SOLUTION: One roofer can put a neew roof on a house three times faster than another. Working together they can roof a house in 4 days. How long would it take the faster roofer working alone?

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Question 98406This question is from textbook beginning and intermediate algebra
: One roofer can put a neew roof on a house three times faster than another. Working together they can roof a house in 4 days. How long would it take the faster roofer working alone? I need to know how to set up the problem then i can work it from there This question is from textbook beginning and intermediate algebra

Found 2 solutions by stanbon, checkley71:
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
One roofer can put a neew roof on a house three times faster than another. Working together they can roof a house in 4 days. How long would it take the faster roofer working alone? I need to know how to set up the problem then i can work it from there
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1st roofer DATA:
Time = x days/job ; Rate = 1/x job/day
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2nd roofer DATA:
Time = 3x days/job ; Rate = 1/3x job/day
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Working-together DATA:
Time = 4 day/job ; Rate = 1/4 job/day
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EQUATION:
rate + rate = together rate
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I'll let you solve it.
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Cheers,
Stan H.

Answer by checkley71(8403) (Show Source):
You can put this solution on YOUR website!
X*3X/(X+3X)=4 X= THE TIME FOR THE FASTER ROOFER.
3X^2/4X=4 NOW CROSS MULTIPLY
3X^2=16X
3X^2-16X=0
X(3X-16)=0
X=0 A NON ANSWER.
3X-16=0
3X=16
X=16/3 DAYS FOR THE FASTER ROOFER.
3*16/3=16 DAYS FOR THE SLOWER ROOFER.
PROOF
16/3*16/(16/3+16=4
(256/3)/[(16+16*3)/3]=4
(256/3)/[(16+48)/3]=4
(256/3)/(64/3)=4
NOW INVERT THE DENOMINATOR & MULTIPLY
(256/3)(3/64)=4 CANCEL OUT THE 3s.
256/64=4
4=4