Help ASAP!
A pirate in search of a treasure hikes three days in a row. He hikes 12.5 km to his destination on the first day, 18 km to his destination on the second day, and 14 km to his destination on the third day. He hikes 1 km/h slower on the first day than on the second day, and 1 km/h faster on the first day than on the third day. The third day's hike takes 30 minutes longer than the second day's hike. How long does the pirate search (across all three days) if he spends 2 hours digging for treasure at each destination?

We can start by expressing the speed of the second trip as x. This means that the first trip must be x-1, and second (x-1)-1=x-2. Next we can form an equation with the information that the third day hike takes 30 min longer than second day hike. Since we now the distance of both days, we can express them in time. So... [tex]\frac{18}{x}+\frac{1}{2}=\frac{14}{x-2}[/tex] ... We can solve by finding a LCM of all three expressions. Doing so we get 2x(x-2). Now multiplying out with the new denominators, we get 36(x-2)+x(x-2)=28x. Solving as a quadratic equation, the solution comes out to be x=6, x=-12. But as speed can't be negative, only solution is x=6. Now we get the speed of each day's hike. First day=x-1=5, second day=6, third day =x-2=4. Finding the time is simple, just do distance/speed. First day=12.5/5=2.5hrs. Second day=18/6=3hrs. Third day=14/4=3.5hrs. Adding total hrs up... 2.5+3+3.5=9. Finally step is accounting for the 2hr digging time per day. 2hrs*3= 6hrs. 6hrs+9hrs=15hrs. Hope this helps!