SOLUTION: a man went ona atrip of 120 miles, travelling at an avg of x miles per hour. several days later he returned over the same route at a rate that was 5 miles per hour faster than his

Algebra ->  Algebra  -> Conversion and Units of Measurement -> SOLUTION: a man went ona atrip of 120 miles, travelling at an avg of x miles per hour. several days later he returned over the same route at a rate that was 5 miles per hour faster than his       Log On

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Question 468963: a man went ona atrip of 120 miles, travelling at an avg of x miles per hour. several days later he returned over the same route at a rate that was 5 miles per hour faster than his previous rate. if the time fo rthe return trip was one third of an hour less than the time for the outgoing trip, which equation can be used to find the value of x?
1.120/(x+5)= 1/3
2.x/120-(x+50)/120-(1/3)
3.120/(x+(x+5)=1/3
4.120/x=120/(x+5)+(1/3)
5.120(x+5)-120x=1/3
Answer by ccs2011(207) About Me  (Show Source):
You can put this solution on YOUR website!
Let d=distance, r=rate or speed, and t = time.
d = r*t
The trip there and back are both the same distance of 120mi.
Therefore using substitution
****************
120+=+x%2At
120%2Fx+=+t
****************
120+=+%28x%2B5%29%2A%28t-%281%2F3%29%29
120%2F%28x%2B5%29+=+t-%281%2F3%29
%28120%2F%28x%2B5%29%29%2B%281%2F3%29+=+t
***************
Setting the times equal to each other
120%2Fx+=+%28120%2F%28x%2B5%29%29%2B%281%2F3%29