SOLUTION: A cyclist has 5 and 1/4 hours to complete today�s training. How long can he ride away from his house if he plans to average 24 mph riding with the wind on the way out and 18 mph ag

Question 937753: A cyclist has 5 and 1/4 hours to complete today�s training. How long can he ride away from his house if he plans to average 24 mph riding with the wind on the way out and 18 mph against the wind on the way back ? How far away from his house will he go ?
Found 2 solutions by TimothyLamb, MathTherapy:
Answer by TimothyLamb(4379) (Show Source): You can put this solution on YOUR website!
x = time downwind
y = time upwind
x + y = 5.25
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z = 1/2 the total distance of the training ride
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s = d/t
t = d/s
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time downwind:
x = z/24
z = 24x
---
time upwind:
y = z/18
z = 18y
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equate z's:
24x = 18y
24x - 18y = 0
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linear system:
x + y = 5.25
24x - 18y = 0
---
put the system of linear equations into standard form
---
x + y = 5.25
24x - 18y = 0
---
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 = time downwind = 2.25
y = time upwind = 3
---
answer:
he will ride 2.25 hours on his out-bound leg ...
z = 24x
z = 24*2.25
z = 54
he will ride 54 miles downwind, at which point he must turn around and head back home
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Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!

A cyclist has 5 and 1/4 hours to complete today�s training. How long can he ride away from his house if he plans to average 24 mph riding with the wind on the way out and 18 mph against the wind on the way back ? How far away from his house will he go ?

Note that the distance traveled away from the house will be the same as the distance traveled back to the house
Let distance traveled away from the house, be D
Then time taken to travel D miles from house =
Time taken to travel D miles, back to house =
Thus, we now have:

3D + 4D = 21(18) ------ Multiplying by LCD, 72
7D = 378
D, or distance traveled away from the house = , or miles
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