Arectangular auditorium seats 1833 people. the number of seats in each row exceeds the number of rows by 8 . find the number of seats in each row.

Let x be the no. of seats in each row and y be the no. of seats in each row.
we have,
Total no. of people in the auditorium = 1833
=> xy = 1833
=> y = 1833/x --- (1)
From the given condition, we get,

y = x+8
=> 1833/x = x+8 [from equation (1)]
=> (1833/x) - x = 8
=> (1833-x^2/x) = 8
=> 1833-x^2 = 8x
=> -x^2 - 8x + 1833 = 0
=> x^2 + 8x - 1833 = 0
=> x^2 + 47x - 39x - 1833 = 0
=> x (x+47)-39 (x+47) = 0
=> (x-39)(x+47) = 0

i.e, either x-39 = 0
=> x = 39

or x+47=0
=>x=-47 [but no. of seats cannot be negative]

therefore x = 39
Hence there are 39 seats in each row.

The third and fourth box.