Question 42401
Let the time it takes the boy alone to be x.  Her time is then x + 5.
The setup for problems like this is
t/A + t/B = 1
where t is the time worked together and A and B are the individual times...so plugging in we get
6/x + 6/(x+5) = 1
now multiply thru by the lowest common denominator, here x(x+5)...and we get
x(x+5)[6/x + 6/(x+5) = 1]
6(x+5) + 6x = x(x+5)
6x + 30 + 6x = x^2 + 5x
now rearrange, factor and solve
x^2 - 7x - 30 = 0
(x - 10)(x + 3) = 0
x = 10 or x = -3
since it cannot be negative time,
x = 10 hours and
x+5 = 15 hours