00:01
All right.
00:01
So in this problem, we have two printers at community college printing out class schedules, and a slower printer takes six hours to complete that job.
00:14
And a faster printer takes four hours.
00:17
So that means that one sixth of the job is completed per hour for the slower printer.
00:26
And one fourth of the job is completed per hour for the slower printer.
00:26
And one fourth of the job is completed per hour.
00:30
For the faster printer.
00:33
They're both going to work for some unknown number of hours.
00:36
So i'll call that h.
00:37
If you use any variable.
00:40
And the work that the first printer does in that amount of time is 1 .6th h.
00:46
Or you could also write that as h over 6.
00:50
Same thing for the faster one, 1 4th h over 4.
00:56
They're working together.
00:58
So i'm going to combine their two separate work.
01:03
And we want them to print three -fourths of the schedules.
01:10
So we don't actually want them to finish the full job.
01:13
We want them to finish three -fourths of the job.
01:19
This is the first time when we don't actually want them to finish the full job, which is rather interesting.
01:24
The next thing i'm going to do is look for the least common multiple of my denominators.
01:31
Six times two is 12.
01:34
Four can also make 12.
01:36
4 cannot make 6, so 12 is the smallest number that i can use...