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April 16, 2014

April 16, 2014

Posted by **cheryl** on Monday, March 19, 2007 at 11:01am.

distance =rate*time

The distance is the same.

y = rate going

z = rate returning

Going: d = y*t = y*5 = 5y

return: d = z*t = z*4 = 4z

we know the return rate was 5 mi/hr faster; therefore, the return rate of z = y+5

set d = d

5y=4z

z=y+5

=========

Solve simultaneously

Post your work if you need further assistance.

many students find distance-rate-time problems really easy if they make a chart

..........│..D..│..R..│..T..│

----------------------------

1st trip..│.5x..│..x..│..5..│

----------------------------

2nd trip..│4(x+5)│x+5│..4..│

clearly the two trips are the same

so 5x=4(x+5).........etc, piece of cake

The chart really IS a good idea.

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