SOLUTION: paul traveled to lexington at 70 mph. on the way back he averages 55 mph. the way back was 86 miles longer and took 2.6 hours longer to drive. how long was the scenic route from Le

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Question 1048246: paul traveled to lexington at 70 mph. on the way back he averages 55 mph. the way back was 86 miles longer and took 2.6 hours longer to drive. how long was the scenic route from Lexington?
Found 2 solutions by Alan3354, advanced_Learner:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
paul traveled to lexington at 70 mph. on the way back he averages 55 mph. the way back was 86 miles longer and took 2.6 hours longer to drive. how long was the scenic route from Lexington?
----------------
d = r*t
r = rate going = 70 mi/hr, d = distance going, t = time going
---
d = 70t
----
d + 86 = 55*(t + 2.6)
---
70t + 85 = 55t + 143
15t = 58
t = 58/15
================
scenic d = 55*(58/15 + 2.6)
= 11*58/3 + 143
=~ 355.67 miles

Answer by advanced_Learner(501) About Me  (Show Source):
You can put this solution on YOUR website!
paul traveled to lexington at 70 mph. on the way back he averages 55 mph. the way back was 86 miles longer and took 2.6 hours longer to drive. how long was the scenic route from Lexington?
how long means? hours or distance? i will solve for both.
To lexington
speed 70
distance d
time t
back
speed 55
distance d +86
time= t+2.6h
70t=55%2A%28t%2B2.6%29
Solved by pluggable solver: EXPLAIN simplification of an expression
Your Result:


YOUR ANSWER


  • This is an equation! Solutions: t=9.53333333333333.
  • Graphical form: Equation 15t=143 was fully solved.
  • Text form: 15t=143 simplifies to 0=0
  • Cartoon (animation) form: simplify_cartoon%28+15t=143+%29
    For tutors: simplify_cartoon( 15t=143 )
  • If you have a website, here's a link to this solution.

DETAILED EXPLANATION

Look at 15%2At=highlight_red%28+143+%29.
Moved these terms to the left highlight_green%28+-143+%29
It becomes 15%2At-highlight_green%28+143+%29=0.
Look at highlight_red%28+15%2At-143+%29=0.
Solved linear equation highlight_red%28+15%2At-143=0+%29 equivalent to 15*t-143 =0
It becomes highlight_green%28+0+%29=0.
Result: 0=0
This is an equation! Solutions: t=9.53333333333333.

Universal Simplifier and Solver


Done!

t is 9.5333
going back time is 12.1333
the required
d=v%2A%2812.1333%29
d=55%2A%2812.1333%29
d=(55)(12.1333)
d=%28667.333%29
d=753.333 is the required scenic distance from lexington..
check that
70*9.5333 is 667.3