SOLUTION: Jeff takes 5 hours longer to build a fence than it takes Bill. When they work together, it takes them 6 hours. How long would it take Bill to do the job alone?
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-> SOLUTION: Jeff takes 5 hours longer to build a fence than it takes Bill. When they work together, it takes them 6 hours. How long would it take Bill to do the job alone?
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Question 326607: Jeff takes 5 hours longer to build a fence than it takes Bill. When they work together, it takes them 6 hours. How long would it take Bill to do the job alone? Found 2 solutions by nyc_function, Alan3354:Answer by nyc_function(2741) (Show Source):
We multiply each term on both sides of the fractional equation by the LCD.
This is done in order to remove all fractions.
1/B * 6B(B + 5) = 6(B + 5)
1/(B + 5) * 6B(B + 5) = 6B
1/6 * 6B(B + 5) = B(B + 5)
We now have the following linear (not fractional) equation in yellow:
6(B + 5) + 6B = B(B + 5)
We now proceed to do basic math to find B.
6B + 30 + 6B = B^2 + 5B
12B + 30 = B^2 + 5B
B^2 + 5B - 12B - 30 = 0
B^2 - 7B - 30 = 0...This is a quadratic equation.
We now factor this quadratic equation using any method of choice. I will use the quadratic formula on paper but you can use other method if it is easier.
At this point, we have two answers for B: B = -3 and B = 10. Of course, we reject B = -3 because we are interested in a positive answer for B in terms of work that is done by two people.
Bill does the job in B = 10 hours. If Jeff does the work in 5 more than 10 hours, then Jeff does the work in 15 hours.
You can put this solution on YOUR website! Jeff takes 5 hours longer to build a fence than it takes Bill. When they work together, it takes them 6 hours. How long would it take Bill to do the job alone?
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Bill does 1/B of the fence per hour
Jeff does 1/J per hour, = 1/(B+5)
Together, they do 1/6 per hour.
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1/B + 1/(B+5) = 1/6
B+5 + B = (1/6)*B*(B+5)
12B + 30 = B^2 + 5B
B^2 - 7B - 30 = 0
(B-10)*(B+3) = 0
B = 10 hours
J = 15 hours
