SOLUTION: A man cycles to a friend's house at 12 mph and arrives 1 hour late.
Next day he travels at 16 mph and arrives 1 hour early.
What speed must he travel to arrive on time?
Algebra.Com
Question 936042: A man cycles to a friend's house at 12 mph and arrives 1 hour late.
Next day he travels at 16 mph and arrives 1 hour early.
What speed must he travel to arrive on time?
Answer by TimothyLamb(4379) (Show Source): You can put this solution on YOUR website!
x = correct travel time
y = distance to friend's house
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s = d/t
t = d/s
---
late:
x + 1 = y/12
12x + 12 = y
---
early:
x - 1 = y/16
16x - 16 = y
---
put the system of linear equations into standard form
---
12x - y = -12
16x - y = 16
---
copy and paste the above standard form linear equations in to this solver:
https://sooeet.com/math/system-of-linear-equations-solver.php
---
solution:
x = correct travel time = 7
y = distance to friend's house = 96
---
answer:
s = d/t
s = 96/7
s = 13.7 mph
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