Question 289919
It takes two laser printers working together 12 minutes to complete a 580 page print job ,if each printer is printing this job alone, the first laser printer takes 10 minutes longer than the second printer . How long does it take each printer to complete the print job alone? What is the speed of each printer. 
I don't even know where to begin with this. Someone please help me step by step?
Thanks!

[see: http://www.algebra.com/algebra/homework/Rate-of-work-word-problems/HOW-TO-Solve-Rate-of-Work-Problems.lesson]

printer A, printer B
A in X minutes, B in Y minutes

time = 1/[1/X + 1/Y]
A takes X minutes to do a job, so 1/X jobs per minute
B takes Y minutes to do a job, so 1/Y jobs per minute
working together: 1/X + 1/Y jobs per minute
one job: 1/[1/X + 1/Y] minutes
1/[1/X + 1/Y] = 12 minutes to do 580 pages (48 1/3 pages/minute)
A would take 10 minutes longer than B
X = Y + 10 --> X - 10 = Y
1/[1/X + 1/(X - 10)] = 12
1/X + 1/(X - 10) = 1/12
(X - 10)/[X(X - 10)] + X/[X(X - 10)] = 1/12 (made common denominators)
[(X - 10) + X]/(X^2 - 10X) = 1/12
(2X - 10)/(X^2 - 10X) = 1/12
(24X - 120)/(X^2 - 10X) = 1 (multiplied numerators both sides by 12)
24X - 120 = X^2 - 10X
0 = X^2 - 34X + 120
0 = (X - 4)(X - 30) (by FOIL this equals above)
X = 4 minutes or X = 30 minutes
since X - 10 = Y, and Y can not be negative minutes,
X = 30 minutes for A, and Y = 20 minutes for B to do the 580 page print job alone
A's speed is 580 pages/30 minutes = 19 1/3 pages/minute
B's speed is 580 pages/20 minutes = 29 pages/minute