all of the terms on one side and leave a 0 on the other.

When we factor, notice that there is a common factor of 3 in all the terms on the right.

Divide both sides by 3.

Factor

Set the factors equal to 0.

*x* + 50 = 0 *x* - 60 = 0

*x* = -50 *x* = 60

Check:

If *x* = 60 then it took her 450/60 = 7.5 hours to get down to Los
Angeles, but if she had to drive 10 mi/hr slower coming back, then
she would have been driving at 50 mi/hr in which case it would take
her 450/50 = 9 hours to get back. Since 9 hours is an hour and a half
longer than 7.5 hours, it checks.

If x = -50 then it took her 450(-50) = -9 hours to get to Los Angeles. That is, if she were driving 50 mi/hr backwards then she would have been in Los Angeles 9 hours ago. 10 mi/hr slower than -50 mi/hr would be -60 mi/hr in which case it would take her -7.5 hours to get back. -7.5 hours is an hour and a half longer than -9 hours, and it checks. Most authors would reject this solution as being too wierd, but how much fun is that?