Question 1198343
<font color=black size=3>
Person A can do the job in 35 minutes if working alone.
Person B (the brother) can do the same job in 14 minutes if working alone.


35 = 5*7
14 = 2*7
LCM = 2*5*7 = 10*7 = 70
or
LCM(x,y) = x*y/GCF
LCM(14,35) = 14*35/7
LCM(14,35) = 70


In short, the LCM of 14 and 35 is 70.
I'll use this value in the example problem below.


Consider the task is to shovel 70 cubic feet of snow.
Person A has a unit rate of 70/35 = 2 cubic feet per minute.
Person B has a unit rate of 70/14 = 5 cubic feet per minute.
If working together, without hindering one another, then their combined rate is 2+5 = 7 cubic feet per minute.


Therefore, the total time taken if working together is 70/7 = <font color=red>10 minutes</font>.


Relevant formulas
(rate)*(time) = amount done
rate = (amount done)/(time)
time = (amount done)/(rate)


-------------------------------------------------------------


Algebraic method:


x = total time taken if working together 
time is in minutes


Person A has a unit rate of 1/35 of a job per minute.
i.e. every 35 minutes, person A does 1 full job. "Job" in this case refers to shoveling the entire driveway.
I'm using the informal idea that
rate = (amount done)/time
rate = (1 job)/time
rate = 1/time


Person B has a unit rate of 1/14 of a job per minute using similar logic as earlier discussed.


The combined unit rate is
1/35 + 1/14
2/70 + 5/70
(2+5)/70
7/70 
1/10


Therefore,
(unit rate)*(time) = amount done
(unit rate)*(time) = 1 job
(1/10 of a job per min)*(x minutes) = 1 job
(1/10)*x = 1
x = <font color=red>10 minutes</font>


Note:
the equation
(1/35 + 1/14)*x = 1
is equivalent to 
1/35 + 1/14 = 1/x
We can multiply both sides by the LCD 70x to clear out the fractions and we'd get
2x+5x = 70
Solving that equation leads to x = <font color=red>10</font>


-------------------------------------------------------------



Answer: <font color=red>10 minutes</font>
This gives plenty of time to make it to campus (since the value is less than 30 minutes, aka half an hour).
</font>