SOLUTION: Hi, I missed school and need some help with HW. Ok heres the problem-- May rides her bike the same distance that Leah walks. May rides her bike 10 km/h faster than Leah walks. If i
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-> SOLUTION: Hi, I missed school and need some help with HW. Ok heres the problem-- May rides her bike the same distance that Leah walks. May rides her bike 10 km/h faster than Leah walks. If i
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Question 998350: Hi, I missed school and need some help with HW. Ok heres the problem-- May rides her bike the same distance that Leah walks. May rides her bike 10 km/h faster than Leah walks. If it takes may 1 hour and leah 3 hours to travel that distance, how far does each travel? (*I need to write and solve an equation*) Thanks! -R Found 2 solutions by stanbon, MathTherapy:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! May rides her bike the same distance that Leah walks. May rides her bike 10 km/h faster than Leah walks. If it takes may 1 hour and leah 3 hours to travel that distance, how far does each travel?
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May bike DATA:
rate = (x + 10) km/hr ; time = 1 hr ; distance = r*t = x+10 km
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Leah walk DATA:
rate = x km/hr ; time = 3 hr ; distance = r*t = 3x km
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Equation:
dist = dist
3x = x+10
2x = 10
x = 5 km/hr (May's rate)
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x + 10 = 15 km/hr (Leah's walk)
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Cheers,
Stan H.
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You can put this solution on YOUR website!
Hi, I missed school and need some help with HW. Ok heres the problem-- May rides her bike the same distance that Leah walks. May rides her bike 10 km/h faster than Leah walks. If it takes may 1 hour and leah 3 hours to travel that distance, how far does each travel? (*I need to write and solve an equation*) Thanks! -R
Let distance be D
Then Leah’s speed is: , and May’s speed = , or D
We then the following SPEED equation:
D = 3D - 30 ------- Multiplying by LCD, 3
D – 3D = - 30
– 2D = - 30
D, or travel distance = , or km
(