Question 1095086
<br>The solution supplied by ikleyn is a perfectly good traditional algebraic one.  I find many students prefer a different method.  Here it is for your problem; give it a try and see if maybe it "works" better for you than the traditional algebraic method.<br>
(1) Take the 10 hours that the one piece of equipment takes to do the job, and the 7 hours that it takes the two pieces together to do the same job.<br>
(2) Consider a period of 70 hours, where 70 is the least common multiple of 10 and 7.  In those 70 hours, the first piece of equipment could do this job 70/10 = 7 times; together the two pieces of equipment could do the job 70/7 = 10 times.  That means the second piece of equipment could do the job 3 times in those 70 hours.  Doing the job 3 times in 70 hours means doing the job once in 70/3 hours.<br>
Answer: 70/3 hours, or 23 1/3 hours.