# Working simultaneously at their respective constant rates, Machine A…

## Solution:

This problem is what we would call a combined worker problem, where

Work (of machine 1) + Work (of machine 2) = Total Work Completed

In this case,

Work (of A) + Work (of B) = 800

We know that machines A and B produce 800 nails in x hours. Thus, the TIME that machine A and B work together is x hours. We are also given that Machine A produces 800 in y hours. Thus, the rate for machine A is 800/y. Since we do not know the rate for machine B, we can label the rate as 800/B, where B is the number of hours it takes Machine B to produce 800 nails.

To better organize our information we can set up a rate x time = work chart:

We now can say:

Work (of A) + Work (of B) = 800

800x/y + 800x/B = 800

To cancel out our denominators, we can multiply the entire equation by yB. This gives us:

800xB + 800xy= 800yB

xB + xy = yB

xy = yB – xB

xy = B(y – x)

xy /(y-x) = B