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Question 125848This question is from textbook Algebra Structure and Method
: I have been having trouble figuring out this problem and I was wondering if anyone could help me? I would deeply appreciate it!!! Please and Thank You!!

Solving Problems Involving Inequalities
Solve.
Randy walked at the rate of 5.2km/h in a straight path from his campsite to a ranch. He returned immediately on horseback at the rate of 7.8km/h. Upon his return, he found that he had been gone for no more than 3.5h. At most how far is it from his campsite to the ranch? This question is from textbook Algebra Structure and Method

Found 2 solutions by josmiceli, checkley71:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
For each trip, d+=+r%2At in km. For both trips the distance is the
same. The time spent each trip will be different
For campsite to ranch:
d+=+5.2%2At%5B1%5D
For ranch to campsite:
d+=+7.8%2At%5B2%5D
5.2%2At%5B1%5D+=+7.8%2At%5B2%5D
Also given is
t%5B1%5D+%2B+t%5B2%5D+=+3.5 hrs
t%5B2%5D+=+3.5+-+t%5B1%5D hrs
5.2%2At%5B1%5D+=+7.8%2At%5B2%5D
5.2%2At%5B1%5D+=+7.8%283.5+-+t%5B1%5D%29
5.2%2At%5B1%5D+=+27.3+-+7.8%2At%5B1%5D
13%2At%5B1%5D+=+27.3
t%5B1%5D+=+2.1 hrs
Plug this back into 1st equation
d+=+5.2%2At%5B1%5D
d+=+5.2%2A2.1
d+=+10.92 km answer
check the other equation:
t%5B2%5D+=+3.5+-+t%5B1%5D
t%5B2%5D+=+3.5+-+2.1
t%5B2%5D+=+1.4
d+=+7.8%2At%5B2%5D
d+=+7.8%2A1.4
d+=+10.92
OK

Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
D=RT
D=5.2X GOING.
D=7.8(3.5-X) RETURNING.
SEEING AS THE DISTANCES ARE THE SAME WE HAVE:
5.2X=7.8(3.5-X)
5.2X=27.3-7.8X
5.2X+7.8X=27.3
13X=27.3
X=27.3/13
X=2.1 HOURS FOR THE GOING TRIP
3.5-2.1=1.4 HOURS FOR THE RETURN TRIP.
PROOF
5.2*2.1=7.8*1.4
10.92=10.92 MILES FROM THE CAMP SITE TO THE RANCH.