Question 1028012
Call the number of hours the faster worker can do the job in, x.  Then the slower worker takes x+3 hours.  The setup looks like this
{{{2/x + 2/(x+3) = 1}}}
where the 1 corresponds to ONE large order, or one job in general.
Now solve by multiplying everything by x(x+3)...we get
2(x+3) + 2x = x(x+3)
2x + 6 + 2x = x^2 + 3x
4x + 6 = x^2 + 3x
0 = x^2 - x - 6
Factoring we get
(x - 3)(x + 2) = 0
and
x = 3 hours
x+3 = 6 hours