Question 978442
Let s = the time required for the older machine to do the job
let the completed job = 1 (production of 100 cans)
:
An old machine and a new, faster machine take 3 hours to produce 100 cans together.
Using the new machine alone it takes 4.5 hours to produce the 100 cans.
How long would it take the old machine to produce the 100 cans by itself?
:
A shared work equation
{{{3/4.5}}} + {{{3/s}}} = 1
each does a fraction of the job, the two fractions add up to 1
Multiply by 4.5s, cancel the denominators and you have
3s + 4.5(3) = 4.5s
13.5 = 4.5s - 3s
13.5 = 1.5s
s = 13.5/1.5
s = 9 hrs hrs for the slower machine to do the job
:
:
Check this
{{{3/4.5}}} + {{{3/9}}} = 
{{{2/3}}} + {{{1/3}}} = 1