5. Linda goes to visit a friend who lives in Lompoc. It is 390 miles

from Linda's house to her friend's house. When she came back, she

didn't stop, so it took her a half an hour less time and her average

speed was 5 miles per hour faster. How fast was she going when she

went down to Lompoc, and how fast was she going when she drove



First we define our unknowns. In this case they are looking for

how fast she was driving when she went down to Lompoc and how

fast she was driving when she drove back. We could set this up with

two unknowns, but since we know that she averaged 5 mi/hr faster

coming back we could


let x = how fast she was driving going down


and then we would have that


x + 5 = how fast she drove on the way back.


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




distance = rate x time table


Once we have two columns filled in we can fill in the third

column by using the appropriate form of the d = rt equation. In this

case we need to fill in the column for the time so we use the equation



t = d/r


Completed d = rt table


Now the sentence "It took her a half hour less time to get back."

can be translated as




This gives us a rational equation. Clear denominators. The

smallest common denominator is 2x(x+5)




Multiply the 2x(x+5) by both terms on the right side.




Cancel and assemble the surviving factors.




Remove parentheses and multiply.




After subtracting 780x from both sides, transpose all the terms

to the left.








and set the factors = 0.





We get two answers. One is 60 mi/hr and the other one is -65

mi/hr. Of the two, 60mi/hr makes more sense. If she was driving to

Lompoc at 60 mi/hr then it would take her 6.5 hours .If she came

back averaging 5 mi/hr faster then it would take her 6 hours which

would be 1/2 hour less time, and the answer checks.


The answer of -65 mi/hr would take a little more imagination

to check. It would mean that she was driving to Lompoc at 65 mi/hr

backwards. If she were driving backwards then she would have been

in Lompoc 6 hours ago. To get her speed coming back, adding 5

mi/hr to -65 mi/hr would give us a speed of -60 mi/hr, and it would

then take her -6,5 hrs which is 1/2 less than -6 hours. However,

most authors would have us reject such an answer as being



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