A roofing crew can roof a house in 10 hours. If a second crew was added, the job would take 6 hours. How long would it take the second crew to roof the house alone?

Working together is like resistors working in parallel:

t1*t2 / (t1 + t2) = 6,
10t2 / (10 + t2) = 6,
Cross multiply:
10t2 = 60 + 6t2,
10t2 - 6t2 = 60,
4t2 = 60,
t2 = 60 / 4 = 15 h.

To solve this problem, let's start by assigning variables. Let's say that the first crew's work rate is represented by "x" houses per hour.

Therefore, we can say that the first crew can complete the job in 1/10 of a house per hour (since they take 10 hours to finish the job).

If we add the second crew, their combined work rate would be (x + y) houses per hour, where "y" represents the work rate of the second crew.

Given that the time taken for the combined crews to finish the job is 6 hours, we can say that the combined crews can complete 1 full house in 6 hours.

Now, we can set up the equation:

1/10 (work rate of the first crew) + 1/y (work rate of the second crew) = 1/6 (combined work rate of the crews)

To solve for "y," the work rate of the second crew, let’s first find a common denominator for 10, y, and 6. The lowest common multiple is 30.

Now, multiply both sides of the equation by 30:

3 + 30/y = 5

Subtracting 3 from both sides, we get:

30/y = 2

To solve for "y," multiply both sides of the equation by "y":

30 = 2y

Dividing both sides by 2, we get:

y = 15

Therefore, the second crew alone would take 15 hours to roof the house.