Mason and nora decided to swim across the river. mason began swimming 8 seconds earlier than nora.

mason swam at a speed of 5 feet per second.
nora swam at a speed of 9 feet per second.
for how many seconds had mason been swimming at the moment when the two swimmers had swam exactly the same distance?

This is a distance = rate * time problem. The easiest way to solve these is to make a table with the information. Since d= rt, we will set up the table like that:

                  distance              rate               time
Mason            d                       5                  t + 8
Nora               d                       9                    t

Let me explain the values in the table. The problem says "...when the swimmers had swam exactly the same distance"; therefore, we put a d there to indicate that, although we have no idea the distance they swam, both distances were the same. The rates are easy; they are self-explanatory. The time could be a little tricky too though. If Mason began swimming 8 seconds earlier than Nora, Nora's time is t and Mason's time is t + 8, which is Nora's time with 8 seconds added to it. Because d = rt, we set up the equations like that: d = 5(t+8), and d = 9t.  Because the 2 d's are the same, we set them equal to each other: 5(t+8) = 9t.  Simplifying that gives you 5t + 40 = 9t and 40 = 4t and t = 10.  Now put that t value of 10 into Mason's time to solve the question they are asking you: t + 8 with the substitution is 10 + 8 = 18. So Mason had been swimming for 18 seconds when they had both swam the same distance.

Idon’t exactly understand but i think dbc=rst

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