SOLUTION: Two workers working together finished a job in 12 hours. Later the first worker made one half of the same job and after it the second worker finished the other half of the job. In

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Question 703894: Two workers working together finished a job in 12 hours. Later the first worker made one half of the same job and after it the second worker finished the other half of the job. In this case they finished the job in 25 hours. How long can they finish the job separately?
I set up a table...
Worker Fraction of job Rate in jobs/hr Time in hrs
A alone 1/2 x
B alone 1/2 25 –x
A + B 1 12
B alone's time is based on fact that x + y = 25 where x and y represent time for 1/2 each of job... so solved to x
then you can rewrite to allow for sum of rates as follows, multiply the first two entries by 2 to give whole jobs:
Worker Fraction of job Rate in jobs/hr Time in hrs
A alone 1 1/2x 2x
B alone 1 1/(50-2x) 2(25 - x)
A + B 1 1/12 12
So knowing that the rate alone must when combined equal 1/12:
1/2x + 1/(50-2x) = 1/12
and then I get lost somewhere in here solving and get stuck... I understand that the introduction of a higher power can result in some unreasonable negatives, but still cannot seem to solve this... Where am I going wrong???

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
You start by defining
x= time it takes one worker to do half the job,
and ended with the equation 1%2F2x+%2B+1%2F%2850-2x%29+=+1%2F12
Nothing wrong, you could have defined x differently, but you got an equation that can lead you to the answer.
In this kind of problems, we always get equations with cumbersome denominators, and we want to get rid of those denominators.
Multiplying everything times 12, we get
6%2Fx+%2B+6%2F%2825-x%29+=+1 which is an equivalent equation, with exactly the same solutions.
Multiplying times x%2825-x%29 would get rid of denominators,
at the cost of risking the introductions of x=0 and/or x=25 as extraneous solutions.
I will not happen, but we do not worry about extraneous solutions at this point,
because we would verify the solutions we get anyway, and we can discard any extraneous solutions at the end.
Multiplying 6%2Fx+%2B+6%2F%2825-x%29+=+1 times x%2825-x%29 we get
6x%2825-x%29%2Fx+%2B+6x%2825-x%29%2F%2825-x%29+=+x%2825-x%29
6%2825-x%29+%2B+6x+=+25x-x%5E2%29
6%2825-x%29+%2B+6x+=+25x-x%5E2%29
150-6x+%2B+6x+=+25x-x%5E2%29
150+=+25x-x%5E2%29 --> x%5E2-25x%2B150=0 an easy quadratic equation

If I had thought about it further, I would have realized that
x%5E2-25x%2B150=%28x-10%29%28x-25%29
and I would have solved the quadratic equation by factoring.

Instead, I applied the quadratic formula:
x+=+%28-%28-25%29+%2B-+sqrt%28+%28-25%29%5E2-4%2A1%2A150+%29%29%2F%282%2A1%29+
x+=+%2825+%2B-+sqrt%28625-600%29%29%2F2
x+=+%2825+%2B-+sqrt%2825%29%29%2F2 --> x+=+%2825+%2B-+5%29%2F2
The solutions are
x+=+%2825+%2B+5%29%2F2 --> x+=+30%2F2 --> highlight%28x=15%29 with highlight%2825-x=10%29 and
x+=+%2825+-+5%29%2F2 --> x+=+20%2F2 --> highlight%28x=15%29 with highlight%2825-x=15%29.
Either way, for one worker it takes 10 hours to do half the job, and for the other it takes 15 hours to do half the job.

At this point, we have to remember that the question was "How long can they finish the job separately?"
For one worker it takes 10 hours to do half the job, and for the other it takes 15 hours to do half the job.
So, the first,faster worker requires 2%2A10=highlight%2820%29 hours to complete the job when working alone.
The other, slower worker needs highlight%2830%29 hours to complete the job when working alone.