Shelby went from her house to the mall at 7mph and came back the same way at 12mph. If the entire round trip took an hour, how far was the distance between the house and the mall?
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Suppose that it took Shelby x hours to walk to the mall. Then, since the round trip was 1 hour, it took her 1 - x hours to walk back.
In the x hours Shelby took to walk to the ball, she walks 7x miles. On the way back, she walks at a rate of 12mph for 1 - x hours, so she travels 12(1 - x) miles in this time.
Since the distance there equals the distance back, we have:
7x = 12(1 - x).
Solving this equation for x yields x = 12/19.
Therefore, plugging x = 12/19 back into either 7x or 12(1 - x) gives the required distance to be 84/19 ≈ 4.42 miles.
I hope this helps!
In the x hours Shelby took to walk to the ball, she walks 7x miles. On the way back, she walks at a rate of 12mph for 1 - x hours, so she travels 12(1 - x) miles in this time.
Since the distance there equals the distance back, we have:
7x = 12(1 - x).
Solving this equation for x yields x = 12/19.
Therefore, plugging x = 12/19 back into either 7x or 12(1 - x) gives the required distance to be 84/19 ≈ 4.42 miles.
I hope this helps!
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Speed to Mall, dx/dt = 7 mi./hr.
Speed from Mall, dy/dt = 12 mi./hr.
Time, t = 1 hr.
Distance = D
Find D:
D / (dx/dt) + D / (dy/dt) = 1
(D / 7) + (D / 12) = 1
84(D / 7) + 84(D / 12) = 84(1)
12D + 7D = 84
19D = 84
D = 84 / 19
D = 4.42
The distance was 4.42 miles.
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Speed from Mall, dy/dt = 12 mi./hr.
Time, t = 1 hr.
Distance = D
Find D:
D / (dx/dt) + D / (dy/dt) = 1
(D / 7) + (D / 12) = 1
84(D / 7) + 84(D / 12) = 84(1)
12D + 7D = 84
19D = 84
D = 84 / 19
D = 4.42
The distance was 4.42 miles.
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