When a trigonometric equation is solvable, it often has infinitely many solutions; and (depending on the context) the first solu
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Let's say if both work together, they can do it in "t" hrs
so, if Latoya can do it by herself in 14hrs, how much can she do in 1hr? well, she can do only 1/14 of the total
if Lisa can do it in 10hrs by herself, how much can she do in 1hr? well, 1/10 of the total
now, let's add their rate together, keeping in mind, their rate added, will be doing in 1hr only 1/t of the total work
thus
so, they'd do it in 5hrs and 5/6 of an hour, which is just 50mins, so, 5hrs and 50mins
so, if Latoya can do it by herself in 14hrs, how much can she do in 1hr? well, she can do only 1/14 of the total
if Lisa can do it in 10hrs by herself, how much can she do in 1hr? well, 1/10 of the total
now, let's add their rate together, keeping in mind, their rate added, will be doing in 1hr only 1/t of the total work
thus
so, they'd do it in 5hrs and 5/6 of an hour, which is just 50mins, so, 5hrs and 50mins