Melissa walked 3 miles to her friend’s house and returns home on a bike. She averages 4 miles per hour faster when cycling than when walking. The total time for both trips is two hours. Find her walking speed
Question
Melissa walked 3 miles to her friend’s house and returns home on a bike. She averages 4 miles per hour faster when cycling than when walking. The total time for both trips is two hours. Find her walking speed
Solution
To solve this problem, we need to set up a system of equations based on the information given.
Let's denote:
- Melissa's walking speed as 'w' (in miles per hour), and
- Her cycling speed as 'w + 4' (since it's 4 mph faster than her walking speed).
We know that the total time for both trips is 2 hours.
The time it takes for each trip can be calculated as the distance divided by the speed. So, the time for the walking trip is 3/w and the time for the cycling trip is 3/(w+4).
According to the problem, the sum of these two times is 2 hours. So, we can set up the following equation:
3/w + 3/(w+4) = 2
To solve this equation for 'w', we can first get rid of the fractions by multiplying every term by 'w*(w+4)':
3*(w+4) + 3w = 2w*(w+4)
Expanding and simplifying gives:
3w + 12 + 3w = 2w^2 + 8w
Rearranging terms gives a quadratic equation:
2w^2 + 2w - 12 = 0
Dividing every term by 2 simplifies this to:
w^2 + w - 6 = 0
This can be factored to:
(w - 2)(w + 3) = 0
Setting each factor equal to zero gives the possible solutions w = 2 and w = -3. Since speed can't be negative, we discard w = -3.
So, Melissa's walking speed is 2 miles per hour.
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