SOLUTION: A motorist travelled to a distant city in 2 hours and returned by another route that is 30 miles longer. On the return trip she travelled 10 miles per hour faster and took 6 minute

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Question 1026741: A motorist travelled to a distant city in 2 hours and returned by another route that is 30 miles longer. On the return trip she travelled 10 miles per hour faster and took 6 minutes longer. Find the length of the shorter route.
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39799) About Me  (Show Source):
You can put this solution on YOUR website!
                   rate             time         distance

short going        r                  2            d

long return       r+10                2%261%2F6        d+30


RT=D basic rule; two-variable linear system;
clear enough?

Answer by MathTherapy(10810) About Me  (Show Source):
You can put this solution on YOUR website!
A motorist travelled to a distant city in 2 hours and returned by another route that is 30 miles longer. On the return trip she travelled 10 miles per hour faster and took 6 minutes longer. Find the length of the shorter route.
Let length of shorter route be D

Then length of longer route = D + 30

Outgoing time: 2 hours, and return time = 2+%2B+6%2F60 -----> 2%261%2F10 -----> 21%2F10 hrs 

We then get the following SPEED equation: D%2F2+=+%28D+%2B+30%29%2F%2821%2F10%29+-+10

D%2F2+=+%28D+%2B+30%29+%2A+%2810%2F21%29+-+10

D%2F2+=+10%28D+%2B+30%29%2F21+-+10

D%2F2+=+%2810D+%2B+300%29%2F21+-+10

21D = 2(10D + 300) - 420 -------- Multiplying by LCD, 42

21D = 20D + 600 – 420

21D – 20D = 180

D, or length of shorter route = highlight_green%28matrix%281%2C2%2C+180%2C+%22miles%22%29%29