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Melissa walks 3 miles to the house of a friend and returns home on a bike. She averages 4 miles per hour faster when cycling than when walking. The total time for the round trip is 2 hours. Find her speed walking and biking. Set up an equation and solve

SHOW WORK PLS!!

User Danyx
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1 Answer

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Answer: Melissa's walking speed is 2mph. Her biking speed is 6mph

Explanation:

Her total distance is 6 miles, because she walks 3 miles, and so the returning distance must also be 3.

Let her cycling speed (in mph) be x

Let her walking speed (in mph) be y

We know that x = y + 4

Since we know that S = D/T:

T = D/S

So her cycling time must be: T = 3/x

and her walking time must be: T = 3/y

since we know that x = y + 4, we can substitute this into her cycling time, and rewrite it as T = 3/y+4

we also know that 3/y+4 + 3/y = 2, because the problem says that the total time taken was 2 hours.

We simply need to solve this equation now:

3/y+4 + 3/y = 2

The LCD for y+4 and y is y² + 4y

so now we have:

(3y/y² + 4y) + (3y+12/y² + 4y) = 2

(6y + 12)/y² + 4y = 2

6y + 12 = 2y² + 8y

2y² + 2y - 12 = 0

Solve this using factoring:

2y² + 2y - 12 = 0

2(y² + y - 6) = 0

(y+3)(y-2) = 0

using ab = 0, then a or b =0:

y = -3, or y = 2

Since speed cannot be negative, our answer is 2 for y. So y, or the walking speed is 2mph. Since x = y + 4, we know that x = 6 mph

Melissa's walking speed is 2mph. Her biking speed is 6mph

User Jamie Marshall
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